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Coloring tensor products of graphs
up vote 3 down vote favorite
Let $ G,H $ are simple finite graphs and $A = G \times H$. Here $ G \times H $ is the tensor product (also called the direct or categorical product) of $ G $ and $ H $.
Let $G$ has smaller chromatic number. Experiments suggest that given a coloring $f$ of $G$ one can color $A = G \times H$. Color the vertices $(a,b)$ of $A$ with $g(a,b)=f(a), \; a \in V(G)$.
Experimentally the coloring is valid.
This was verified for 1000 random graphs and for $\{ \text{Petersen graph}, K_2,K_6, C_5,\text{Star graph 6} ,\text{Random graph of order 14}, \} \times \\\\ \{\text{All graphs up to 7 vertices}\}$
This is related to Hedetniemi's Conjecture which states $ \chi(G \times H) = \min \{ \chi(G), \chi(H) \} $.
1. Any counterexamples to this coloring?
2. Is it possible to prove this is valid coloring for certain $G$ or $H$?
3. What types of graphs are potential counterexamples?
graph-theory graph-colorings
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1 Answer
active oldest votes
The only way that two vertices $(u,v)$ and $(u',v')$ end up getting the same color is if $f(u)=f(u')$. But then there is no edge between $u$ and $u'$ in $G$ so there is no edge
between $(u,v)$ and $(u',v')$ in $G\times H$. So your conjecture is true for all graphs.
up vote 6 down In fact, as explained in the page you linked to above, a coloring with $n$ colors is simply a graph homomorphism to $K_n$, the complete graph on $n$ vertices. Since we have a
vote accepted homomorphism $G\times H\to G$ we can simply compose this with a homomorphism $G\to K_n$ whenever $\chi(G)=n$. This gives your construction. It also proves that $\chi(G\times H)\le \
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1. Lie algebras and graphs (Nick)
- describe Dynkin diagram associated to a finite-dimensional simply laced Lie algebra [Sc09a, Section A.1]
- give the ADE classification [Sc09a, Section A.1], [McG05. Section 1]
- define the (simply laced) Kac-Moody algebra associated to a graph [Sc09a, Section A.2]
- real and imaginary roots
2. Representations of quivers (Rob)
- define quivers and their representations [Sa06, Section 4]
- Gabriel's theorem: ADE classification of quivers of finite representation type [McG05. Section 2], [Sc09a, Theorem 3.7], [Sa06, Theorem 4.5]
- Kac's theorem: generalizing Gabriel's theorem to arbitrary quivers [Sc09a, Theorem 3.13], [Sa06, Theorem 4.6]
3. Hopf algebras and quantum groups (Josiah and John)
- definition of a Hopf algebra [CP, Section 4.1]
- modules and tensor products
- examples: U(g) [CP, Example 4.1.8] and C[G] [CP, Example 4.1.6]
- topological duality between U(g) and C[G] [CP, Example 4.1.17]
- special case: graded duality between U(n) and C[N]
- definition of a quantum Kac-Moody algebra [Sc09a, Section A.4]
- maybe a little bit about its representation theory?
- maybe something about the quantum Yang-Baxter equation?
4. Hall algebras (Josiah and Michael)
- define the Hall algebra of a quiver (or of a finitary category) [Sc09a, Lecture 1.3], [TL09]
- show that it is naturally a Hopf algebra [Sc09a, Lecture 1], [Gr95, Theorem 1]
5. Ringel's construction (David and Joe)
- check Serre relations and prove injectivity, along with surjectivity in finite type [Sc09a, Lecture 3.3], [Ri93], [Sa06, Section 5]
- explain how to do it directly over C with constructible functions [Lu91, Prop 10.20], [Lu91-ICM, Section 19], [TL09]
6. The canonical basis and crystals (AJ and Dylan)
- PBW basis = {characteristic functions} = {simple modules} = {constant sheaves on orbits} depends on orientations [Sc09a, Lecture 3.4]
- some vague words about perverse sheaves; define canonical basis = {simple perverse sheaves}
- the canonical basis is independent of orientations [Lu90], [Lu91-ICM, Section 19]
- Kashiwara's construction of the canonical basis [Ka91, Section 6]
- introduction to the theory of crystals as motivation for the canonical basis [Ka91], [Sc09b, Lecture 4.1]
7. The preprojective algebra and the semicanonical basis (Nick)
- give Lusztig's construction of U(n) [Lu91, Theorem 12.13] and [Sa06, Section 6.1]
- show that it is not sensitive to changes in the orientation of Q [Lu91, Theorem 12.15]
- prove that the components of the nilpotent variety index a basis [Lu91, Conjecture 12.14], [Lu00, Theorem 2.7]
- still descends to a basis of every finite-dimensional simple module [Lu00, Section 3]
8. Nakajima's construction (Matt and Dan)
- define quiver varieties and their cores [Na94]
- explain why core components index the semicanonical basis of an irrep [Na94, Section 10], [Sa06, Section 6]
9. Multiplication formulas for the dual semicanonical basis (Dylan and Michael)
- explain the GLS formulas for products of dual semicanonical basis elements [GLS05, Theorem 1], [GLS07a, Theorem 1.1]
- prove that the canonical and semicanonical bases differ [GLS07a, Sections 7 and 13]
- a few words about 2CY categories [GLS07a, Sections 1.5, 7.1, 8]
10. Prepare problem set
[CP] Chari and Pressley, A guide to quantum groups
[GLS05] Geiss, Leclerc, and Schröer, Semicanonical bases and preprojective algebras
[GLS07a] Geiss, Leclerc, and Schröer, Semicanonical bases and preprojective algebras II: a multiplication formula
[Gr95] Green, Hall algebras, hereditary algebras and quantum groups
[Ka91] Kashiwara, On crystal bases of the q-analogue of universal enveloping algebras
[Lu91] Lusztig, Quivers, perverse sheaves, and quantized enveloping algebras
[Lu91-ICM] Lusztig, Intersection cohomology methods in representation theory
[Lu00] Lusztig, Semicanonical bases arising from enveloping algebras
[McG05] McGerty, Quivers and lattices (video and notes)
[Na94] Nakajima, Instantons on ALE spaces, quiver varieties, and Kac-Moody algebras
[Ri93] Ringel, The Hall algebra approach to quantum groups
[Sa06] Savage, Finite-dimensional algebras and quivers
[Sc09a] Schiffmann, Lectures on Hall algebras
[Sc09b] Schiffmann, Lectures on canonical and crystal bases of Hall algebras
[TL09] Toledano Laredo, Hall algebras (video and notes)
Note: The references listed above are (almost) a subset of those listed here. | {"url":"http://pages.uoregon.edu/njp/CB.html","timestamp":"2014-04-16T07:18:18Z","content_type":null,"content_length":"5765","record_id":"<urn:uuid:df91fa66-e965-42e5-8ceb-1b28c6c83e63>","cc-path":"CC-MAIN-2014-15/segments/1397609521558.37/warc/CC-MAIN-20140416005201-00070-ip-10-147-4-33.ec2.internal.warc.gz"} |
he C
"Algebra 2 Common Core (textbook) is, in other words, a typical, old-fashioned algebra textbook. It's a highly efficient engine for the creation of math rage: a dead scrap heap of repellent
terminology, a collection of spiky, decontextualized, multistep mathematical black-box techniques that you must practice over and over and get by heart in order to be ready to do something
interesting later on, when the time comes."
-Nicholson Baker, Harper's
Algebra 2 as the "Decider"
by C.J. Westerberg
It seems as if we've been seeing many more articles about WHAT students should learn in school perhaps as an offshoot of discussions about The Common Core, the new standards being adopted by states
of what our students should learn year-by-year. Yet we tend to tip-toe around
what we should eliminate
or make as an elective while we keep layering more knowledge and skill requirements into student backpacks.
One of the subjects up for discussion is Algebra II inspired by Nicholson Baker's article,
Wrong Answer - The Case Against Algebra II
magazine. Curiously enough, I found it by chance after having a lengthy discussion with some friends about advanced high school math relevance and sure enough the rationale voiced was how Algebra 2
is necessary for students to "learn how to learn" or to "learn how to think" or to "develop the brain". As if nothing else can? But back to the larger point.
Curiously, what was not mentioned by anyone in our conversation that day is how important Algebra 2 is for college entry as a gatekeeper "subject". These were the same parents who are willing and
able to spring for SAT test prep and the like but it struck me how decontextualized even our conversations about education are in reality (unless the adults were in some form of denial, maybe even by
In any event, the aforementioned Baker skewers this algebra obsession throughout his lively and lengthy article:
"We've once again gotten ourselves caught up in a
strangely self-destructive cold war with other high-achieving countries.
The recruits are young teenagers,
their ammunition the little bubbles on standardized tests.
America's technological future hinges, say the rigorists, on
whether our student population can plus-and-chug the binomial theorem better than, say, Korean or Finnish or German or Chinese students.
The childishness of this hypernationalistic mentality depresses me,
and I want it to end,
and I am not alone."
Ouch, them those (sic) are fighting words. As you can tell, the provocative Baker has a flair for the dramatic with his writing so even a topic like Algebra can have sparks flying. After reading his
full missive (which is under lock so you either have to buy it or visit library), I thought that so much of what Baker was REALLY arguing about was how poorly Algebra 2 is taught, so no wonder why so
many students hate it. An excerpt (parens added mine):
Cornell's Steven Strogatz, a mathematician of crowds and swarms and oscillating bridges, told me that he agreed with much of what Hacker wrote (Is Algebra Necessary?). "As someone who is
working on the front lines, it's alarming to me, and discouraging, that year after year I see such a large proportion of people really not learning anything - and just suffering while they're
doing it."
We need less math for the average kid, Strogatz said, but more meaningful math.
"We spend a lot of time avalanching students with answers to things that they wouldn't think of asking."
This line of attack, along with the quote at the header of this post, are not so much a case against Algebra 2 but how it is taught, learned, and not learned. Baker's insistence about eliminating
Algebra 2 for all is not quite as convincing, especially since he relegates to non-algebra students limited career options, not exactly an ideal situation for a student to mis-calculate his or her
life path as early as high school.
He gives us an interesting history lesson how the same is-algebra-worth-it arguments arose in the early 1900s to the point that by 1950, only 25% of high school students were taking algebra. Who
knew? He goes on to say that "In the misty childhood days of IBM's Louis Gerstner (who would later co-found Achieve) and of a thousand other brilliant businessmen, inventors, engineers, and
innovators, algebra was a nonexistent force in the lives of the majority of high school students." And in spite of this low percentage, "Dictaphones were king and food engineers gave us mashed-potato
flakes, when GM was designing the Chevy small-block V-8 engine, when missile silos held freshly minted hydrogen bombs and Admiral Hyman Rickover's nuclear-powered submarines patrolled the waves . .
." An illuminating argument on one hand but in our increasingly technological and global world, not sure ALL
the parallels hold up. One just needs to
visit Conrad Wolfram's site, TED video, and argument for a wake-up call
on this angle about math education.
I would have, instead, preferred more of these conversations and arguments:
• make math a gateway, not a gatekeeper to college, and
• that perhaps substituting other logic-inducing courses such as coding which may appeal to more hands-on learners who would like to produce something at the end of a day instead of just a grade
score -see video below from Code HS (disclosure: no business relationship w/TDR), or
One more excerpt from Nicholson's piece that may give you a laugh since it is one of those time-travel blasts from the past:
Dean was a former engineer and a graduate of MIT who had taught math for years: his column, called Your Boy and Your Girl, was full of compassion and good sense. In an item published on March 27,
1930, Dean wrote:
I cannot see that algebra contributes one iota to a young person's health or one
grain of inspiration to his spirit . . It is the one subject in the curriculum that has kept children from finishing high school, from developing their special interests and from enjoying
much of their home study work. It has caused more family rows, more tears, more heartaches and more sleepless nights than any other school subject.
One final thought - there will always be the voices that say, "Oh, let students just suck it up like we did when we were in high school," or "students just don't try hard enough (ie. not enough
grit)." Those casual dismissals are a cop-out. Teachers are often caught in the middle.
be an
amazing time for math in this country.
We can do better. | {"url":"http://www.thedailyriff.com/articles/lately-ive-been-reading-a-1147.php","timestamp":"2014-04-20T11:28:41Z","content_type":null,"content_length":"43776","record_id":"<urn:uuid:8d562d69-6fe8-4cca-8c24-61420c7ed839>","cc-path":"CC-MAIN-2014-15/segments/1397609538423.10/warc/CC-MAIN-20140416005218-00193-ip-10-147-4-33.ec2.internal.warc.gz"} |
Polynomial with no common roots with its first and second derivatives
up vote 0 down vote favorite
I'm sorry if the question is too vague but maybe someone can help me. I am working with a polynomial $f(x)$ with distinct roots $x_1,\ldots,x_d$. I am able to solve the problem I am currently working
on if it turns out that $f''(x_i)=0$ for a certain $i\in{1,\ldots,d}$, i.e. $f(x)$ shares a root with its second derivative.
Then I'd like to know what I can say about $f(x)$ knowing that $f'(x_i)$, $f''(x_i)\neq 0$ $\forall i=1,\ldots,d$.
Any suggestion or deduction would be helpful since I fear I am overlooking something or perphaps this condition is too weak to be useful.
EDIT: I already have some algebraic relations among the roots (any cross-ratio lies in a finite set) and I'd like to obtain informations about the coefficients of $f$; of course if I could find an
algebraic relation for every single root it would be over.
2 You can check if two polynomials share a root by taking their resultant. – John Wiltshire-Gordon Nov 4 '12 at 22:11
2 What do you WANT to say about $f$? – Alexandre Eremenko Nov 5 '12 at 0:48
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1 Answer
active oldest votes
The resultant $R$ of two polynomials (a polynomial in the coefficients of the two polynomials) vanishes iff the two polynomials share a common root. Thus $R(f,f'')=0$ and $R(f,f')\ne 0$
are your conditions. Thus the polynomials that you want form a hyper surface in a Zariski open domain.
up vote 5
down vote Looking at $R(f,f'')$ one sees that as a polynomial in one coefficient $f_i$ it is of order $d-1$ in $f_i$ (keeping the other coefficients fixed). So, if you vary only the constant term
$f_0$, say, you find $d-1$ intersection points of this line with $R(f,f'')=0$, disregarding the requirement that $R(f,f')\ne 0$. This you regain by wiggling the line to a (complex) curve
which avoids the hyper surface $R(f,f')=0$.
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Re: Expectation
From the definition of expectation..
bobbym wrote:
That is not right either. Actually, the problem is related to a CCP but we have not covered that yet.
What is CCP?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Expectation
We have not come to it yet but we soon will. Do you know another way for the last problem?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Expectation
I do not.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Expectation
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Expectation
anonimnystefy wrote:
From the definition of expectation..
bobbym wrote:
That is not right either. Actually, the problem is related to a CCP but we have not covered that yet.
What is CCP?
Did you not see the first sentence of this post?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Expectation
Yes, I did and I said we have not got to it yet. CCP is short for Coupon Collector Problem.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Expectation
No, I didn't mean that. The text before the quote is on its own. It is not conected to the rest of the post. It was an answer to your previously asked question.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Expectation
I see, but I think I already answered the question with post #104.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Expectation
Post #104 told me sometying I had already known
Last edited by anonimnystefy (2012-11-09 09:12:10)
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Expectation
Well then, when I asked for another way you should have posted it prior to post #104
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Expectation
I did! It is in the first part of post #101!
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Expectation
Yes, I know but it is nice to see the formula actually in front of you...
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Expectation
Do you have another problem?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Expectation
I have many problems, social, psychological and math. Which do you want to try and solve?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Expectation
Let's try math first.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Expectation
What is the expected number of times a fair die must be rolled until all scores appear at least once?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Expectation
Hi bobbym
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Expectation
Yes! Very good! Although it does not look like it that is a CCP.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Expectation
Oh, but it does look like a CCP.
Any more?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Expectation
First, how did you solve the last one?
Just one or two more of these lightweight problems and we can move on to tougher ones.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Expectation
I solved it using absorbing Markov chains.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Expectation
Hmmm, mind If I see it? Then I can show you another way.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Expectation
It's a pretty large one to do by hand.
Last edited by anonimnystefy (2012-11-10 01:25:14)
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Expectation
You did forget the labels but here is another way.
The expected wait for all scores to appear = expected wait for one score +
expected wait for second score + ... + expected wait for sixth and final score.
The probabilities of these events are, respectively, 6/6, 5/6, 4/6, 3/6, 2/6, 1/6.
Therefore the expected wait for all scores to appear?
It's a pretty large one to do by hand.
No one but a lunatic would want to.
When you computed the answer you should have got a big surprise.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Expectation
Why 6/k?
Why should the anwser be a big surprise?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment | {"url":"http://www.mathisfunforum.com/viewtopic.php?pid=239427","timestamp":"2014-04-20T23:28:19Z","content_type":null,"content_length":"39509","record_id":"<urn:uuid:dae81110-c20c-4be5-9163-499134875cc5>","cc-path":"CC-MAIN-2014-15/segments/1398223202774.3/warc/CC-MAIN-20140423032002-00374-ip-10-147-4-33.ec2.internal.warc.gz"} |
Integrating a Discrete Function in MATLAB
Trapz implements the trapezoid rule--if this is what you're looking for, that's great (but since you're not putting in the t-values, you're just getting the sum of x, which may or may not be what
you're looking for):
If you want to integrate only a portion of the data, put in only a portion of the vectors!
For instance, to extract the 1st through 5th values of x, you'd type in:
>> extract=x(1:5)
More on array indexing (including the colon operator used above):
As per the examples under trapz above, as long as the two input vectors are the same size, trapz should produce something--not sure what the error message is, but if you'd post it, we may be able to
help you. | {"url":"http://www.physicsforums.com/showthread.php?t=537004","timestamp":"2014-04-19T15:17:55Z","content_type":null,"content_length":"29684","record_id":"<urn:uuid:270e32a0-faaa-4bf3-b967-808f33e580de>","cc-path":"CC-MAIN-2014-15/segments/1397609537271.8/warc/CC-MAIN-20140416005217-00383-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Scatter search and path relinking: advances and applications.
(English) Zbl 1041.90074
Glover, Fred (ed.) et al., Handbook of metaheuristics. Boston, MA: Kluwer Academic Publishers (ISBN 1-4020-7263-5/hbk). Int. Ser. Oper. Res. Manag. Sci. 57, 1-35 (2003).
Summary: Scatter search (SS) is a population-based method that has recently been shown to yield promising outcomes for solving combinatorial and nonlinear optimization problems. Based on formulations
originally proposed in the 1960s for combining decision rules and problem constraints, SS uses strategies for combining solution vectors that have proved effective in a variety of problem settings.
Path relinking (PR) has been suggested as an approach to integrate intensification and diversification strategies in a search scheme. The approach may be viewed as an extreme (highly focused)
instance of a strategy that seeks to incorporate attributes of high quality solutions, by creating inducements to favor these attributes in the moves selected. The goal of this paper is to examine SS
and PR strategies that provide useful alternatives to more established search methods. We describe the features of SS and PR that set them apart from other evolutionary approaches, and that offer
opportunities for creating increasingly more versatile and effective methods in the future. Specific applications are summarized to provide a clearer understanding of settings where the methods are
being used.
90C59 Approximation methods and heuristics
90B40 Search theory (optimization) | {"url":"http://zbmath.org/?q=an:1041.90074","timestamp":"2014-04-20T21:05:42Z","content_type":null,"content_length":"21265","record_id":"<urn:uuid:7bd6bc62-c302-4cbb-a710-c7d6ee1189e7>","cc-path":"CC-MAIN-2014-15/segments/1397609539230.18/warc/CC-MAIN-20140416005219-00366-ip-10-147-4-33.ec2.internal.warc.gz"} |
graphs about a function
Let g(x) be a function where g(0)=0, g(2)=0 and g'(4)=0. Which of the following is a possible graph of g(x)?
Tell us what you think the answer is, then someone will either confirm your answer or help you correct it.
Well, right away I eliminated choice C) because it doesn't satisfy g(0)=0. And, from here Im not sure what to do. I dont really understand what g '(4)=0 exactly means. Any feedback would be helpful
$g'(x)$ is the slope (or gradient) of $g(x).$ $g'(4)=0$ tells you that the gradient is zero when $x=4.$ | {"url":"http://mathhelpforum.com/calculus/208753-graphs-about-function.html","timestamp":"2014-04-21T16:30:17Z","content_type":null,"content_length":"43207","record_id":"<urn:uuid:aa8865cc-2431-403c-a2a4-37adb87d387f>","cc-path":"CC-MAIN-2014-15/segments/1397609540626.47/warc/CC-MAIN-20140416005220-00243-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Digits IN SPACE! (Puzzle Agent 2) - Telltale Community
I'm stumped on this puzzle.
Can you figure out the next two digits in the sequence?
31 41 59 26 ??
Hint 1: The Solution cannot be derived from the numbers.
Hint 2: The Sequence refers to a mathematical constant.
Hint 3: Try to imagine a decimal point after the 3.
I have absolutely no idea how to figure this one out and I need help!
12 Comments - Linear Discussion: Classic Style
• SPOILER WARNING:
This comment may contain spoilers!
I thought it was funny that Telltale expects people to know the digits to pi beyond the first couple. I did, but I also for some reason memorized it to 100 places at one point in school.
• My calculator doesn't give enough digits! I had to guess the correct rounding as well.
• There is a couple of puzzles that requires a knowledge of maths. Annoying. I knew it was pi but still had to look up the solution. Didn't help that there was only one question mark, I thought it
just wanted the next number not two?
For next update this puzzle should have two question marks making plain that the answer requires a two digit answer.
• I think they should have asked for all the next digits. This was just half-assed.
@daro2096 said: There is a couple of puzzles that requires a knowledge of maths. Annoying. I knew it was pi but still had to look up the solution. Didn't help that there was only one
question mark, I thought it just wanted the next number not two?
For next update this puzzle should have two question marks making plain that the answer requires a two digit answer.
The rules stated the answer was a two digit number (and all the numbers before it were two-digit).
This puzzle got me HARD. I should have realized what it was. I was feeling so smart after figuring out the other puzzle in Korka's house.
@daro2096 said: I think they should have asked for all the next digits. This was just half-assed.
• SPOILER WARNING:
This comment may contain spoilers!
Having memorised the first 10 digits of pi back in high school, it was really obvious to me. Of course, I had to consider whether the puzzle would round the numbers or not...
• SPOILER WARNING:
This comment may contain spoilers!
@chalon said: I also for some reason memorized it to 100 places at one point in school.
christ almighty.
Going to assume this is exaggeration, but that's still pretty incredible
• After the dot hint it became apparent. Then it was crtl+alt and googling pi.
Can't say I would have figured it out without that one though... so many math puzzles. I failed hard on pretty much all of them...
• SPOILER WARNING:
This comment may contain spoilers!
@Molokov said: Having memorised the first 10 digits of pi back in high school, it was really obvious to me. Of course, I had to consider whether the puzzle would round the numbers or not...
Lolz. You're right about it! Usually, students would say that their math lessons are unnecessary in our daily lives. But, here it is, that day where we need to memorize the 10digits of pi came in
• SPOILER WARNING:
This comment may contain spoilers!
@Irishmile said: Hint
do you like pie?
Answer if my hint was not clear 53
Uuuuh, thank you SO MUCH!
I never would have got this in a million years.
One of the hints was something like "imagine there is a decimal point after the 3" so I was thinking "ok so it's 3.1 4.1 5.9 2.6 and then..... OH FFS". I'm worse than I thought at following
instructions :rolleyes: | {"url":"http://www.telltalegames.com/community/discussion/comment/663757","timestamp":"2014-04-18T09:40:44Z","content_type":null,"content_length":"36644","record_id":"<urn:uuid:591c5d2d-f96a-41cd-b5d3-7f399c783b28>","cc-path":"CC-MAIN-2014-15/segments/1397609538824.34/warc/CC-MAIN-20140416005218-00621-ip-10-147-4-33.ec2.internal.warc.gz"} |
Expected minimum face angle of random convex polyhedron in $\mathbb{R}^3$
up vote 7 down vote favorite
Let $P_n$ be a "random convex polyhedron" in $\mathbb{R}^3$ of $n$ vertices, where "random" could follow any one of a number of models: (1) the convex hull of $n$ points randomly and uniformly
distributed on a sphere; (2) the convex hull of $N>n$ points randomly and uniformly distributed in a sphere; (3) analogous definitions but using different distributions, or replacing "sphere" by "a
given convex body." I think my question is largely independent of the precise model:
Does the expected measure of the minimum face angle $\theta_{\min}$ over all faces of $P_n$ go to zero as $n \rightarrow \infty$?
I am hoping there is a succinct argument that avoids computing the precise expectation of $\theta_{\min}$, which might be difficult, and would certainly depend on the model. I have seen many papers
on properties of random convex hulls, but none that I've found address my specific question. Thanks for ideas/pointers, under any model!
add comment
3 Answers
active oldest votes
The answer is YES. (I am assuming you mean the angle between two adjacent edges on a common face. (The dihedral angles all go to $\pi$.)) The easy and brief reason is that, in a
large random point set, everything (that depends on local conditions) happens almost surely.
Here is a sketch of a proof for your model (1).
1. Fix $\varepsilon>0$ arbitrarily, and take a (small) triangle $abc$ on the sphere with smallest angle $\varepsilon$.
2. Construct its circumcircle $K_0$.
3. Let $K$ be a concentric circle twice as large as $K_0$.
4. Construct some small neighborhoods $A,B,C$ around $a,b,c$ such that any triangle with vertices taken from these neighborhoods
1. has smallest angle $<2\varepsilon$.
2. has its circumcircle within $K$.
5. Now we let the number $n$ of points go to infinity. For each $n$:
up vote 1 down vote 1. Construct a scaled-down copy of the configuration $A',B',C',K'$ of $A,B,C,K$ (but still on the sphere) such that the expected number of points that falls into $K'$ is 3.
accepted (The area of $K'$ is a $3/n$ fraction of the whole sphere.)
2. Now, the probability that exactly 3 points fall into $K'$ is at least some positive probability $p_0$, independend of $n$. ($p_0$ is not so small, the number of points is
essentially Poisson-distributed with mean 3.)
3. The probability that
one point each falls into $A'$, $B'$, and $C'$ but no other point falls into $K'$
is at least some (small) constant $p_1>0$ (independent of $n$). The reason is that $A'$, $B'$, $C'$ have some (almost) constant fraction of the area of $K'$.
1. If this event happens, there will be a face angle smaller than $2\varepsilon$.
2. Now, place const$\cdot n$ disjoint copies $A',B',C',K'$ on the sphere. Then these copies behave essentially like independent Bernoulli experiments with success probability
$p_1$. As $n\to\infty$, the probability of having at least one "success" approaches 1.
Thank you, Günter! That is a clever and careful argument. – Joseph O'Rourke Jan 30 '13 at 11:45
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Section 8.2.4 of
Rolf Schneider, Wolfgang Weil: Stochastic and Integral Geometry, Springer Verlag 2008
up vote 2
down vote may be a good place to start. Roughly, there they select $n$ random points in a given convex body (say the unit ball) and they describe the large $n$ behavior of support function of the
expected convex hull. There are lots of references and historical remarks following this subsection and maybe you get lucky.
p.315: "The asymptotic behavior of such expectations, as $n \rightarrow \infty$, depends heavily on the boundary structure of $K$." Thanks for the pointer! – Joseph O'Rourke Feb 1 '12
at 15:24
add comment
For i.i.d. points chosen in a bounded subset of $\mathbb{R}^3$ (or $\mathbb{R}^d$) it seems to me that $\theta_\min(n)\to 0$ is ensured when the support of the distribution has a smooth
up vote boundary. This covers the case of the uniform distribution on an Euclidean ball, and a uniform spherical distribution as well. (I'm not quite sure about how to state a converse).
1 down
Let $\theta_\max(P_1,\dots,P_n)$ be the maximum face angle of the convex hull of the $n$ points $P_1,\dots,P_n$ in $\mathbb{R}^3$. A useful geometric lemma should be the following: "Let $A
\subset \mathbb{R}^3$ be a bounded open set with smooth boundary. Then, for any $\epsilon > 0$ there exists $\delta > 0$ such that for any finite set of points $P_1,\dots,P_n$ in $A$, if $
\theta_\max(P_1,\dots,P_n)\ge\delta $ then $\operatorname{Vol}(A\setminus \operatorname{co}(P_1,\dots,P_n)\ge \epsilon.$" (In words: if the convex hull of the $P_i$'s almost fills $A$, the
face angles must be small). – Pietro Majer Feb 1 '12 at 15:21
Thanks, Pietro! That seems eminently reasonable, especially with the smooth boundary assumption. – Joseph O'Rourke Feb 1 '12 at 15:28
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Not the answer you're looking for? Browse other questions tagged pr.probability mg.metric-geometry convex-geometry discrete-geometry reference-request or ask your own question. | {"url":"http://mathoverflow.net/questions/87230/expected-minimum-face-angle-of-random-convex-polyhedron-in-mathbbr3/87237","timestamp":"2014-04-16T22:05:09Z","content_type":null,"content_length":"64747","record_id":"<urn:uuid:3243ca64-c70e-4508-8be9-9c7222916176>","cc-path":"CC-MAIN-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00288-ip-10-147-4-33.ec2.internal.warc.gz"} |
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hi nicboo
Welcome to the forum.
This is how you can devise your own formula for this:
Think of a very easy example. Say you want to pack 1000 envelopes and you know you can do 200 per hour.
That's two hundred in the first hour, another two hundred in the second hour, ............, another two hundred in the fifth.
Can you see this will take 5 hours?
So what is the connection between the three numbers ?
So now put words in place of the numbers:
This time will be in hours. Times by 60 to change to minutes. | {"url":"http://www.mathisfunforum.com/post.php?tid=18029&qid=228413","timestamp":"2014-04-16T07:24:03Z","content_type":null,"content_length":"19376","record_id":"<urn:uuid:77a2fe83-2db5-437d-8d15-c35c7cf2f38d>","cc-path":"CC-MAIN-2014-15/segments/1397609521558.37/warc/CC-MAIN-20140416005201-00424-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Can anyone help me answer these questions? I just really suck at inequalities and don't know how to answer these. Please help? 1.Explain why and when you need to switch the inequality sign when
solving an inequality. 2.How can you tell which direction to shade the graph of an inequality? 3.What is the difference between graphing conjunctions and disjunctions? 4.What is one difference
between an inequality with one variable and one with two variables? 5.How do “closed and open circles” relate to “dashed and solid lines”?
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I'm not going to do all 5. I'm only doing one.
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That's fine. I'll take all the help I can get.
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Alright, next time, post one question at at time, not five in a question.
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Let's say we have: \[-a>b\]If we rearrange the equation we get: \[0>b-a\]And if we rearrange again we get:\[-b>a\] or \[a<-b\]So, when do swap the inequality? Do you know?
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when it's only a negative
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Negative and when we are multiplying and dividing. This is something you just have to memorize.
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thank you
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|dw:1356903237059:dw|open circle means the number is not include and dashed line means the numbers on the line are not included solid circle means the number is included and solid line means the
numbers are included
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|dw:1356903359427:dw|3 in not a solution but any number bigger than 3 is (example 3.1, 3.2, 4, 5, 100, etc)
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|dw:1356903434970:dw|now 3 is also part of the solution
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|dw:1356903467129:dw|only the area under the dotted line is the solution
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|dw:1356903509899:dw|now the line is also part of the solution
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There are rules for inequalities and once you learn these, you will be answering others peoples questions about how to do them. Rule one ...Adding or subtracting the same number on both sides,
even if the numbers are negative, WILL NOT change the sign. Rule two ...Multiplying and dividing by a POSITIVE number, WILL NOT change the sign. Rule three ...Multiplying and dividing by a
NEGATIVE number, WILL change the sign. Rule 4 ...When switching the different sides, you WILL change the signs. example : x > 1 is also 1 < x Which direction to shade the graph .. When y is less
then, you shade below the line, and when y is greater then, you shade above the line. If x is less then, you shade to the right of the line, and when x is greater then, you shade to the left of
the line. conjunction - contains the word "and" . Is true if both inequalities are true disjunction - contains the word " or ". Only one of the inequalities is true. A conjunction is the
combination of two inequalities on a number line. example : x > 0 and 3 > x can be written as 0 < x < 3 when graphing this....your number line is 0 - 3...you have open circles on 0 and 3 with
everything shaded in between. A disjunction is going to have two inequalities on the same number line. example : x <1 or x > 3 when graphing this .....lets make your number line from 0 - 4...you
have an open circle on 1 shaded to the left and an open circle on 3 shaded to the right. An open circle means that the number is not part of the solution. example : x < 3 : x > 5....neither of
these inequalities are going to have closed circles because the 3 and the 5 are not part of the solution. A closed circle means that the number is part of the solution. example : x >= 4 (greater
then or equal) This one would have a closed circle because 4 is part of the solution. If there is an equal sign, then the number is part of the solution. Dashed line means numbers on the line are
not included. Solid line means the numbers on the line are included.
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Wow thank you so much!
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_+3=5+7=_: it depends what 'equals' equals
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Pop quiz! Answer the following before reading further:
A: 6+9=__+4
B: __+8=12+5
C: __+3=5+7=__
D: True or False: 6+8=3+11
E: 160=___
I've become painfully aware recently how sloppy communication can be. I am surprised how many times I have to reply to an email asking for clarification because of unclear writing or simply not
taking the time to think things through. And that's with other scientists.
The GK-12 program I run at the University of Nebraska places graduate students in upper elementary, middle and high schools to work with teachers on improving math and science education for their
students. We spend an entire day of the week-long orientation discussing communication. I break it down roughly into "scientific" communication and "normal person" communication. That's probably
not a fair breakdown, but we really have to emphasize to the students that, although it is perfectly OK to reply to a scientist's idea with "here's why that won't work", it's a death knell for the
relationship if you do that with a teacher (or, often, a spouse).
At the end of the year, one of my students made an observation I wholeheartedly endorse. "I like scientific communication better," she said, "It's just faster." And clearer, I would add.
Although Jennifer and I come from very different disciplinary backgrounds, I think one of the reasons we've hit it off is that we share the trait of wanting to use words properly. Jennifer
recognizes that scientists and mathematicians use words and symbols to convey very specific meanings. If I use the word "velocity", she's likely to ask if there's a reason I didn't say "speed".
(Speed is a scalar, velocity is a vector. Sometimes it makes a difference, sometimes not.)
Nowhere, perhaps, is the specificity of symbols more rigorous than in mathematics. My mother was a graduate student in math, then economics, while I was in elementary and middle school. I remember
seeing her scribblings filling up scads of yellow legal pads and asking her once "when do I get to learn this language?" And math is definitely its own language. One of the biggest problems
teaching (or communicating) science and math is that sometimes words mean different things in the discipline than they do everyday life.
But the equal sign should be an easy one, right? It means, well, equal.
Apparently, American students have a much less clear idea of the meaning of the equal sign than their Chinese, Korean and Turkish compatriots. A study by Capraro, et al in Psychological Reports (106
(1), 49-53 (2010)), which draws on their previous data in Li, et al. (Cognition and Instruction, 26, 195-217 (2008) compares 6th grade students from different countries. Both papers originate from
the research group of Mary Margaret Capraro and Robert M. Capraro at Texas A&M University. Incidentally, "et al." translates literally in Texan to "and them".
The results from the first two questions I posed from their study (6+9=__+4 and __+8=12+5) were surprising/appalling. Only 28.6% of American students got these questions right. The Chinese and
Korean rates were in the 90+% range and the Turkish rates were 61% and 79% respectively.
As is often the case, we learn more by looking at the wrong answers than the right ones. The first two problems were designated "Type A" and "Type B" - similar, except the missing numbers are on
opposite sides of the equal sign. The third problem - the one that lent itself to the title of this blog - is classified as "Type C" and provides a slightly different probe of the understanding (or
For the "Type C" problem __+3=5+7=__, American students got the first blank right 23.8% of the time, while the rates for other students were 98.6% (Chinese), 86.5% (Korean) and 60.2% (Turkish).
Interestingly, the correct rates for the second blank were much more comparable: 86.7% (American), 97.9% (Chinese), 93.3% (Korean) and 86.0% (Turkish).
What this strange disparity between the first and second blanks tells us, the authors argue, is that American students disproportionately don't understand the equals sign. The most common wrong
answer for the first blank was "2". It is true that 2+3 = 5, but 2+3 definitely doesn't equal 5+7. Almost 90% of the students recognized that "5+7=12", but a significant number of those got the
first blank wrong. This is apparently a common misconception among American students that isn't seen nearly as much in student from the other countries studied: the belief that the answer is the
number immediately following the equal sign.
In their previous study, the authors used the True/False question "6+8=3+11?" to test understanding of the reflexive property of the equal sign. Reflexive, which I had to look up, means a=a. The
popular phrase "it is what it is" embodies the mathematical philosophy of reflexivity. The educators doing the study, though, realized that students with the misconception I mentioned above - the
answer is the number immediately after the equals sign - would get this question wrong for that reason and not because they don't understand that a=a.
In their new study, they replaced that question with "160=___", and expected the blank to be filled in with "160". But a number of students put an operation in that blank, like 80*2 or 40+120. Those
answers are not wrong, but (strictly speaking), but they indicate that those students look at the equal sign as indicating that a mathematical operation is required and not solely as a representation
of equality.
I've always thought of the equal sign as the pivot on a see saw. Whatever is on the left has to balance with whatever is on the right. If I fill in the blank with a 6, there's only 9 on the left and
12 on the right, so the see saw isn't balanced. OK, I can't draw a picture for the multiple equals signs on one lin e, but you get the idea. Who ever thought that something as seemingly simple as
'equal' could be so complicated?
At this point, you might be thinking that this seems like a bunch of quibbling over a precise definition of interest only to the highly mathematical. One of the original motivations for this study
was the low performance of American students relative to their international counterparts on standardized tests like the TIMSS and the PISA tests. There apparently isn't much explicit attention to
the equal sign in middle school curricula and, the authors (along with other math educators) believe that not understanding the equal sign puts students at a distinct disadvantage when it comes time
to learn algebra. If you put 'x' in place of the blank, you realize you're actually doing algebra answering the questions I posed at the start of the blog.
These misconceptions carry over to physics. For example, consider the force on an object falling: F = mg. The force gravity exerts on a ball falling through the air is equal to the product of its
mass times the acceleration due to gravity (in the absence of air resistance - sorry, I felt compelled as a professor to add that. I couldn't help myself.)
Students identify gravity as a force, but a significant number of them also identify a force "F" in the above equation as distinct from the force of gravity. The problem gets worse when there are
multiple terms on the right-hand side of the equation due to multiple forces.
It is amazing that something so seemingly fundamental can so impact a student's education. One of the (many) reasons I am looking forward to Jennifer's book is that I speak fluent calculus. Imagine
trying to explain to someone how to walk. That's what me teaching calculus is like. One of the best reasons for using peer teaching (students teaching each other) is that they explain things in
ways I wouldn't have thought to use. Listening to them explaining how they understand an idea helps me realize how I can explain it better. It's research like this that reminds me that sometimes
the better part of teaching is listening.
ADDED 8/13/10: An interesting study notes the need for better prepared mathematics teachers, as well as a significantly strengthened math curriculum. Jennifer and I have been talking a lot here
recently about stereotypes. Although we've focused on those in the media, this article, by researchers at the University of Chicago in PNAS, suggests a scary chain: Female first and second grade
teachers who are anxious about math pass that anxiety along to their female students. More female students are likely to agree with the suggestion that boys are better than girls at math after being
exposed to this anxiety, and the female students who did agree with this stereotype performed worse in math as the year went on. Great article and PNAS makes the full text publicly available.
AND: A Christopher E. Granade speaks on the topic of 'equal' - a very nice post focusing on the importance of relationships and how that is really at the base of math and science.
Thank you for this - I'll be teaching physics this fall, and I will be using some variation of this mini-quiz to pinpoint these reasoning errors of the students.
Thank you for this very interesting article. It drives me insane when my students write something like 3+5=8-7=1. I think we take for granted that they will understand the meaning of "=", when
according to the figures quoted above they certainly do not! I might use those five questions at the start of the article with my students this year.
This is clearly a 'hot topic' at moment! I agree with you about precision in the use of language (actually the impact of teaching language on the learning of Mathematics is my main research
interest!). I have already made comments about this on two other blogs: http://castingoutnines.wordpress.com/ Robert Talbert on "calculator syndrome" (as I call it) and http://
georgewoodbury.wordpress.com/2010/08/11/thoughts-on-equal-signs/#comment-261 George Woodbury's thoughts on simplification of algebraic expressions. Your readers may find them interesting too.
Oh, and since you mentioned speed... does it annoy you when people talk about "driving at a high rate of speed"? Surely they mean acceleration, don't they? Or, do they just mean driving very fast...
Colin: Yes! I am an advocate for talking simple: "I was driving fast" is accurate and works just fine.
I like: "I was turning the corner and I wasn't even accelerating..." and anything Rusty Wallace says when he's doing commentary during a race.
I wouldn't characterize this as a difficulty with understanding equals so much as a difficulty with operator precedence. If you placed parentheses, making, for example, (_+3)=(5+7), I'll bet the
failure rates change dramatically.
As a long-time programmer and computer scientist, whenever I write code that is meant to evaluate such an equation, I put in the parentheses, it's just simpler. I will furthermore add that the sort
of thing that cox_dan points out above makes perfect sense if you are working on a calculator. It's exactly the steps you might take in a calculation: you might press '3' then '+' then '5' then '=',
showing '8' on screen, then '-' then '7' then '=' again showing '1'.
Actually the translation for et al. into Texan is et y'all.
not all of us are observant enough to notice this, if you're noticing what I think you're noticing. the difference between boolean equals and the other kind. boolean equals is noted as "==" or an
equals with three lines instead of two. this should be a simple and obvious topic because it has been studied for how many decades or hundreds of years? but it certainly isnt common curriculum. the
most basic definitions of our most basic math operator.
I agree with Doctor Jay, the problem seems to be understanding of syntax, not semantics.
I was coaching a boy who had a bit of trouble with equations. So I got him to stand up, arms outstretched, and pretend to be a balance.
Then I asked him to add, say +5, to one side, and then asked what's happening? Then to ask him to add +5 to the other side, and asked again. This seemed to help.
An online friend sent me this example of abuse of the equals sign from a textbook. This problem is propagating right from the top.
I guess that I had good math teachers in high school because they always taught that the "=" symbol read "true" not equals.
Yes! This couldn't be more important. Thank you. Having tutored HS math students, many read = as the enter key on a calculator--no understanding of the symbol having meaning.
Interesting, but the conclusion about the 160= part really baffles me. Mind you: i'm just somebody who finished high school and who does nothing with maths. When I did the 'test' I also finished it
with something like 160=120+40. Not because I expect some kind of operation after the blank, but because I thought 'well of course 160=160! That is so bloody obvious. But this test obviously requires
me to do some addition tricks, so, if the testers want me to do a trick, I'll do it'
I think it's extremely logical to fill it out like that. When you do a test, you're always second-guessing the people who put it together. It's not because I thought the equal sign requires an
operation, it's because I thought the testers required an operation. It has nothing to do with understanding the equal sign.
My answers:
A. False
B. False
C. False
D. True
E. False
Did I get them right?
[Note: Given that the author of this paean to precision amusingly failed to actually ask a question, I substituted in my own:
"Assume __ = 0. Are the following equations true or false?"]
Maybe I'm just weird, but when I saw __+3=5+7=__, I initially assumed that the definition of + was what was really at stake, that some kind of cyclic group or modular arithmetic or something was
involved. (Upon further reflection, that doesn't work either, because it would be mod 3, but there are numbers larger than 3 in the equation.) It did not occur to me that the same value (represented
in the problem's symbology by an underscore) could stand in for different numbers on the same line of equation.
Of course, it's possible the problem's instructions (which I didn't see) hinted at that, with wording like "fill in each blank with whatever number will make the equation true", or somesuch.
2+3=5+7=12 makes perfect sense to someone who is using a calculator. Since American kids use calculators for even the most trivial arithmetic, it's no wonder they would see this as a reasonable math
"sentence". Kids in many places around the world still do arithmetic manually, and seldom see this kind of strung-together expression. | {"url":"http://twistedphysics.typepad.com/cocktail_party_physics/2010/08/_357_-it-depends-what-equals-equals-.html","timestamp":"2014-04-24T11:18:48Z","content_type":null,"content_length":"58281","record_id":"<urn:uuid:78c7f99c-108f-499c-be24-b2a0f0500bf2>","cc-path":"CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00130-ip-10-147-4-33.ec2.internal.warc.gz"} |
A question about closed curves
up vote 3 down vote favorite
Does three dimensional Euclidean space contain unbounded closed curves that do not cross themselves? It does not seem possible to find examples of such anmd yet it is not clear just what is standing
in the way. More precisely, let n be any positive integer not less than 3 and let E(n) be n-dimensional Euclidean space. Does there exist a subset S of E(n) with the following properties that is not
compact? (1)S is closed, connected and locally connected. (2) If any point of S is removed, the resulting space is still connected and is homeomorphic to a straight line. (3) If any distinct pair of
points of S are removed, the resulting space is no longer connected and has two components. Finally can S be a non-compact subset of a separable and infinite dimensional Hilbert space?
add comment
2 Answers
active oldest votes
up vote 6 Note that $S$ is a connected 1-dimensional manifold. Since it is not compact we get that $S$ is homeomorphic to $\mathbb R$, a contradiction.
down vote
Thanks for both of your illuminating responses. Anton, that one-line demonstration is very nice! I wonder whether the no answer would still hold if I changed condition (2) to state
that S minus any one of its points was still connected but did not stipulate that it should be homeomorphic to a straight line. Condition (1), of course, implies that S is arc-wise
connected. It seems also that bringing Hilbert space into the picture doesn't change the situation in any way. – Garabed Gulbenkian Mar 29 '13 at 19:35
add comment
Let $S$ be as assumed.
Choose distinct $p,q \in S$, and consider the parametrizations
$$ \psi : (0,1) \to S \setminus \{q\} $$ and $$ \phi : (0,1) \to S \setminus \{p\}. $$
Let $p$ correspond to time $t_p \in (0,1)$ under $\psi$.
Claim: for $t < t_p$, $\phi^{-1}(\psi(t))$ is a monotone function of $t$, and similarly for $t > t_p$.
proof: clearly $\phi^{-1}(\psi(t))$ is defined and continuous for $t \in (0,t_p)$. Moreover, it is injective as a map into $(0,1)$, by hypothesis on $\phi$ and $\psi$. Thus, it is monotone.
Ditto $t \in (t_p,1)$.
up vote 1
down vote Next claim: as $t \to t_p^-$, $\phi^{-1}(\psi(t)) \to 0$ (after reflecting $\phi$ if necessary), and as $t \to t_p^+$, $\phi^{-1}(\psi(t)) \to 1$.
proof: as $\phi^{-1}(\psi(t))$ is monotone for $t < t_p$, it has a limit in $[0,1]$ as $t \to t_p^-$. If this limit lies in $(0,1)$, we obtain a contradiction that $p \notin \text{im }\phi$
(since $\psi(t) \to p$ as $t \to t_p^-$ and $\phi$ is continuous as a map into $\mathbb{R}^3$). Thus $\phi^{-1}(\psi(t)) \to 0$ or $1$ as $t \to t_p^-$, and reflecting $\phi$ if necessary
we may assume it is 0. Similar reasoning shows $\phi^{-1}(\psi(t)) \to 0$ or $1$ as $t \to t_p^+$, and it cannot approach the same limit as when $t \to t_p^-$ by injectivity/connectedness/
continuity/etc. The claim follows.
Note that it follows (by connectedness and continuity) that $\phi^{-1}(\psi(t))$ maps $(t_p - \epsilon, t_p) \cup (t_p, t_p + \epsilon)$ onto (a set containing) $(0,\delta) \cup (1 - \
delta, 1)$ for some $\epsilon, \delta > 0$. Hence for $t$ near 0 or 1, $\phi(t)$ is near $p$. But clearly for $t$ bounded away from 0 and 1, $\phi(t)$ is bounded. Thus $\phi$ is a bounded
map, contradicting the unboundedness of $S$.
add comment
Not the answer you're looking for? Browse other questions tagged gt.geometric-topology or ask your own question. | {"url":"http://mathoverflow.net/questions/125850/a-question-about-closed-curves","timestamp":"2014-04-19T12:25:16Z","content_type":null,"content_length":"55615","record_id":"<urn:uuid:a7c50da0-b454-41fd-9616-0a904c277df2>","cc-path":"CC-MAIN-2014-15/segments/1397609537186.46/warc/CC-MAIN-20140416005217-00483-ip-10-147-4-33.ec2.internal.warc.gz"} |
Laplace equation
How am I ever gonna get [tex]u(2,\theta)[/tex], wich contains an [tex]ln 2[/tex], to equal [tex]\sin^2 \theta[/tex]?
You aren't. Better set A0 = 0.
Basically, you are free to choose any combination of the coefficients you need to match the boundary conditions. There are some integrals you can do to solve this analytically, but in most cases that
you are likely to see in homework or on a test, you can just pick out the coefficients by inspection. | {"url":"http://www.physicsforums.com/showthread.php?t=141014","timestamp":"2014-04-19T09:39:01Z","content_type":null,"content_length":"37285","record_id":"<urn:uuid:d115d4a4-14e7-4b4a-9e90-b665dcc41d8c>","cc-path":"CC-MAIN-2014-15/segments/1397609537097.26/warc/CC-MAIN-20140416005217-00337-ip-10-147-4-33.ec2.internal.warc.gz"} |
November 2
I can't believe it took so long for Choose Your Own Adventure to show up on Youtube. There has to be some justification for showing it in a math classroom. Every film has two options? Exponential
growth? I'll think of something.
Edit: Here is the file you can download and edit.
(Alternate Title: What is a Radian?)
What I wanted them to grok:
Discussed before they started: what is pi? Could you cut a string exactly pi inches long? Most of them said "no". They thought it was theoretically possible, but couldn't be measured precisely
enough. I promised them that they'd walk out of class with a string of length pi in their pocket.
The Activity:
Materials: Circular object (collected from around my room: a CD, a water bottle, paper plates, cardboard mailing tube, etc etc)
Metric ruler
Strip of Paper (I purchased adding machine paper at big box office supply store)
1. Measure the diameter of
your circular object. Calculate the radius to the nearest 10th of a centimeter and record: r = _______ cm
2. Cut a strip of paper at least 8 times longer than your radius, and tape it to a flat surface.
3. With a marker, make a mark on the edge of your circular object. Make a corresponding mark near the end of your strip of paper. Line up the two marks, then roll your circular object one
complete revolution. Make a second mark on the paper at exactly one complete revolution. What does the distance between the two
marks on the paper represent? Verify this value by using the radius to calculate
4. Cut a length of string the same length as the distance between the marks. Set it aside.
5. Using one of the marks on your strip of paper as "zero", use your ruler and a pencil to mark off intervals along the paper of the same length as your radius. About how many radii before you
get to the one-revolution mark? This value should be the same for all of the various circles. Why?
6. Take your length of string, fold it in half, and cut it exactly in half. This string is precisely pi long. You and your partner can walk around all day with pi in your pocket! The question
Pi whats?
At the appropriate time we discussed this, and came around to "pi radii of the circle". Many of them still didn't like this and insisted it "wasn't really pi", so we had to have the discussion about
operating in the perfectly abstract realm...some of them noticed that if their circle had a radius of exactly one centimeter, then the string would be pi centimeters...that seemed to help. Then I had
them convert a bunch of degree measures to radians and back, and come up with formulas.
The scary thing was, most of them didn't recognize it from last year until they started writing down a formula. But I'm hoping that more of them actually know what the heck a radian is, and this
makes the unit circle a slightly easier mountain to climb. We shall see.
I'm sure I'm not the first one to think of this, but this lesson was pleasingly effective. In past years graphing inequalities in two variables was something that seemed easy for the kids in class,
but they could never remember what was up when we reviewed it later.
This year, to start off, I had the kids pick the coordinates of three different points (x, y) on the plane, test them in the given inequality, and mark the point with a closed circle if it came out
to be true, and an open circle if it came out to be false. After a few minutes I had to say, "If you only got true points, find a point where it comes out to be false." We ended up with a screen that
looked like this:
We discussed the significance of the boundary between the blue and white spots, and also tried a few non-integral points. Then I had them try a different one on their own. They developed a very
serviceable method for indicating the "true region". One class wanted to write the words "True" and "False" instead of shading, which isn't bad, but I warned them that eventually we'd need to know
where two of them overlapped, so they shaded.
This seems like a stunningly obvious approach now that I've tried it, but I thought I'd share since it took me 4 years to think of it. Does anyone do anything similar, or have a better way?
I know lots of other teachers are veterans, too. I am spending today writing marking period 1 comments, going to see madagascar 2, and grumbling about civilians. So just a quick thanks. And one of my
favorite videos from 116, showing the lengths people will go to to keep each other sane in close quarters. They are having so much fun, they almost make me nostalgic for being at sea. Almost.
Well it looks cool, but I honestly don't know how much good its doing. I haven't carved enough time out of the curriculum to really make the vocab a part of the course. I feel like we're just
slapping copied definitions on the wall. Anyway:
And just for fun, here are some other pictures of my room: | {"url":"http://function-of-time.blogspot.com/2008_11_01_archive.html","timestamp":"2014-04-16T13:09:20Z","content_type":null,"content_length":"181482","record_id":"<urn:uuid:0af9b976-0d19-4f07-aa18-3b1575e3d57c>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00286-ip-10-147-4-33.ec2.internal.warc.gz"} |
Homework Help
Posted by christina on Monday, January 21, 2013 at 9:00pm.
9/x + 9/x-2= 12
• alg/trig - Katie, Monday, January 21, 2013 at 9:09pm
The first thing you want to do is get rid of x in the denominator, because that simplifies things quite a bit. Think of the easiest common denominator between 9/x and 9/(x-2), which would be x
You know that x-2/x-2 = 1, regardless of the value of x; the same with x/x, so you therefore can multiply 9/x with (x-2)/(x-2) (which is the same thing as multiplying by one) and 9/(x-2) by x/x.
This yields
9(x-2)/[x(x-2)] and 9x/[x(x-2)], giving an equation of
9(x-2)/[x(x-2)] + 9x/[x(x-2)] = 12
Because you have a common denominator, you can condense the two terms on the right side of the equation to:
[9(x-2) + 9x]/[x(x-2)] = 12.
Multiply both sides of the equation by x(x-2), eliminating the denominator:
9(x-2) + 9x = 12x(x-2)
Distribute the terms:
9x-18 + 9x = 12x^2 - 24x
18x-18=12x^2 - 24x
Move all terms to the left side of the equation (subtract 18x from both sides and add 18 to both sides):
0= 12x^2 - 24x - 18x + 18
0=12x^2 - 42x + 18
You now have a quadratic formula and can now do one of two things: apply the quadratic formula or factor. I'll just factor it out (you can undoubtedly plug the numbers into the quadratic formula
on your own time, should you so desire):
The values that would make this equation true would be
2x-1=0 or x-3=0, giving x-values of 0.5 and 3.
Plug in the x-values into your equation. Do you get 12?
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Lecture 7: Chinese Rings Puzzle
Lecture 7: Chinese Rings Puzzle
3443 views, 2 ratings - 01:22:03
More information about this course:
Licensed under Creative Commons Attribution ShareAlike 2.0:
Learn about the Chinese Rings Puzzle and how it relates to Discrete Math.
• What is an algorithm for the Tower of Hanoi problem?
• How do you write a recursive program for solving the Towers of Hanoi problem?
• What is the seven rings problem and how do you solve it?
• What is the Chinese rings puzzle and how do you solve it?
• How can you remove all the rings in the seven rings problem using recursion?
• What are the two possible moves in the seven rings problem?
• What does the seven rings problem look like using six binary digits?
• How do you get the hardest ring off of the Chinese rings puzzle?
• What is the time complexity for the Chinese rings puzzle?
• How can you use induction to find the number of steps needed to solve the seven rings problem?
• Why does it take 42 moves to solve the Chinese rings puzzle with six rings?
• How can you set up a spinner with 0s and 1s so if the spinner lands on a boundary between two regions, it picks either the right or the left region is selected?
• What is a gray code?
This video starts off with an explanation of some ideas behind how to solve a variation of the Towers of Hanoi problem. Then, the meat of this lecture deals with the Chinese rings puzzle or seven
rings problem. Theory behind this problem and analysis of the problem using binary place-holder digits. This discussion really explains clearly how to solve this puzzle. A really long and hairy
time complexity computation is done in depth using a recurrence equation, some algebra, and induction. Some applications of the Chinese rings puzzle are the shown. | {"url":"http://mathvids.com/topic/20-discrete-math/major_subtopic/70/lesson/617-lecture-7-chinese-rings-puzzle/mathhelp","timestamp":"2014-04-17T12:29:43Z","content_type":null,"content_length":"80926","record_id":"<urn:uuid:627c0b5b-85a0-43a7-b1a5-c44fadd98e73>","cc-path":"CC-MAIN-2014-15/segments/1397609530131.27/warc/CC-MAIN-20140416005210-00064-ip-10-147-4-33.ec2.internal.warc.gz"} |
RE: st: gologit2
[Date Prev][Date Next][Thread Prev][Thread Next][Date index][Thread index]
RE: st: gologit2
From Mike Lacy <Michael.Lacy@colostate.edu>
To statalist@hsphsun2.harvard.edu
Subject RE: st: gologit2
Date Thu, 17 Apr 2008 13:04:48 -0600
In an earlier response in this thread,
Richard Williams <Richard.A.Williams.5@ND.edu> remarked:
>My experience is that it is rare to have a model where the
>proportional odds assumption isn't violated! Often, though, the
>violation only involves a small subset of the variables, in which
>case gologit2 can be useful. You might also want to consider more
>stringent alpha levels (e.g. .01, .001) to reduce the possibility of
>capitalizing on chance. You can also try to assess the practical
>significance of violations, e.g. do my conclusions and/or predicted
>probabilities really change that much if I stick with the model whose
>assumptions are violated as opposed to a (possibly much harder to
>understand and interpret) model whose assumptions are not violated.
I would sound in to support the idea that the Brant test commonly detects departures from proportional odds that are so small as to be uninteresting. In fact, I would suggest as a conjecture that, if
the sample size
is large enough to trust the asymptotic p-values from the Brant test, then the sample size is large enough that trivial departures from prop. odds will achieve small p-values. I would suggest instead
approaching this specification problem by looking at the relative increase in the pseudo-R^2 value associated with moving to a non-proportional odds model. My own experiments on using such measures
to address the related problem of variable choice ordinal logit models shows that one measures is about as good as the next. (see my comment in http://www.stata.com/statalist/archive/2008-03/
msg00249.html for a brief discussion of this point and a citation.)
Now, I admit that there is a problem in knowing exactly how big a *relative* change in R^2 (10%?) warrants a more complicated model, but I don't think this is worse than to p-values as the sole
Mike Lacy, Assoc. Prof.
Soc. Dept., Colo. State. Univ.
Fort Collins CO 80523 USA
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
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* http://www.ats.ucla.edu/stat/stata/ | {"url":"http://www.stata.com/statalist/archive/2008-04/msg00765.html","timestamp":"2014-04-18T01:00:52Z","content_type":null,"content_length":"7400","record_id":"<urn:uuid:8ed7990c-56b9-45f6-92a6-155ce603741c>","cc-path":"CC-MAIN-2014-15/segments/1397609532374.24/warc/CC-MAIN-20140416005212-00632-ip-10-147-4-33.ec2.internal.warc.gz"} |
Cubic and Quartic Functions Problems: Part 3
April 26th 2011, 09:10 PM #1
Nov 2010
I am having trouble with a couple of questions in terms of the proccess of solving the questions:
4 A cuboid (recatangular prisim) has dimensions x metres, h metres and 4x metres. The cuboid is made of 640m of wire
a Find h in terms of x
b Find the volume, Vm^3 of the cuboid in terms of x
c Find V when x = 11
d Find the possible values of x for the cuboid to exist
e Find the possible values of x when V = 60,000 (give 2 decimal places)
f Find the max volume of the cuboid and the value of x for which it occurs
That is a long list of questions, and I will be greatly apprecieated if any of you would have the time to help and explain these questions for me.
Last edited by Chris L T521; April 26th 2011 at 11:16 PM. Reason: Splitting up original thread.
I am having trouble with a couple of questions in terms of the proccess of solving the questions:
4 A cuboid (recatangular prisim) has dimensions x metres, h metres and 4x metres. The cuboid is made of 640m of wire
a Find h in terms of x
b Find the volume, Vm^3 of the cuboid in terms of x
c Find V when x = 11
d Find the possible values of x for the cuboid to exist
e Find the possible values of x when V = 60,000 (give 2 decimal places)
f Find the max volume of the cuboid and the value of x for which it occurs
That is a long list of questions, and I will be greatly apprecieated if any of you would have the time to help and explain these questions for me.
1. Draw a sketch! (see attachment)
2. The cuboid contains the edges x, 4x and h four times each. Since all edges together have a length of 640 m you already know:
$4 \cdot x + 4 \cdot 4x + 4 \cdot h = 640~\implies~5x+h=160$
3. Now you're able to answer all your questions.
April 27th 2011, 04:45 AM #2 | {"url":"http://mathhelpforum.com/algebra/178716-cubic-quartic-functions-problems-part-3-a.html","timestamp":"2014-04-16T20:16:39Z","content_type":null,"content_length":"35775","record_id":"<urn:uuid:f59fc1b0-59b9-488f-a291-87caab765d47>","cc-path":"CC-MAIN-2014-15/segments/1397609539066.13/warc/CC-MAIN-20140416005219-00195-ip-10-147-4-33.ec2.internal.warc.gz"} |
from The American Heritage® Dictionary of the English Language, 4th Edition
• n. A person skilled or learned in mathematics.
from Wiktionary, Creative Commons Attribution/Share-Alike License
• n. An expert on mathematics.
from the GNU version of the Collaborative International Dictionary of English
• n. One versed in mathematics.
from The Century Dictionary and Cyclopedia
• n. One who is versed in mathematics.
• n. An astrologer.
from WordNet 3.0 Copyright 2006 by Princeton University. All rights reserved.
• n. a person skilled in mathematics
Of Middle French mathematicien, from mathematique, of Latin mathematicus. (Wiktionary)
Log in or sign up to get involved in the conversation. It's quick and easy.
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(if they disagree with the 'odd' number, reply with 'prove it'
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irrationalirratiirratirrairririan anasuggest suggestsuggessuggesuggsugsusI I | {"url":"https://www.wordnik.com/words/mathematician","timestamp":"2014-04-16T08:14:57Z","content_type":null,"content_length":"38497","record_id":"<urn:uuid:984d791b-faa0-4e9e-8cd5-d7323836020e>","cc-path":"CC-MAIN-2014-15/segments/1397609521558.37/warc/CC-MAIN-20140416005201-00566-ip-10-147-4-33.ec2.internal.warc.gz"} |
limit of sequences and series-real analysis
October 18th 2012, 05:17 PM
limit of sequences and series-real analysis
please help me to find the answer for the following problem immediately...
limit (sin(sigma k from 1 to infinity)(-1)^k+1/2k-1)) as n goes to infinity.
basically find the limit of SIN(Leibniz formula) as k goes to infinity.
October 18th 2012, 05:38 PM
Re: limit of sequences and series-real analysis
using the alternating series test,I have found that the Leibnitz series converges to zero.but is it sufficient for us to say that the sin of that series converges to zero???
October 18th 2012, 07:58 PM
Re: limit of sequences and series-real analysis
I think this is what you are asking.
$\lim_{k \to \infty} \sin \left( \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{2k-1} \right)$
Since the sine function is continous, the limit can be moved \lim_{k \to \infty} inside the function.
The sum inside it the infinite series for the arctangent function evaluated at 1.
The $\tan^{-1}(1)=\frac{\pi}{4}$
$\sin \left(\lim_{k \to \infty} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{2k-1} \right)=\sin\left( \frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$ | {"url":"http://mathhelpforum.com/advanced-math-topics/205635-limit-sequences-series-real-analysis-print.html","timestamp":"2014-04-21T16:11:27Z","content_type":null,"content_length":"6077","record_id":"<urn:uuid:3b685e1c-9927-470f-990f-948cc0a586a0>","cc-path":"CC-MAIN-2014-15/segments/1397609540626.47/warc/CC-MAIN-20140416005220-00607-ip-10-147-4-33.ec2.internal.warc.gz"} |
Zero-Error Classical Channel Capacity and Simulation Cost Assisted by Quantum Non-Signalling Correlations
Seminar Room 1, Newton Institute
We study the one-shot zero-error classical capacity of quantum channels assisted by quantum non-signalling correlations, and the reverse problem of exact simulation. Both lead to simple semi-definite
programmings whose solutions can be given in terms of the conditional min-entropies. We show that the asymptotic simulation cost is precisely the conditional min-entropy of the Choi-Jamiolkowski
matrix of the given channel. For classical-quantum channels, the asymptotic capacity is reduced to a quantum fractional packing number suggested by Harrow, which leads to an operational
interpretation of the celebrated Lovasz function as the zero-error classical capacity of a graph assisted by quantum non-signalling correlations. This talk is based on a joint work with Andreas
Winter (UAB).
The video for this talk should appear here if JavaScript is enabled.
If it doesn't, something may have gone wrong with our embedded player.
We'll get it fixed as soon as possible. | {"url":"http://www.newton.ac.uk/programmes/MQI/seminars/2013100314001.html","timestamp":"2014-04-17T18:36:20Z","content_type":null,"content_length":"6515","record_id":"<urn:uuid:555f359c-e4f5-4c1b-8e3d-3afc8e2e3719>","cc-path":"CC-MAIN-2014-15/segments/1397609530895.48/warc/CC-MAIN-20140416005210-00176-ip-10-147-4-33.ec2.internal.warc.gz"} |
Round Robin Tournament Scheduling - 9 golfers - 4 games
YaBB Newbies
Posts: 5
Re: 9 golfers - 4 games
Reply #9 - 04/15/11 at 10:03:39
Thanks Ian. I am also curious about close solutions for the number of players between 22 and 33.
For instance, according to the schoolgirl problem, 18 does not meet the "perfect" requirements of 6n+3, but for a total of 18 players you can complete 8 rounds and only be missing one opponent. Here
is the table:
{"day 1: ", "ABC", "DEF", "GHI", "ahf", "dbi", "gec"}
{"day 2: ", "Abc", "Def", "Ghi", "aHF", "dBI", "gEC"}
{"day 3: ", "abC", "deF", "ghI", "AHf", "DBi", "GEc"}
{"day 4: ", "aBc", "dEf", "gHi", "AhF", "DbI", "GeC"}
{"day 5: ", "ADG", "BEH", "CFI", "aei", "bfg", "cdh"}
{"day 6: ", "Adg", "Beh", "Cfi", "aEI", "bFG", "cDH"}
{"day 7: ", "adG", "beH", "cfI", "AEi", "BFg", "CDh"}
{"day 8: ", "aDg", "bEh", "cFi", "AeI", "BfG", "CdH"}
The only pairs that don't play together are Aa, Bb, Cc, Dd, etc. So I would consider that a very close solution.
The same is true for 24 players. You can complete 10 normal rounds and in the last round have 6 foursomes play to make everything "perfect". See below:
{"day 1: ", "ABC", "DEF", "GHI", "JKL", "MNO", "PQR", "STU", "VWX"}
{"day 2: ", "JOT", "ESB", "API", "DML", "HCV", "QWN", "FKU", "GRX"}
{"day 3: ", "DVK", "SFO", "JQI", "EHL", "PTG", "WRC", "BMU", "ANX"}
{"day 4: ", "EGM", "FBV", "DWI", "SPL", "QKA", "RNT", "OHU", "JCX"}
{"day 5: ", "SAH", "BOG", "ERI", "FQL", "WMJ", "NCK", "VPU", "DTX"}
{"day 6: ", "FJP", "OVA", "SNI", "BWL", "RHD", "CTM", "GQU", "EKX"}
{"day 7: ", "BDQ", "VGJ", "FCI", "ORL", "NPE", "TKH", "AWU", "SMX"}
{"day 8: ", "OEW", "GAD", "BTI", "VNL", "CQS", "KMP", "JRU", "FHX"}
{"day 9: ", "VSR", "AJE", "OKI", "GCL", "TWF", "MHQ", "DNU", "BPX"}
{"day 10: ", "GFN", "JDS", "VMI", "ATL", "KRB", "HPW", "ECU", "OQX"}
{"day 11: ", "AFMR", "BNHJ", "CPOD", "EQVT", "GKSW", "IXUL", , }
So here is my question. I'd like to create tables for an upcoming tournament. I suspect that there will be between 22 and 33 players show up and I need to be prepared ahead of time to just fill in
the blanks. It appears that there are better solutions that exist for plugging in teams rather than just creating a bye.
For instance, on game day if I have 26 players show up, is there a better or closer solution that exists rather than using the 27 player table and just making one of the players a BYE? The thing that
helps me is it is unlikely that all rounds will be completed, so an odd round like I have shown above will not affect anything. I know ahead of time that we will not complete all the rounds for a
"perfect" solution. What I am interested in is that we complete at least say 8 rounds with "perfect" threesomes and no repeats. | {"url":"http://www.devenezia.com/round-robin/forum/YaBB.pl?num=1160084931/1","timestamp":"2014-04-19T22:06:34Z","content_type":null,"content_length":"78277","record_id":"<urn:uuid:bc1ec38e-cc17-4018-85cc-96928b9e51ab>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00006-ip-10-147-4-33.ec2.internal.warc.gz"} |
Mixing and hitting times for finite Markov chains
Roberto Imbuzeiro Oliveira (IMPA)
Let $0<\alpha<1/2$. We show that that the mixing time of a continuous-time Markov chain on a finite state space is about as large as the largest expected hitting time of a subset of the state space
with stationary measure $\geq \alpha$. Suitably modified results hold in discrete time and/or without the reversibility assumption. The key technical tool in the proof is the construction of random
set $A$ such that the hitting time of $A$ is a light-tailed stationary time for the chain. We note that essentially the same results were obtained independently by Peres and Sousi.
Full Text: Download PDF | View PDF online (requires PDF plugin)
Pages: 1-12
Publication Date: August 27, 2012
DOI: 10.1214/EJP.v17-2274
• List of open problems from AIM Workshop on Algorithmic Convex Geometry. http://www.aimath.org/WWN/convexgeometry/convexgeometry.pdf. Compiled by Navin Goyal (2009).
• Aldous, David J. Some inequalities for reversible Markov chains. J. London Math. Soc. (2) 25 (1982), no. 3, 564--576. MR0657512
• Aldous, D. and Fill, J.A.: Reversible Markov Chains and Random Walks on Graphs. Book draft available from http://www.stat.berkeley.edu/~aldous/.
• Aldous, David; Lovász, László; Winkler, Peter. Mixing times for uniformly ergodic Markov chains. Stochastic Process. Appl. 71 (1997), no. 2, 165--185. MR1484158
• Levin, David A.; Peres, Yuval; Wilmer, Elizabeth L. Markov chains and mixing times. With a chapter by James G. Propp and David B. Wilson. American Mathematical Society, Providence, RI, 2009.
xviii+371 pp. ISBN: 978-0-8218-4739-8 MR2466937
• Lovász, László; Winkler, Peter. Mixing of random walks and other diffusions on a graph. Surveys in combinatorics, 1995 (Stirling), 119--154, London Math. Soc. Lecture Note Ser., 218, Cambridge
Univ. Press, Cambridge, 1995. MR1358634
• Peres,Y.: Personal communication (2011).
• Peres, Y. and Sousi, P.: Mixing times are hitting times of large sets, arXiv: 1108.0133.
This work is licensed under a
Creative Commons Attribution 3.0 License | {"url":"http://www.emis.de/journals/EJP-ECP/article/view/2274.html","timestamp":"2014-04-19T20:13:13Z","content_type":null,"content_length":"18790","record_id":"<urn:uuid:8fa81150-88bd-40b0-9ba4-8393faad28da>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00595-ip-10-147-4-33.ec2.internal.warc.gz"} |
Physics Forums - View Single Post - Charge distribution of a uniformly charged disk
1. The problem statement, all variables and given/known data
The problem can be found in Jackson's book, I think in chapter 1 problem 3 or something like this.
I must determine the charge distribution of a uniformly charged disk of radius R in spherical coordinates (I've done it in cylindrical coordinates and had no problem). The total charge is Q.
I've found a solution on the internet but the answer is different from mine.
I forgot to mention that I have to use Dirac's delta.
2. Relevant equations
[itex]\int _{\mathbb{R}^3} \rho (\vec x )=Q[/itex].
3. The attempt at a solution
Since the charges are over a 2d surface, there will be 1 Dirac's delta in the expression for rho, the charge density. I will use Heaviside's step function because the surface is limited.
Let [itex](r, \theta , \phi )[/itex] be the coordinates. I make the ansatz/educated guess that [itex]\rho[/itex] is of the form [itex]C \delta \left ( \theta - \frac{\pi }{2} \right ) \Theta (r \sin
\theta -R)[/itex].
Integrating this distribution in all the space, I reach that C is worth [itex]\frac{3Q}{2\pi R^3}[/itex].
Therefore [itex]\rho (r, \theta )=\frac{3Q\delta \left ( \theta - \frac{\pi }{2} \right ) \Theta (r \sin \theta -R)}{2\pi R^3}[/itex].
However the solution provided on the internet is [itex]\rho (\vec x )=\frac{q }{\pi R^2r} \delta \left ( \theta - \frac{\pi }{2} \right ) \Theta (r -R)[/itex].
Are they both equivalent (I doubt it), if not, did I do something wrong? If so, what did I do wrong? Thanks a lot! | {"url":"http://www.physicsforums.com/showpost.php?p=3838599&postcount=1","timestamp":"2014-04-21T14:53:33Z","content_type":null,"content_length":"10409","record_id":"<urn:uuid:3435e120-9fe4-4ca0-8038-986a313ad552>","cc-path":"CC-MAIN-2014-15/segments/1397609540626.47/warc/CC-MAIN-20140416005220-00542-ip-10-147-4-33.ec2.internal.warc.gz"} |
Haskell' - class aliases
John Meacham john at repetae.net
Fri May 2 06:47:52 EDT 2008
On Fri, May 02, 2008 at 11:24:11AM +0100, Simon Peyton-Jones wrote:
> | The more I think about it, I think 'superclass' is just the wrong
> | terminology for dealing with class aliases. Superclass implies a strict
> | partial order on classes, which just isn't the case for class aliases,
> | for instance
> |
> | > class alias Foo a => Foo a = Bar a where ...
> Crumbs! I have no idea what that means! Did you really mean to repeat "Foo"? According to your expansion in type signatures
> f :: (Foo a) => ...
> expands to
> f :: (Foo a, Bar a) => ...
> which presumably expands again. I'm totally lost here
Yes I did, because I wanted to make the differences between class alias
contexts and superclasses very clear, the above context is valid, if
vacuous. the expansion goes as follows .
1. Foo a --> reduce(Foo a,Bar a)
-- Foo a expanded
2. reduce(Foo a,Bar a) --> (Foo a,Bar a)
-- no entailment reduction possible, reduction is unchanged from H98
3. (Foo a,Bar a) -> reduce(Foo a,Bar a,Bar a)
-- Foo a expanded
4. reduce(Foo a, Bar a, Bar a) -> (Foo a, Bar a)
-- reductino removes duplicates
5. we notice we are the same as in step #2. fixed point reached, we stop
6. we remove all class aliases from result:
(Foo a, Bar a) -> Bar a
7. 'Bar a' is our final result.
informal proof of termination:
each step adds a new class or class alias to the context, there are a
finite number of classes or class aliases, therefore we must eventually
reach a fixed point.
> Have a look at my last message, which gives a variant of your
> desugaring that IMHO greatly clarifies the meaning of (what I
> understand by) aliases.
I think the difference in what we mean is that I intend class aliases to
be a true bijection in all contexts (isomorphism?) between a single
alias and a set of classes. This is opposed to superclasses which are a
one directional implication.
One of my main motivations is being able to mix unchanged H98 and H'
code (with different numerical hierarchies, and both calling each other)
without modifications or prefered treatment for either. this means
instances for H' must silently and transparently create instances for
H98 classes and vice versa, moreso, type signatures should be
As in, the H' specification should be able to make absolutely no
reference to H98 and vice versa, yet class aliases allow one to write a
compiler that seamlessly allows mixing code from the two without
compromising the design of either.
John Meacham - ⑆repetae.net⑆john⑈
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Hyperdimensional Hurricanes, Part Three: The Physics
Hyperdimensional Hurricanes?
Part Three: The Physics
By Richard C. Hoagland
© 2004 The Enterprise Mission
So, why could I still not find the source data for MSNBC’s amazing Ivan promo …?
I had all my “scouts” out looking too … like David Wilcock, who in addition to searching the complexity of NOAA’s websites, was sending out e-mails to his own extensive network. One of those
resulted in some stunning “hurricane hunter” aircraft stills (below) from a “Peter Thompson” – which shows the remarkable vorticular activity in Ivan’s eye on Saturday, September 11^th -- but still
no satellite shots of the remarkable geometry shown on MSNBC on Friday night, the 10^th.
Cheryll, despite her network and meteorological contacts … was also coming up with nothing.
Then, as Ivan finished his trek across the Gulf of Mexico and was approaching dreaded landfall on the southern coast of the United States – with the whole world watching this, now at Category 4 – the
“Ivan promo” suddenly was back on MSNBC!
On Wednesday night, September 15^th, I was showing Robin and two more Enterprise associates – Nick Skouras and Dana Balaban -- comparisons between this amazing (if still incredibly elusive) MSNBC
footage, the live MSNBC coverage of Ivan (even then closing in on southern Alabama) … and the NOAA computer image of Isabel’s astonishing geometry in 2003....
When, suddenly I noticed something that simply “didn’t fit”:
The digital numerals indicating elapsed time superimposed over the image in the MSBNC “loop” (below -- left) precisely matched the time listed on the NOAA Isabel image (below -- right)!
Now -- what were the odds that two separate hurricanes, in two separate oceans, at two separate latitudes -- would both reach Category 5 status … and display interior hyperdimensional geometry …
exactly one year after one another?—
AT EXACTLY THE SAME TIME?!
There weren’t any such odds ….
The two images were THE SAME … of Isabel -- NOT Ivan! Despite MSNBC’s totally misleading “supers,” both images were from September 12, 2003!
Comparison of the details of the outlying “rain bands” circling both storms (above) cinched it: MSNBC had simply taken an old “loop” from Isabel from a year ago … and pasted new “Ivan icons” on it!
A bit more searching on the NOAA sites (now that I knew what I was looking for!) revealed a stunning “rapid scan” GOES movie of Isabel’s amazing eye geometry, from 2003 – taken about once a minute!
The stability of the interior geometry of Isabel over almost half an hour (below) – in a supposedly wildly turbulent environment of slashing wind and rain – is truly astonishing ... as is the
subsequent, totally hyperdimensional geometric evolution which manifests over the next four and half hours, covered in the remainder of NASA/NOAA video….
The closest three-dimensional analogy to what we believe is creating this “geometric control” of the clouds within Isabel’s “eye” is given by the science of “Cymatics”: the term comes from the Greek
kyma (“great wave”), and is the modern study of “wave phenomena and its interaction with a variety of fluids.”
This relatively arcane field was pioneered in the 1950s by the late Hans Jenny, a Swiss medical doctor who followed the work of German physicist and acoustician, Ernst F. F. Chladni. Chladni, in the
late 18th century, created intricate sand patterns by resonating a steel plate with an ordinary violin bow. In his extensions of Chladni’s work, Jenny used an electronic sine wave
hi-fi loud speaker to vibrate various powders, pastes and liquids – thereby succeeding in making visible the ability of “standing waves of sound” to create striking geometric order in a range of
fluid materials (below).
In our Hyperdimensional Model, it isn’t “sound” in the middle of a hurricane which creates the regular geometry seen in the NOAA images; it is the interaction of the energy/information from higher
dimensions with our own -- via the intervening medium of a massless aether – which creates similar standing wave patterns (a wave ... is a wave … is a wave) when the hyperdimensional “gate” is
established … causing visible matter (in its mobile forms …) to conform to the shape of these otherwise invisible “aetheric, hyperdimensional patterns.”
In the images from Dr. Jenny's work above, water is being vibrated by precisely regulated sound waves, with tiny particles (colloids) suspended inside of it. These "colloids" gather together in the
areas of highest pressure, making the otherwise invisible geometry visible.
In the Hyperdimensional Model, matter itself is similarly formed by "aetheric, hyperdimensional" energy waves, which behave like a fluid as they flow into and out of our own "3-D reality"... creating
what we recognize as "matter" and "energy" along the way. Put simply, Jenny's work tells us, "Any fluid, once vibrated at a steady rate by an external energy source, will create regular geometric
patterns -- be it in a laboratory ... or in the aetheric fabric of three-dimensional Reality itself."
These hidden geometric patterns ultimately trace back to a topological hyperdimensional analysis called the Schlafli Double-Six (below) – a mathematical projection of the literal geometry of a “gate
between dimensions.” A projection which, in turn, resolves
into the familiar circumscribed tetrahedral geometry (below) which, in a rotating planetary sphere, predicts the locations of the largest, stationary hyperdimensional energy transfer locations (
upwellings) – relative to the planet’s spin axis – at ~19.5 degrees.
Around 200 B.C., the Greek geometer Euclid proved that there are only five regular polyhedra (technically called “polytopes” by topologists) in three spatial dimensions -- the tetrahedron, cube,
octahedron, icosahedron, and the dodecahedron. Fast forward the film: in 1901, Ludwig Schlafli showed that there are only six regular polychora1 (or polytopes) in four dimensions.
Therefore, if a “gate” were to be resonantly opened between dimensions, according to Schlafli’s pioneering topological calculations of over a century ago, it could only manifest (in our three spatial
dimensions) as the geometry of one of the five regular “Platonic solids” (below).
It is this energy transfer process, we propose – as three-dimensional aetheric waves, resonantly mirroring the invisible force structure of these topologies between dimensions -- which created the
startling, evolving “Platonic” geometry of Isabel’s “eye” -- as the storm attained a Category 5 (below) ….
Meanwhile, keep in mind the very similar formations that have been observed in the perpetual storms occurring in the polar regions of Jupiter and Saturn:
As time advanced, the striking five-sided “eye” initially captured by the NOAA GOES satellite scans did indeed evolve (below) -- into progressive, two dimensional “slices’ of the less complex
Platonic solids ....
A final observation:
The hyperdimensional hurricane model makes one major physical prediction: the decreased inertia of the “eye’s” central spinning clouds (the “DePalma Effect”) -- due to their high rotational velocity
-- should significantly enhance vertical convection in those clouds – leading to higher than average thunderstorms within the “eye.”
Look at the comparison images (below) -- taken at the beginning and the end of Isabel’s remarkable geometric presentation; the image on the left was taken as the “controlling” geometry was at its
height; the image on the right was taken about five hours later, as the “geometry” had almost disappeared.
Notice the striking shadow changes ….
In the image on the left (maximum modeled hyperdimensional control), not only is the blatant Platonic geometry apparent (with multiple, parallel delineations!) ... this phase displays (as the model
specifically predicts) the maximum cloud heights! The “eye” is a five-sided pentagonal plateau!
Five hours later, in the image on the right, these towering convective clouds have all but disappeared … replaced with a “normal” hurricane’s central “eye” depression ….
Clearly, whatever hyperdimensional forces were at work here … were acting precisely as the model would predict! | {"url":"http://www.enterprisemission.com/hurricane3.htm","timestamp":"2014-04-16T07:13:52Z","content_type":null,"content_length":"52346","record_id":"<urn:uuid:791d916d-12fb-4257-a25e-a14324f878a5>","cc-path":"CC-MAIN-2014-15/segments/1397609521558.37/warc/CC-MAIN-20140416005201-00454-ip-10-147-4-33.ec2.internal.warc.gz"} |
Help me with my Geometry! ;3
September 3rd 2007, 01:53 PM #1
Sep 2007
Siiigh...I am so horrible at math.. I'm a class behind everyone lol
Anyway, Heres my problem:
Line segment AC in bisected by ray BD at point B. Ray BT bisects angle DBC.
a. If angle DBC= 25x+7 and angle TBC= 12x+5, find the degree measures of angle TBC, DBT and DBC
b. If angle CBT= 10x-8 and angle TBA=15x-12, find the degree measures of both angles.
I drew a diagram and everything, but I am still utterly confused.
<33 Buyo
Siiigh...I am so horrible at math.. I'm a class behind everyone lol
Anyway, Heres my problem:
Line segment AC in bisected by ray BD at point B. Ray BT bisects angle DBC.
a. If angle DBC= 25x+7 and angle TBC= 12x+5, find the degree measures of angle TBC, DBT and DBC
b. If angle CBT= 10x-8 and angle TBA=15x-12, find the degree measures of both angles.
I drew a diagram and everything, but I am still utterly confused.
<33 Buyo
someone responded to your question here.
Please do not double post, it is against the rules
September 3rd 2007, 03:02 PM #2 | {"url":"http://mathhelpforum.com/geometry/18433-help-me-my-geometry-3-a.html","timestamp":"2014-04-16T13:05:29Z","content_type":null,"content_length":"34130","record_id":"<urn:uuid:b0aa334d-287e-4d74-bdec-fd56d93c4e45>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00045-ip-10-147-4-33.ec2.internal.warc.gz"} |
Summary: On the Implementation Complexity of
Specications of Concurrent Programs
Paul C. Attie 1
College of Computer Science, Northeastern University, Boston, MA
MIT Laboratory for Computer Science, Cambridge, MA
Abstract. We present a decision algorithm for the following problem:
given a specication, does there exist a concurrent program which both
satises the specication and which can be implemented in hardware-
available operations in a straightforward manner, i.e, without long cor-
rectness proofs, and without introducing excessive blocking and/or cen-
tralization? In case our decision algorithm answers \yes," we also present
a synthesis method to produce such a program. We consider specica-
tions expressed in branching time temporal logic. Our result gives a way
of classifying specications as either \easy to implement" or \diĘcult to
implement," and can be regarded as the rst step towards a notion of
\implementation complexity" of specications.
1 Introduction
One of the major approaches to the construction of correct concurrent programs | {"url":"http://www.osti.gov/eprints/topicpages/documents/record/819/3742917.html","timestamp":"2014-04-19T12:43:29Z","content_type":null,"content_length":"8240","record_id":"<urn:uuid:5007a6d1-3c12-438c-9754-70fbd3364c4e>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00562-ip-10-147-4-33.ec2.internal.warc.gz"} |
FOM: Significance and significant people
Torkel Franzen torkel at sm.luth.se
Thu Mar 19 05:05:27 EST 1998
Neil Tennant says:
>Not so; the epochal advance intimated at this
>stage is that for any proposed large cardinal axiom, Friedman's method
>can be adjusted to provide a natural combinatorial principle that
>turns out to be provably equivalent to the consistency of that large
OK, so what you regard as an epochal advance is a result "intimated"
but not stated. It would less misleading, I think, to speak of
an "intimated" or "hypothetical" epochal advance.
Setting aside the matter of "epochal advance", which surely is of
minor importance, you explain further how you look at the importance
of the stated result:
>Imagine writing down, in primitive notation,
>(B) Con(ZFC + existence of appropriate large cardinals)
>so that its combinatorial character is explicit. Would the result
>strike you as (i.e. look to you like) the kind of claim that ought,
>if true, to be true for simple reasons? I think not, since you know
>its provenance as, precisely, a consistency claim about an extremely
>powerful extension of currently accepted mathematics.
To this I can only reply that written in primitive notation, this
statement wouldn't strike me as anything at all, since I wouldn't be
able to make head or tail of it. As to whether (B) "ought, if true, be
true for simple reasons", I don't see any grounds for holding this.
You contrast (B) with (A):
>It would be several orders of syntactic magnitude easier to write
>down, in similarly primitive notation, Friedman's combinatorial
>principle (A). Of this statement I would say that it looks to me like
>the kind of claim that ought, if true, to be true for simple
>reasons. Again, this is because I know its provenance as a claim about
>inserting nodes in finite trees according to certain clearly stated
You emphasize the "provenance" of (A), by which I assume you mean that
you know Friedman to have arrived at this statement, not in the course
of thinking about large cardinals, but in the course of thinking about
inserting nodes in finite trees. I suppose some such distinction also
explains why you are not equally impressed by an equivalent of B in
the form of a statement of the form "the Diophantine equation
p(k1,..kn)=0 has no solution". Would you say that such a statement
"ought, if true, to be true for simple reasons" if it has arisen in
the context of Diophantine studies, but not necessarily if it has been
first brought to our attention via recursion theory? I don't see
any justification for such a view.
You comment, finally:
>THEN one looks at Friedman's result, and learns that (A) implies
>(B). One realizes further that, if one were ever to prove the
>combinatorial principle (A), it could only be by dint of assuming the
>existence of a large cardinal even larger than those whose consistency
>is asserted by (B).
I don't realize anything of the kind. What is your justification for this
Torkel Franzen
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Moving Averages
A moving average is the average price of a contract over the previous n-period closes. For example, a 9-period moving average is the average of the closing prices for the past 9 periods, including
the current period. For intra-day data the current price is used in place of the closing price.
The moving average is used to observe price changes. The effect of the moving average is to smooth the price movement so that the longer term trend becomes less volatile and therefore more obvious.
When the price rises above the moving average, it indicates that investors are becoming bullish on the commodity. When the prices falls below, it indicates a bearish commodity. As well, when a moving
average crosses below a longer term moving average, the study indicates a down turn in the market. When a short term moving average crosses above a longer term moving average, this indicates an
upswing in the market. The longer the period of the moving average, the smoother the price movement is. Longer moving averages are used to isolate long term trends.
There are many variations of the moving average available, such as the moving average of the high prices and the low prices represented in a channel called the Moving Average High/Low channel, this
is also known as the Jake Bernstien's high/low channel. There is also the Moving Average Percent Channel. The first argument (X) is the x-day moving average of the closing price and the second
argument (Y) is used as (Y/10,000*Price) plotted as a channel around over and under the result of the x-day moving average.
The Exponential Moving Average assigns a weight to the price data as the average is calculated. The more recent the price the heavier the weighting. The oldest price data in the exponential moving
average is never removed from the calculation, but its weighting is decreased the further back it gets in the calculations.
As an example, the calculations for a 10 period exponential moving average are as follows.
• First, go back to the beginning of trading or back 1 year or anything consistent. The longer the period, the more accurate the result.
• Add up the closing prices for the first 10 periods and divide by 10. This is the result for the 10th period (there are no results for periods 1 thru 9).
• Then take 9/10 of the 10th period result plus 1/10 of the 11th period close. This is the 11th day result, etc, etc.
Barchart.com uses the classical exponential smoothing formulas described by H. Wells Wilder in his book "New Concepts in Technical Analysis". This defines the smoothing factor as 1/days or 1/3 for a
3 day exponential moving average study. The result of the study will then be 2/3 of the old value plus 1/3 of the new. Others have developed their own formulas, the most notable being Trade Station.
In the Trade Station and some other look-alike formulas, the smoothing factor is defined as 2/(days+1), which for the 3 day study produces 2/4 or 1/2. This gives a result of 1/2 of the old plus 1/2
of the new. 1/2 smoothing will give "faster" results than 1/3 smoothing. You might get an equivalent result if you used a 2-day smoothing factor on the barchart calculations. Alternatively, if you
want a 1/3 smoothing on a site using the Trade Station logic, you might try a 5 day factor, 2/(5+1)= 2/6 = 1/3.
The Offset Moving Average is a simple moving average offset by moving the average "x" periods to the right, where "x" is the second argument. The first argument is used to calculate the simple moving
average of the price, and the second argument determines the number of offsets to the right, hence shifting the moving average "x" periods to the right. The Exponential Moving Average is the same
except it uses the exponential moving average in the calculation.
The Offset MidPoint Average is a simple moving average calculated from the average of the high and low for the period, offset by moving the average "x" periods to the right, where "x" is the second | {"url":"http://acs.barchart.com/sample/support/studies/movave.htm","timestamp":"2014-04-18T02:58:12Z","content_type":null,"content_length":"5025","record_id":"<urn:uuid:a7531fb8-ae44-4bf4-a756-975dda71f5ae>","cc-path":"CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00577-ip-10-147-4-33.ec2.internal.warc.gz"} |
Adjointness in foundations
Adjointness in foundations, with author commentary
F. William Lawvere
Originally published in:
Dialectica, 23 (1969), used by permission of Blackwell Publishing.
2000 MSC: 00A30, 14F20, 18B25, 18B30, 18B40, 55U40, 81P05
Keywords: Formal-conceptual duality, cartesian-closed categories, algebraic logic, globalized Galois connections
Republished in:
Reprints in Theory and Applications of Categories, No. 16 (2006), pp 1-16 (revised 2006-10-30)
Version originally republished 2006-05-21 at: | {"url":"http://www.emis.de/journals/TAC/reprints/articles/16/tr16abs.html","timestamp":"2014-04-19T01:55:21Z","content_type":null,"content_length":"1688","record_id":"<urn:uuid:e011b5c1-2265-406b-804a-9ad5b1e1840f>","cc-path":"CC-MAIN-2014-15/segments/1397609535745.0/warc/CC-MAIN-20140416005215-00254-ip-10-147-4-33.ec2.internal.warc.gz"} |
John Fyfe
John Fyfe is an internationally recognized researcher into climate variability and change and is a Review Editor of the Intergovernmental Panel on Climate Change (IPCC) Fifth Assessment Report to be
published in 2013. He was a Lead Author of the IPCC Fourth Assessment Report (2007), and contributed to the Nobel Peace Prize that was awarded to the IPCC in 2007. He was an author of the Arctic
Climate Impacts Assessment (2005), and has received a number of awards including the President's Prize from the Canadian Meteorological and Oceanographic Society for his contributions to the
understanding of climate variability and change, especially in polar regions.
John Fyfe is a Senior Scientist in the Canadian Centre for Climate Modelling and Analysis of Environment Canada, and is an Adjunct Professor at the University of Victoria in the School of Earth and
Ocean Sciences.
Dror Bar-Natan
I believe math is too deep. Rather than making it deeper, a better use of my time would be to make some deep ends easier and more accessible. I believe math is too abstract, or at least appears to be
too abstract, for much of what may be computed hardly ever is. Thus, whenever I can, I code. Yet I have sinned a few times and written on deep math that was not accompanied with programs. I usually
work on knot theory and its surprising relationship with algebra, geometry and quantum field theory. I got my Ph.D. at Princeton, did time at Harvard, Hebrew U., Berkeley and MSRI, and I now work at
the University of Toronto.
Lisa Jeffrey
Lisa Jeffrey obtained her D.Phil. in mathematics in 1992 from Oxford University under the supervision of Michael Atiyah. Her first academic appointment in Canada was in the Mathematics Department at
McGill University. Since 1998 she has been Professor in the Mathematics Department of University of Toronto. Her research is on mathematical physics and symplectic geometry.
Marius Junge
Marius Junge, born in 1962, obtained his PhD and habilitation from the Christian-Albrechts Universität Kiel, Germany, after the study of mathematics, computer science and philosophy. After visiting
Isreal (92) and Denmark (98), he moved to the University of Illinois in 1999 and become professor in 2007.
His research interests are centered around functional analysis and include operator algebras, in particular operator spaces, noncommutative probability, noncommutative harmonic analysis, and more
recently quantum information theory. Many of Junge's work falls into the Grothendieck program for operator algebras. He was awarded the title of Doob Scholar from the mathematics department, and
received the Romano Professorial Scholarship from the College of Liberal Arts and Sciences at the University of Illinois. Alongside with Charles Feffermann he is one of the leaders of a laboratory at
the ICMAT, a Spanish research institute.
Ulrike Tillmann
Ulrike Tillmann received her PhD from Stanford in 1990 and her Habilitation from Bonn in 1996. She has held positions at Cambridge and Oxford and is interested in algebraic topology and its
applications, in particular to the study of moduli spaces and quantum field theories.
Professor Tillmann was an EPSRC Advanced Fellow 1997-2003; an invited speaker at the ICM 2002; a recipient of the LMS 2004 Whitehead Prize and the Bessel Research Award 2008 of the Humboldt Stiftung.
She is a Fellow of the Royal Society.
Margaret Walshaw
Margaret Walshaw is based at Massey University in New Zealand. She is Research Director at her College and co-director of the Centre of Excellence for Research in Mathematics Education. She is
Associate Editor of the Journal of Mathematics Teacher Education and a past chief editor of Mathematics Education Research Journal.
Margaret’s main research interest is in making connections between social theory and mathematics education and has developed through two edited books: ‘Mathematics Education within the Postmodern’,
and ‘Unpacking Pedagogy: New Perspectives for Mathematics Classrooms’. She is the author of the book ‘Working with Foucault in Education’ and co-author of ‘Are our standards slipping? Debates over
Literacy and Numeracy Standards’ and of ‘Effective Pedagogy in Mathematics: Best Evidence Synthesis Iteration’, written for the New Zealand Ministry of Education.
Douglas Farenick
Douglas Farenick obtained his PhD from the University of Toronto in 1990 under the direction of Chandler Davis. After two years at the Centre de recherches mathématiques in Montreal, he joined the
faculty at the University of Regina, the institution at which he did his undergraduate studies. His areas of research interest are operator theory, functional analysis, and linear algebra.
Don Stanley
Don Stanley completed his PhD in 1997 with Paul Selick at the University of Toronto. After three years working in Germany and France, he returned to Canada in 2000 and in 2003 moved to the University
of Regina. His main areas of research are homotopy theory and derived categories. | {"url":"http://cms.math.ca/Events/summer12/biographies","timestamp":"2014-04-20T23:33:41Z","content_type":null,"content_length":"16135","record_id":"<urn:uuid:1ab3e138-2773-4435-a564-deca456cab78>","cc-path":"CC-MAIN-2014-15/segments/1397609539337.22/warc/CC-MAIN-20140416005219-00223-ip-10-147-4-33.ec2.internal.warc.gz"} |
Calculating the Expected Cost. Help!
February 13th 2011, 06:34 AM #1
Feb 2011
Calculating the Expected Cost. Help!
Hi guys,
I have been set the following question and I'm really struggling and need some advice. I think I may have completed it, but it seems too simple, so I am just after some confirmation if it's right
or not.
The question is as follows:
A particular type of electronic component for use in PCs is mass produced and subject to quality control checks since it is known that 4% of all components produced in this way are defective. The
quality of a day's output is monitored as follows. A sample of 15 components is drawn from the day's output (which may be assumed to be large) and inspected for defective components. If this
sample contains 0 or 1 defectives the day's output is accepted , otherwise it is rejected. If it contains more than 2 defectives the output is rejected. If the sample contains 2 defectives a
second sample of 15 is taken. If this sample contains 0 defectives the output is accepted, otherwise it is rejected.
So far I have worked out the following:
P(0 defectives) = 0.542
P(1 defective) = 0.339
P(2 defectives) = 0.099
I have then worked out the probably of 2 defectives and then 0 defectives in the second sample is 0.099 x 0.542 = 0.054. However I'm not sure this is correct.
I have then worked out the probability of the accepted output to be 0.542 + 0.339 + 0.054 = 0.935
I have then been asked the following:
Suppose that it is estimated that it costs £100 to inspect a sample of 15. What is the expected cost of a day's sampling?
I have simply done 100 x 0.935 = £93.50. Cost of daily sampling = £93.50.
I appreciate any help given on this
Last edited by mr fantastic; February 13th 2011 at 10:34 AM. Reason: Restored deleted question.
If I read the question correctly, your final answer is not correct.
The probabilities (P(0 defectives), P(1 defective), P(2 defectives)) you calculated are correct, but your follow-up steps seem hinky. It is said that it costs 100 pounds to inspect a sample. You
will atleast inspect one sample a day. Hence the expected cost should be at least 100 pounds.
So the next question is: In which case do you sample more than once in a day?
Thanks for the response. However, I've asked for the thread to be deleted as it is related to some coursework that I am doing, and I was unaware that I couldn't post questions that are related to
any academic work. I appreciate your help.
Why does this tread require deleting? Your duplicate post you also required deleting as well. It makes sense to have one deleted but both? The only possible reason you want that could be that
what Pim wrote allowed you to fix your mistake and this is an assignment to be graded so you are detroying the evidence.
If you don't have to turn it in for a grade, you can ask for help.
The work doesn't count towards my overall grade, but it is work that we have to complete for lessons and we get assessed on this. I've asked for them both to be deleted as I thought they
infringed the forum's rules.
February 13th 2011, 09:43 AM #2
Dec 2008
The Netherlands
February 13th 2011, 09:47 AM #3
Feb 2011
February 13th 2011, 09:49 AM #4
MHF Contributor
Mar 2010
February 13th 2011, 09:50 AM #5
MHF Contributor
Mar 2010
February 13th 2011, 09:53 AM #6
Feb 2011 | {"url":"http://mathhelpforum.com/statistics/171104-calculating-expected-cost-help.html","timestamp":"2014-04-16T12:06:25Z","content_type":null,"content_length":"45249","record_id":"<urn:uuid:9c6086b2-cd48-47d3-8ed4-88503c0c395a>","cc-path":"CC-MAIN-2014-15/segments/1397609523265.25/warc/CC-MAIN-20140416005203-00199-ip-10-147-4-33.ec2.internal.warc.gz"} |
Ordered Dynamics in Genetic Networks
Steve E. Harris, Bruce Sawhill, Andrew Wuensche, Stuart Kauffman.
Derrida plots made with DDLab
These plots were made for random Boolean networks, n=1000, for k=3, 4, and 5, using the old+new data. The C values are higher in all cases.
k=3, C=0.795
k=4, C=0.708
k=5, C=0.649
There are two diagrams for each k.
left: an overall plot showing rules at random, and with canalizing set. The normalized Hamming range is 0 to 0.5. Each Hamming distance is plotted for a sample of 25 initial pairs of states.
right: details of the initial Hamming range (0 to 0.01) to calculate the Hamming coefficient based on the first 5 steps. The average slope between these points and the origin is taken as the initial
slope. If the initial slope = x degrees, then the Derrida coefficient, Dc=log2(tan(x)). A slope of 45 degrees (Dc=0) indicates the dynamics is balanced between order and chaos. A slope above the 45
degree diagonal (positive Dc) is in the chaotic regime, below (negative Dc) is in the ordered regime. If Derrida plot is based on more than 1 iteration, the slope above or below the diagonal is
amplified, giving a clear indication where it lies relative to the 45 degree diagonal.
The plots how 4 curves. The top curve is for rules set at random. The 3 lower curves are for canalizing set, for 1, 2 and 3 iterations. Each Hamming distance is plotted for a sample of 1000 initial
pairs of states. These are typical plots, and the variance for other networks with the same parameters was found to be low.
Comments: the curves no longer lie on each other, but become shallower with greater k. The k=3 curve in the old paper looks too shallow compared with new k=3 curve, which is steeper despite having
greater C.
Click on each image to see it full size.
k=3, n=1000. Steep curve: random rules. Shallow curve: C=0.795. Derrida coefficient. Steep curve: random rules, Dc=+0.58
Shallow curve, 1 itteration, Dc=-0.9
k=4 n=1000. Steep curve: random rules. Shallow curve: C=0.708. Derrida coefficient. Steep slope: random rules, Dc=+1.31
Shallow curve, 1 itteration, Dc=-0.25
k=5 n=1000. Steep curve: random rules. Shallow curve: C=0.649. Derrida coefficient. Steep curve: random rules, Dc=+1.33
Shallow curve, 1 itteration, Dc=-0.39
back to the DDLab home page
Last modified: June 2001 | {"url":"http://uncomp.uwe.ac.uk/wuensche/derrida_plots.html","timestamp":"2014-04-20T06:24:17Z","content_type":null,"content_length":"3842","record_id":"<urn:uuid:a759da66-d5fe-4b13-a761-c351016a11e3>","cc-path":"CC-MAIN-2014-15/segments/1397609538022.19/warc/CC-MAIN-20140416005218-00319-ip-10-147-4-33.ec2.internal.warc.gz"} |
Hybrid Automata: Introduction, First-Order Approach, and Approximation Techniques
Alberto Casagrande - Dipartimento di Matematica e Informatica (Università di Udine)
Data e ora
martedì 7 novembre 2006 alle ore 17.30 - caffè, tè & C. ore 17.00
Ca' Vignal 3 - Piramide, Piano 0, Sala Verde
Referente esterno
Data pubblicazione
20 ottobre 2006
Systems having a mixed discrete-continuous evolution are called hybrid systems. Since hybrid systems cannot be studied by either dynamical system techniques or finite state system approaches only,
specific formal tools, hybrid automata, were introduced to model them.
Intuitively, a hybrid automaton is a ``finite-state'' automaton with continuous variables which evolve according to a set of continuous laws. Such automata have been widely used to demonstrate the
validity of hybrid system properties and, even if it is proved that many simple verification problems,
such as reachability, are not in general decidable over them, various model checking techniques have been proposed in the literature.
After a brief introduction, this talk will present both theoretical and computational approaches to model checking of hybrid automata.
For the theoretical based approach, we relate first-order theories and analytical results on multi-valued maps and we reduce the bounded reachability problem for hybrid
automata whose continuous laws are expressed by inclusions to a decidability problem for first-order
formulae over the reals. Then, we present two classes of hybrid automata for which the reachability problem can be decided.
For the approximation based approach, we introduce a new software package, called Ariadne, for the verification of hybrid automaton properties and we show that, since it relies on a rigorous
analysis framework to represent geometric objects, it is capable to achieve provable approximation bounds along the computation. | {"url":"http://www.di.univr.it/?ent=seminario&id=679&idC=70&lang=it","timestamp":"2014-04-21T07:25:07Z","content_type":null,"content_length":"12089","record_id":"<urn:uuid:23a71a03-4169-4840-ac7c-e87dd20ddfe6>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00499-ip-10-147-4-33.ec2.internal.warc.gz"} |
Hangman 1
Re: Hangman 1
Hmmmm, so is there an M...?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof. | {"url":"http://www.mathisfunforum.com/viewtopic.php?pid=269933","timestamp":"2014-04-19T09:30:16Z","content_type":null,"content_length":"33061","record_id":"<urn:uuid:7164b624-d26c-44ea-ad21-c0cc9d66d415>","cc-path":"CC-MAIN-2014-15/segments/1397609537097.26/warc/CC-MAIN-20140416005217-00249-ip-10-147-4-33.ec2.internal.warc.gz"} |
Ralated rates
A stone is dropped into a deep, dark mine. A clunk is heard 7 seconds later. Estimate the depth of the shaft in feet. Ignore air resistance. Take the speed of sound as 1000 feet/second.
Velocity of sound= 1000 ft/sec
d = 16t^2
can anyone help me to solve this question?
Thank you. | {"url":"http://mathhelpforum.com/calculus/22988-ralated-rates.html","timestamp":"2014-04-20T12:46:37Z","content_type":null,"content_length":"46999","record_id":"<urn:uuid:7eb277b9-4cd8-484e-8f8a-15afa5960aad>","cc-path":"CC-MAIN-2014-15/segments/1397609539493.17/warc/CC-MAIN-20140416005219-00560-ip-10-147-4-33.ec2.internal.warc.gz"} |
Re: Diatonic notation system
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Re: Diatonic notation system
From: Hans Aberg
Subject: Re: Diatonic notation system
Date: Tue, 9 Dec 2008 13:41:14 +0100
On 9 Dec 2008, at 13:26, Graham Breed wrote:
No, because Lilypond also preserves the number of scale steps. At
least, it should.
I attach what I wrote for E36. There seems to be two systems, but they keep
the ratio M/m = 2.
Where are the transpositions?
I attach an example file.
I think it is the commented out part (long time ago), which for some reason only works with E24. Therefore, the keys must be written explicitly.
Let me know how to do E53.
Yes. That would be chromatic transposition. I can't see a way to
specify it in Lilypond. So what does Lilypond actually do?
Do not know.
Then why are you asking for changes on the developer list?
I have not asked for changes, have I?
First compute the scale degree p + q, and compute the octave and note
by dividing by 7: octave is the fraction, remainder the note name. Then subtract octave and note name from p m + q M; the result is of the form (-r)m + r M, where r if > 0 is the
number of sharps, and if < 0, the
of flats is -r.
Or it could do what it already does.
So what does it do. Does it generalize?
It records a nominal and a rational alteration. It generalizes to
different alterations.
But that is tied to E12, right?
Yes, so change the init file. Why are we going around in circles here?
Perhaps you are stuck to the same idea, and repeating.
I'm stuck with that idea because it works.
In certain primitive settings.
Then don't use E24 for Arabic music.
You already do.
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• Re: Diatonic notation system, (continued)
☆ Re: Diatonic notation system, Hans Aberg, 2008/12/07
○ Re: Diatonic notation system, Graham Breed, 2008/12/07
○ Re: Diatonic notation system, Hans Aberg, 2008/12/07
○ Re: Diatonic notation system, Graham Breed, 2008/12/07
○ Re: Diatonic notation system, Hans Aberg, 2008/12/07
○ Re: Diatonic notation system, Graham Breed, 2008/12/07
○ Re: Diatonic notation system, Hans Aberg, 2008/12/08
○ Message not available
○ Re: Diatonic notation system, Graham Breed, 2008/12/08
○ Re: Diatonic notation system, Hans Aberg, 2008/12/09
○ Re: Diatonic notation system, Graham Breed, 2008/12/09
○ Re: Diatonic notation system, Hans Aberg <=
○ Re: Diatonic notation system, Graham Breed, 2008/12/10
○ Re: Diatonic notation system, Hans Aberg, 2008/12/10
○ Re: Diatonic notation system, Graham Breed, 2008/12/10
○ Re: Diatonic notation system, Hans Aberg, 2008/12/10
○ Re: Diatonic notation system, Graham Breed, 2008/12/10
○ Re: Diatonic notation system, Hans Aberg, 2008/12/10
• (attn doc team) Re: Diatonic notation system, Graham Percival, 2008/12/10
• Re: (attn doc team) Re: Diatonic notation system, Hans Aberg, 2008/12/10
• Re: (attn doc team) Re: Diatonic notation system, Hans Aberg, 2008/12/10
• Re: (attn doc team) Re: Diatonic notation system, Graham Breed, 2008/12/10 | {"url":"http://lists.gnu.org/archive/html/lilypond-devel/2008-12/msg00172.html","timestamp":"2014-04-18T00:24:39Z","content_type":null,"content_length":"11771","record_id":"<urn:uuid:3f546533-f02b-4ede-ae5b-5e7d2c0a3852>","cc-path":"CC-MAIN-2014-15/segments/1397609532374.24/warc/CC-MAIN-20140416005212-00376-ip-10-147-4-33.ec2.internal.warc.gz"} |
Continuous Output - The sigmoid function
Given Summed Input:
x =
Instead of threshold, and fire/not fire,
we could have continuous output y according to the sigmoid function:
Note e and its properties.
As x goes to minus infinity, y goes to 0 (tends not to fire).
As x goes to infinity, y goes to 1 (tends to fire):
At x=0, y=1/2
More threshold-like
We can make this more and more threshold-like, or step-like, by increasing the weights on the links, and so increasing the summed input:
More linear
Q. How do we make it
step-like (more linear)?
For any non-zero w, no matter how close to 0, ς(wx) will eventually be asymptotic to the lines y=0 and y=1.
Is this linear? Let's change the scale:
This is exactly same function.
So it's not actually linear, but note that within the range -6 to 6 we can approximate a linear function with slope.
If x will always be within that range then for all practical purposes we have linear output with slope.
Or try this:
Is this linear? Let's change the scale:
This is exactly same function.
Approximation of Linear with slope
In practice, x will always be within
So we can always get, within that range, an
of many different linear functions with slope.
e.g. Given x will be from -30 to 30:
Approximation of any linear function so long as y stays in [0,1]
And centred on zero. To centre other than zero see below.
Linear y=1/2
The only way we can make ς(wx)
linear is to set w=0, then y = constant 1/2 for all x.
Change sign
We can also, by changing the sign of the weights, make large positive actual input lead to large negative summed input and hence no fire, and large negative actual input lead to fire.
This is of course a threshold-like function still centred on zero. To centre it on any threshold we use:
y = ς(x-t)
where t is the threshold for this node. This threshold value is something that is learnt, along with the weights.
The "threshold" is now the centre point of the curve, rather than an all-or-nothing value.
General case: use ς(ax+b)
Can we have linear output?
Can y be linear? Not if it has slope. Must stay between 0 and 1.
Can be linear constant y=c, c between 0 and 1. We already saw y=1/2. Can we have other y=c?
By setting a=0, y=ς(b) constant for all x
By varying b, we can have constant output y=c for any c between 0 and 1.
d/dx (fg) = f (dg/dx) + g (df/dx)
Quotient Rule
d/dx (f/g) = ( g (df/dx) - f (dg/dx) ) / g^2
Properties of the sigmoid function
Slope = y (1-y)
The slope is greatest where? And least where?
To prove this, take the
derivative and look for where it equals 0:
d/dy ( y (1-y) )
= y (-1) + (1-y) 1
= -y + 1 -y
= 1 - 2y
= 0 for y = 1/2
This is a maximum. There is no minimum.
Slope of ς(ax+b)
For the general case:
y = ς(ax+b)
a positive or negative, fraction or multiple
b positive or negative
y = ς(z) where z = ax+b
dy/dx = dy/dz dz/dx
= y(1-y) a
if a positive, all slopes are positive, steepest slope (highest positive slope) is at y = 1/2
if a negative, all slopes are negative, steepest slope (lowest negative slope) is at y = 1/2
i.e. Slope is different value, but still steepest at y = 1/2 | {"url":"http://www.computing.dcu.ie/~humphrys/Notes/Neural/sigmoid.html","timestamp":"2014-04-21T15:41:39Z","content_type":null,"content_length":"11040","record_id":"<urn:uuid:9b08b830-3d19-46d5-92b6-f1a1279491ac>","cc-path":"CC-MAIN-2014-15/segments/1397609540626.47/warc/CC-MAIN-20140416005220-00223-ip-10-147-4-33.ec2.internal.warc.gz"} |
Gear Train design for the wooden clock
The gears used in all of my clocks have geometry based on the standard gear profile formulae with some adjustment to the tooth profile to thin them down a little to make it a bit less sensitive to
the inaccuracies inherent when using hand cutting methods to produce them.
When designing clock gears I generally use the metric system to define the teeth but both metric and imperial can be used interchangeably. The chart shown below shows the formulae I use for gear
For the teeth to engage with each other they must both be calculated using either the same Diametral Pitch (DP) or the same Module these two terms relate to the way the teeth are spaced around the
Pitch Circle Diameter (PCD). Once you have decided on the DP or Module you are going to use that to calculate all the other features using the formulae in the chart above.
There are two main gear trains in a clock, the minutes train that requires a total ratio of 60:1 this runs between the shaft carrying the minute hand and the escapement wheel. The escape wheel
normally turns once a minute and it connects through the train with the shaft holding the minute hand, which turns once an hour, hence the 60:1 ratio.
The second train runs between the minute hand and the hour hand which requires 12: reduction. This ratio normally has to 2 sets of gears with ratio's of 3:1 and 4:1 and the centre distance of each
are normally the same as the gears are mounted on the same pairs of shafts, I always use a 8 teeth and 32 teeth pair along with a 10 teeth and 30 teeth pair, as it allows the shafts to be shared.
With the 60:1 ratio there are many more options open to you and will to a large extent depend on the design you are trying to achieve, so I must leave that up to you to decide. Having said that a
simple set of 3 gears using 3 pair sets with ratio's of 3:1 4;1 and 5:1 works quite well as when multiplied together they will give you a 60:1 ratio.
Having decided on the gear arrangement I normally try to determine the Module value I am going to use and that relates to the size of the gears I want. To determine that I would normally work out on
CAD or on paper what size I want the largest gear to be, so in my case this is usually a 60 tooth gear. As an example I have decided the gear needs to be about 125mm (5ins) diameter so from the chart
above I can work out a value for the module.
Module = Outside Diameter mm / ( Number of teeth +2)
This gives a value for the Module of 2.016 so round that down to 2.0 and you can now use that value to calculate the sizes for all of your gears.
If you are an organised person you would use a spread sheet to enter all your values to work out the relevant gear sizes, I never got round to doing this so finish up working them out on a
The chart below shows a typical gear pairing for a 3:1 gear ratio and a Module of 2 using a 60 tooth and a 20 tooth gears. The tooth thickness is slightly less than the normal 50% of the CP value I
have used between 44% and 48% of that value depending on what looks to work best in the CAD simulations.
1 comment:
1. Thanks for this write up! Just what I was looking for.
How do you determine the gearing for the hanging weight? | {"url":"http://brianlawswoodenclocks.blogspot.com/2013/02/gear-train-design-for-wooden-clock.html","timestamp":"2014-04-19T15:22:24Z","content_type":null,"content_length":"72960","record_id":"<urn:uuid:2c170b08-e799-43c3-8c63-9385ac5da546>","cc-path":"CC-MAIN-2014-15/segments/1398223211700.16/warc/CC-MAIN-20140423032011-00161-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Generalized Characteristic Polynomials
John F. Canny
EECS Department
University of California, Berkeley
Technical Report No. UCB/CSD-88-440
August 1988
We generalize the notion of characteristic polynomial for a system of linear equations to systems of multivariate polynomial equations. The generalization is natural in the sense that it reduces to
the usual definition when all the polynomials are linear. Whereas the constant coefficient of the general characteristic polynomial is the resultant of the system. This construction is applied to
solve a traditional problem with efficient methods for solving systems of polynomial equations: the presence of infinitely many solutions "at infinity". We give a single-exponential time method for
finding all the isolated solution points of a system of polynomials, even in the presence of infinitely many solutions at infinity or elsewhere.
BibTeX citation:
Author = {Canny, John F.},
Title = {Generalized Characteristic Polynomials},
Institution = {EECS Department, University of California, Berkeley},
Year = {1988},
Month = {Aug},
URL = {http://www.eecs.berkeley.edu/Pubs/TechRpts/1988/6040.html},
Number = {UCB/CSD-88-440},
Abstract = {We generalize the notion of characteristic polynomial for a system of linear equations to systems of multivariate polynomial equations. The generalization is natural in the sense that it reduces to the usual definition when all the polynomials are linear. Whereas the constant coefficient of the general characteristic polynomial is the resultant of the system. This construction is applied to solve a traditional problem with efficient methods for solving systems of polynomial equations: the presence of infinitely many solutions "at infinity". We give a single-exponential time method for finding all the isolated solution points of a system of polynomials, even in the presence of infinitely many solutions at infinity or elsewhere.}
EndNote citation:
%0 Report
%A Canny, John F.
%T Generalized Characteristic Polynomials
%I EECS Department, University of California, Berkeley
%D 1988
%@ UCB/CSD-88-440
%U http://www.eecs.berkeley.edu/Pubs/TechRpts/1988/6040.html
%F Canny:CSD-88-440 | {"url":"http://www.eecs.berkeley.edu/Pubs/TechRpts/1988/6040.html","timestamp":"2014-04-19T09:26:42Z","content_type":null,"content_length":"5957","record_id":"<urn:uuid:9372a21b-f52c-4107-afad-0ddfd63fb7b9>","cc-path":"CC-MAIN-2014-15/segments/1398223206147.1/warc/CC-MAIN-20140423032006-00657-ip-10-147-4-33.ec2.internal.warc.gz"} |
Differential Equations, Linear Algebra and Vector Calculus
Clickable Calculus: Differential Equations, Linear Algebra and Vector Calculus
"Clickable Calculus" refers to Maple's remarkable set of user-interface features that makes common mathematical operations as easy as pointing and clicking. Unique among professional math tools,
Maple provides both computational power and a new-generation user interface that allows students and new users to become proficient with Maple without the burden of learning commands and their
related syntax. The result is that instructors can spend their time teaching mathematics rather than the tool.
Overall, teaching and learning become more efficient and effective. By solving a spectrum of standard (and not-so-standard) problems drawn from differential equations, linear algebra, and vector
calculus, this session will demonstrate the potential of "Clickable Calculus" to enrich the mathematical experience. | {"url":"http://www.maplesoft.com/demo/streaming/ClickableCalculus-DiffEquations.aspx","timestamp":"2014-04-17T18:24:54Z","content_type":null,"content_length":"15574","record_id":"<urn:uuid:f2b4ffbf-f135-4bf2-9065-b0e5dd21b875>","cc-path":"CC-MAIN-2014-15/segments/1397609530895.48/warc/CC-MAIN-20140416005210-00163-ip-10-147-4-33.ec2.internal.warc.gz"} |
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can ya go through the steps for me?
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There are a couple of ways to do this. The quickest is factoring, if you are good at seeing factors. This is a little tricky with factoring, so when in doubt you can always use the quadratic
formula (the second way) which is actually my favorite way.\[x = \frac{ -b \pm \sqrt{b ^{2} - 4ac} }{ 2a }\]where you have an equation in the form: ax^2 + bx + c = 0 So you have a=4, b=-36, and c
=81 So, you substitute those into the formula.
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ughh i just dont understand this!!
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Well, let's look at factoring for a bit instead. As for factoring (the other way), you can rewrite the equation into: (2x)^2 - 36x + 9^2 You will see that the first term is a square and the third
term is a square which might sugeest that the factors are squared.
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That means that we can partially set up the solution: (2x )(2x ) But the third term is a square also, so going a bit further and just leaving out the signs: (2x 9)(2x 9) Now this all presupposes
you can work the distributive law well enough: (a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd Tell me before we go any further, can you deal with that last equation?
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That's great! That really is. Don't worry if it's a little rocky. You're not supposed to "see it all" right away, so we'll just go along until it gets murky for you. np. So, continuing where we
left off with the factoring, writing it again: (2x 9)(2x 9) We have a good start, but we don't have signs yet. We know that 81 is positive so the signs are both negative or both positive. Since
the 36x has a negative sign, we go with negative signs: (2x - 9)(2x - 9)
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Now, we are goingto multiply this out to check our work: (2x)(2x - 9) - 9(2x - 9) = 4x^2 - 18x - 18x + 81 = 4x^2 - 36x + 81
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So, that's the same as our original expression with a slight change in order of the terms. So, that's 2 ways to do this. Ok, your turn, ask questions.
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ok. i understand most of it i just dont understand what happened to the #36
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In the checking of the solution that I did, the -36x came from the -18x -18x
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i get it!!!!!
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Believe me. If really get this or most of it, or even half. That's real progress! For anyone! I suggest going over this a couple times to consolidate your understanding
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And don't give up on the quadratic formula because that is actually much more important.
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ok i already copied and pasted it to a word document. and ive used the equations for more than one of the other problems i have!
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You very nice to work with because you really want to learn (all students should be that way) and you are a nice person.
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thank you! i appreciate that. maybe its because they are mostly in high school. im in nursing school and am relearning everything and harly remember it all. i personallt DO NOT want to have to
repay for this class. hs was 5 years ago so its a little more than rocky.
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and its an online class. which is way harder than on campus
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This is good what you just said. You are approaching this from a perspective that is more mature than high school. And the world needs good nurses. My sister is a nurse. She was also a math whiz,
so she breezed through math. But just get through math since you already have a higher goal. And good luck in both math but especially the nursing!
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thank you!
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You are entirely welcome!
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did i mention i hate rational expressions :)
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lol then go with the irrational ones :-)
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A little math humor. There's so little material from which to get humor in math!
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haha. oh i am aware!
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I think the developers of all these math formulas are secretly laughing in their graves at all the misery they caused!
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I know! Maybe one day you can get a math PhD. as a patient! Payback!
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hahahahaha. thats funny!!! i wouldnt doubt it!! all i know is this is the only subject that i curl up in the fetal position and cry over ;)
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Math is tough because it is so unforgiving and as a field of study, it is so vertical. Can't go too far ahead without getting the basics.
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uh oh. got to another one....wanna help?
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vertical as in concepts building upon one another. Sure, I've got time for one more. Then I curl up and cry.
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\[\frac{ x ^{2}-25 }{ x+5 }\]
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Oh, you'll like this one because it's easier than the other one.
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oh god i hope so lol
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dont you get rid of the division and turn it to multipication
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Sort of. I'm going to factor the numerator:\[\frac{ (x + 5) \times (x - 5) }{ (x + 5) }\]Now, the division concept comes into play insofar as the x + 5 factor cancels out of the numerator and the
denominator and you are left with: x - 5 as your answer.
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that i comletely understand!
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Either the website or my computer just went goofy.
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This is the testimonial you wrote.
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Right angle: you know all 3 angles and length a, not b or c?
June 18th 2008, 08:33 AM #1
Jun 2008
Right angle: you know all 3 angles and length a, not b or c?
Could someone help me with this? I have a right angle triangle, 90, 30 and 60 degrees. Say the length of A is 30 inches. What would B and C be? Any help would be appreciated. thank you, johnr.
OK, say I know the length of one side of the triangle, but need to find the length of the hypotenuse and the other side of the triangle. john r.
If you know trigonometry this is easy. If you don't know trig then you have to memorize something. I'll assume you don't know the trig.
You have a 30, 60, 90 triangle. We can construct the ratios of these sides. They are always:
$\frac{a}{c} = \frac{1}{2}$
$\frac{b}{c} = \frac{\sqrt{3}}{2}$
Note carefully that we need to know which angle is opposite which side in order to finish this problem. Without knowing that side a is opposite the 30 or 60 degree angle, I cannot proceed
further. (But you should be able to take it from here.)
Can't draw that fast Tops!
Ok, let's recall a couple of rules about a 30-60-90 triangle. The hypotenuse is twice the length of the shortest side (the side opposite the 30 degree angle). The side opposite the 60 degree
angle is the long side and it is equal to the short side times the sqare root of 3.
Look at my diagram. If the short side is 30 in. That makes the hypotenuse 60 inches, and the other leg $30\sqrt3$ in.
We know that a = c/2. So what is b? According to Pythagoras:
$b^2 = c^2 - a^2$
$b^2 = c^2 - \frac{c^2}{4} = \frac{3c^2}{4}$
$b = \frac{\sqrt{3}}{2} \cdot c$
thank you all very much. I haven't had much use for geometry since college. thanks. and so quick, thankk you. john r.
You're very welcome, John. Good luck. Glad we could help.
I was amazed how quickly I found this website and how quickly I received multiple answers. Thanks for helping a 36 year old figure out something I knew by heart 15-20 years ago. john r
sine 30d=.5
Drop the perpendicular from the top of an equilateral triangle. The angle is bisected to 30d. By Pythagorean theorem, the perpendicular= .5*3^2. This 30-60-90 triangle is similar to yours. If the
base is 30, then the hypotenuse is 60 and the third side is .5*3(^.5)*60=30*(3^.5)
June 18th 2008, 08:51 AM #2
A riddle wrapped in an enigma
Jan 2008
Big Stone Gap, Virginia
June 18th 2008, 09:02 AM #3
Jun 2008
June 18th 2008, 09:16 AM #4
June 18th 2008, 09:18 AM #5
A riddle wrapped in an enigma
Jan 2008
Big Stone Gap, Virginia
June 18th 2008, 09:30 AM #6
June 18th 2008, 09:49 AM #7
Jun 2008
June 18th 2008, 09:53 AM #8
A riddle wrapped in an enigma
Jan 2008
Big Stone Gap, Virginia
June 18th 2008, 01:59 PM #9
Jun 2008
July 4th 2008, 06:38 PM #10
May 2008 | {"url":"http://mathhelpforum.com/geometry/41896-right-angle-you-know-all-3-angles-length-not-b-c.html","timestamp":"2014-04-16T07:41:04Z","content_type":null,"content_length":"65209","record_id":"<urn:uuid:f1971591-0df8-4c89-b273-d3ad7821e1b9>","cc-path":"CC-MAIN-2014-15/segments/1397609521558.37/warc/CC-MAIN-20140416005201-00639-ip-10-147-4-33.ec2.internal.warc.gz"} |
Embedding Effect on the Mechanical Stability of Pressurised Carbon Nanotubes
Journal of Nanomaterials
Volume 2013 (2013), Article ID 767249, 9 pages
Research Article
Embedding Effect on the Mechanical Stability of Pressurised Carbon Nanotubes
^1Division of Engineering and Policy for Sustainable Environment, Faculty of Engineering, Hokkaido University, Sapporo 060-8628, Japan
^2Department of Physics, Faculty of Science, Hokkaido University, Sapporo 060-0810, Japan
^3Department of Environmental Sciences & Interdisciplinary Graduate School of Medicine and Engineering, University of Yamanashi, 4-4-37 Takeda, Kofu, Yamanashi 400-8510, Japan
Received 6 March 2013; Revised 15 April 2013; Accepted 26 April 2013
Academic Editor: Teng Li
Copyright © 2013 Motohiro Sato et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any
medium, provided the original work is properly cited.
We elaborate on the cross-sectional deformation of carbon nanotubes embedded into a self-contracting host medium. The continuum elastic approach is used to formulate the mechanical energy of both the
embedded nanotubes and the self-contracting outer medium with finite thickness. Our formula allows us to evaluate the critical radial pressure applied on the interface between the embedded nanotube
and the outer contracting medium as well as the deformation mode that arises immediately above the critical pressure. An interesting mechanical implication of the embedding effect, that is, the
power-law dependence of the critical pressure on the elastic modulus of the medium, is deduced by the theoretical approach established.
1. Introduction
The salient structural feature of a carbon nanotube is its self-repairing behaviour that arises under high-energy beam irradiation [1, 2]. When the kinetic energy transferred from the incident beam
to the constituent carbon atoms is sufficiently large, the atoms are pushed away from the original equilibrium positions, leaving vacancies in the host hexagonal lattice [3, 4]. In typical solids,
such irradiation-induced vacancies survive without curing as time passes. However, this is not the case in carbon nanotubes; the removal of carbon atoms from the purely hexagonal lattice leads to a
local reconstruction that acts to maintain its coherent network structure with cylindrical geometry. For this reason, carbon nanotubes are often referred to as self-repairing (or self-healing)
materials [5–7].
This self-repairing nature provides beneficial effects for manipulating the carbon nanotube morphology, especially when combined with heat treatment. For instance, fine-tuning of electron beam
irradiation makes it possible to synthesize multiwall carbon nanotubes (MWNTs) with reduced interwall spacings [8]; the spontaneous shrinkage in the radial direction is a result of the knock-on
collision of carbon atoms followed by annealing reconstruction of the vacancies. This experimental finding implies that when the outermost carbon walls of an MWNT are eroded selectively by
irradiation, the self-contraction of the outermost walls exerts high pressure on the encapsulated, undamaged innermost walls [9–11]. Application of high pressure may then trigger a novel class of
cross-sectional transformations of the inner walls [12], similar to the case of pristine (irradiation-free) MWNTs under hydrostatic pressure [13, 14]. Another class of radial contraction has been
observed in MWNTs synthesized in the presence of nitrogen [15]. The yielded nanotubes showed polygon-shaped cross-sections rather than ordinary circular ones, a phenomenon that is partly attributed
to the interwall thermal contraction upon cooling, as verified numerically by molecular dynamics simulations [16].
From a nanoengineering perspective, the tunability of cross-sectional geometry may be useful for developing nanofluidic [17–19] or nanoelectrochemical devices [20] based on carbon nanotubes, because
both utilize the hollow cavity within the innermost tube. A very interesting issue from an academic viewpoint is the effect of the core tube deformation on the physicochemical properties of
intercalated molecules confined in the hollow cavity. It is indeed known that various types of intercalated molecules can fill the innermost hollow cavities of nanotubes [18] and exhibit intriguing
behaviours that are distinct from their macroscopic counterparts [21–23]. These distinct behaviours originate from the similarity between the intermolecular spacings and the linear dimension of the
nanoscale compartment. Therefore, cross-sectional deformation that breaks cylindrical symmetry will provide a clue to improving the performance of nanotube-based devices.
In this paper, we establish the continuum elastic approach that describes the cross-sectional deformation of carbon nanotubes surrounded by a self-contracting host medium. The mechanical energy of
both the pressurized nanotubes and the contracting medium with finite thickness are formulated using thin-shell theory. The obtained formula allows us to evaluate the critical radial pressure applied
on the interface between the inner nanotube and the outer medium. Our numerical calculations have unveiled a power-law dependence of the critical pressure on the elastic modulus of the medium that is
independent of the medium thickness.
2. Methodology
2.1. Relevant Energy Components
Figure 1 illustrates the self-contraction process of MWNTs subjected to high-energetic beam irradiation. Under irradiation, the induced beam kicks off a portion of carbon atoms at outer walls,
causing vacancies followed by spontaneous shrinkage as marked by color in Figure 1(b). Even during the outer-walls self-contraction, the inner walls remain undamaged by irradiation, and thus they
tend to keep their initial tube radii. As a result, the contracting outerwalls exert a high pressure, designated by , on the inner undamaged walls (see Figure 1). A possible consequence of the
radiation-induced high-pressure application is a circumferentially wrinkling structure, called radial corrugation [13], in the inner undamaged walls embedded in the eroded region, as examined later
in an approximation based on the thin-shell theory.
The stable cross-sectional shape of the embedded tube is obtained by minimizing its mechanical energy per unit axial length [13] as The first term with the definition represents the deformation
energy of the embedded nanotubes. and are, respectively, in-plane and bending-induced strains of the th wall, and is the circumferential coordinate. For (2), we supposed that each th wall had a
radius prior to cross-sectional deformation and that the deformation caused a displacement of a volume element of the th wall at in the polar coordinate representation. The two strain terms and are
described in terms of the displacement components by [14] where .
For quantitative discussions, the elastic coefficients and need to be carefully determined. In conventional thin-shell theory for macroscopic objects, and are related to the Young’s modulus of the
wall and its thickness as However, for carbon nanotubes, the macroscopic relations for and noted earlier fail because there is no unique way of defining the thickness of the graphene wall [24]. Thus,
the values of and should be evaluated ab initio from direct measurements or computations of carbon sheets, without reference to the macroscopic relations. In actual calculations, we substitute nN/
nm,nN·nm for carbon nanotubes, and from prior work [25] based on the density functional theory.
The second term, , in (1) accounts for the van der Waals (vdW) interaction energy, which determines the equilibrium distance between adjacent concentric walls. Thus far, several continuum models for
the vdW interactions have been proposed. Expressions for the wall-wall interaction proposed in [26] were based on the surface integration of the vdW force and its derivative over the cylindrical
walls, while disregarding the vectorial nature of the force. The significance of the vectorial nature of the force was addressed in [27], where analytical expressions for the wall-wall interaction
were obtained by considering only the component of the vdW force normal to the wall. In accordance with the result of [27], we define the interaction energy by where the coefficients are derived
through a harmonic approximation of the interwall force [28] associated with the vdW intermolecular potential with the definitions of meV and nm the same as those in [29]. Equation (5) takes into
account correctly the normal-to-wall component of vdW forces, and it is valid for infinitesimal deformation, which we address in the present work.
The final term in (1) is the negative of the work done by during cross-sectional deformation; it can be written as [13] Note that all three terms are functions of and of the th wall under .
The remaining term, , in (1) is the elastic energy of the eroded medium surrounding the inner part of the nanotubes. To derive it, we assume that the medium has finite thickness of , where is the
outmost radius of the cylinder-shaped surrounding medium and is the outmost tube radius of the embedded nanotube. In addition, the medium is assumed to be homogeneous and isotropic with Young’s
modulus and Poisson’s ratio . The validity of the latter assumption depends on the following two effects of irradiation on the mechanical stiffness of MWNTs. Irradiation reduces the axial stiffness
because it creates vacancies [30, 31], and it simultaneously enhances the radial stiffness, owing to the production of covalent bonding between adjacent walls [32]. The possible value of ranges from
100GPa (for amorphous carbon) [33, 34] to much less. An explicit form of is presented in the next section.
2.2. Corrugation Mode Analysis
Our objectives are to determine (i) the optimal displacements and that minimize under a given value of and (ii) the critical pressure above which the circular cross section of the embedded nanotube
is elastically deformed into a noncircular one. These are accomplished by decomposing the displacement just after buckling as follows: In (8), the term describes a uniform radial contraction at , at
which the cross section remains circular and thus the displacement is independent of . The other term describes a deformed (noncircular) cross-section observed immediately above . Note that the
superscript attached to differentiates it from the -dependent displacement observed at . As to , we can write because no circumferential displacement arises at (i.e., ).
Applying the variation method to with respect to and , we obtain a system of linear differential equations with regard to and . To derive the 2 linear differential equations, quadratic or cubic terms
in and are omitted since we consider elastic deformation with sufficiently small displacements. In addition, the terms consisting only of and are also omitted; the sum of such terms should be equal
to zero since represents an equilibrium circular cross-section under . In fact, the function form of is determined by the fact that the sum of those terms equals zero. The differential equations can
be solved using the Fourier series expansions wherein we took into account that , , and their derivatives are periodic in . Substituting the expansions into the differential equations results in the
matrix equation ; the vector consists of and with all possible and , and the matrix involves one variable and other material parameters. It should be noted that, due to the orthogonality of and , the
matrix can be expressed by a block diagonal matrix of the form Here, for arbitrary integer is a submatrix that satisfies , where is a -column vector composed of and . As a result, the secular
equation is rewritten by Solving (12) with respect to , we obtain a sequence of discrete values of . Among these values, the minimum one serves as the critical pressure that is associated with a
specific integer . From the definition, the associated with a specific allows only and be finite, but it requires and . Immediately above , therefore, the cross section of embedded nanotubes becomes
radially deformed as described by where the value of is uniquely determined by the one-to-one relation between and .
3. Mechanical Energy of the Embedding Medium
3.1. Total Energy Cost
In polar coordinates, the radial and circumferential components of normal stress in the medium are denoted by and , respectively, and the shear stress is denoted by ; all three quantities are
functions of and . Then, is determined by and at as where and describe the corrugation amplitudes of the outermost wall of the embedded nanotube; see (10). The superscripts and indicate the
quantities corresponding to uniform contraction and radial corrugation, respectively. In other words, represents the energy required for uniform radial contraction of the surrounding medium keeping
in contact with the embedded nanotube, and represents the energy required for radial corrugation with mode index .
3.2. Stress-Strain-Displacement Relation
The mechanics of an elastic medium is governed by the stress function , which satisfies the so-called compatibility equation [35] where and . Once is obtained, we can deduce the stress components as
follows: By definition, the strain components , , and are given by the matrix form where and are, respectively, the radial and circumferential displacements of a volume element in the elastic medium.
The general solution of (17) is given by The zeroth component represents a uniform contraction of the circular cross section, thus corresponding to the energy that we have introduced in (15). The
first one, , implies a rigid-body translation, which is irrelevant to our consideration. Other components, for , describe radial corrugations with mode index , thus providing the energy given by (16
). In the following, we set [36] in order to obtain physically relevant solutions of , , and that decay with increasing . Without loss of generality, we set in (21) according to our assumption of
cosine-type radial displacement ; see (10). We emphasize that the hypothesized solution forms of and differ from those in our earlier work [12]; in the present study, we introduce the positive power
terms of in (22) to take into account correctly the boundary condition at , whereas these terms were omitted in [12].
3.3. Energy Cost under Uniform Contraction
We now evaluate the explicit form of the energy . It results from the uniform contraction of the medium described by the -dependent medium displacement . The elastic nature of the embedding medium
assures that with the -dependent stiffness coefficient . In addition, complete contact at the interface of the medium and the embedded nanotube implies Hence, the previous expression of , given in (
15), can be rewritten as The remaining task is, therefore, to represent in terms of already-known material parameters such as , and .
To accomplish this task, we consider a specific solution of (17) that has the form , and then we substitute it back into (18) to obtain and . The coefficients and are determined by imposing the
boundary conditions of at and as follows: Eliminating and from (27), we have The obtained and lead to the displacements and at the medium region (), as confirmed by (19). By comparing (23), (28), and
(29), we conclude that
3.4. Energy Cost under Radial Corrugation
Next, we consider the energy that corresponds to the radial corrugation of the th order. A similar procedure to the case of yields leading to the results with and . Of the four coefficients , , , and
involved in the previous equations, two of them are eliminated by considering the boundary conditions and .
Owing to the complete contact condition, we have Besides, the elastic nature of the medium at the contact interface implies the relations From (35), the stiffness coefficients are given by Note that
a set of two normalization conditions of which is indicated by (36), allows us to describe the two remaining undetermined coefficients, from among , , , and , in terms of materials parameters such as
,, and ; see the statement immediately below (33). Hence, and derived as mentioned are those expressed by already-known parameters, though the explicit forms are not shown here to save space.
Eventually, we obtain by substituting the results into (16).
4. Results and Discussions
Figure 2 shows the critical pressure required for the cross-sectional deformation of embedded carbon nanotubes. Different types of curves in the plots (dashed, dashed-dotted, and solid) correspond to
different values of Young’s modulus ratio , where TPa is assumed to be the nanotube’s modulus. We found that, independent of the value, the curves are upward convex as functions of the medium radius
in units of . The growth of is rapid for , and then, it saturates for larger . The rapid growth in indicates a “hardening effect” caused by the surrounding medium, that is, an enhancement in the
radial stiffness of the embedded nanotube by encapsulation. This hardening effect disappears with a further increase in ; the results imply that the surrounding medium with thickness no longer
enhances the hardening effect and thus can be identified as a medium with infinitely large thickness , which is the case considered in our earlier work [12].
Figure 3 provides the index of radial deformation modes observed immediately above . Successive transformation of deformation modes with increasing medium thickness was confirmed, as a result of the
fact that the energy required to deform the surrounding medium needed to be responsible for determining the stable corrugation pattern of the embedded nanotube. Again, we found that the mode
variation disappears for , and a larger value of arises for a higher modulus ratio . The nonmonotonic variance in the corrugation mode observed within is what can be clarified for the first time by
the present work; it is applicable to the condition , which lay beyond the scope of our earlier work [12].
Figure 4 shows the dependence on the modulus ratio , where the radius ratio is fixed to be or . For every choice of and , the curves in this plot obey a power law represented by We have confirmed
that the power-law behaviour holds for any values of and within the ranges of and , respectively. Recall here that, for , becomes almost constant even with increasing . These facts lead to the
conjecture that the power law described by (38) is universal to all embedded nanotubes, at least under the numerical conditions that we have addressed. The power law of represented by (38) is an
interesting manifestation of the embedding effect of nanotubes into an elastic medium.
It is worthy to note that the two-third power law of has been also proposed for the buckling of macroscopic “tunnels" embedded in an infinite soft ground [36]. The radial collapse of tunnel liners,
which is one of the major civil engineering disasters, was analysed to reveal a closed form of ; several approximations led to the conclusion of a power-law of similar to (38), though an infinitely
large medium and only the case of one cylindrical shell (i.e., ) are hypothesized. In this context, our results suggest that the approximation theory proposed for macroscopic tunnels [36] is valid
for nanoscopic cylinders, even though they consist of more than one concentric walls.
The previous discussion may pose a further question: does the power law of hold true for macroscopic counterparts of “many”-walled nanotubes? In fact, macroscaled pipe-in-pipe structures (i.e., a
pipe inserted inside another pipe) are known to be an efficient design solution for subsea-pipeline developments in deep water [37, 38], wherein buckling resistance to large amounts of external
hydrostatic pressure is a key structural design requirement. Addressing these problems is expected to shed light on the development of multiple-cylindrical structures from multidisciplinary
viewpoints and will be considered in our future work.
5. Conclusion
We have demonstrated a continuum elastic approach that describes the cross-sectional deformation of carbon nanotubes surrounded by a self-contracting host medium. The approach enables quantitative
discussions of the critical pressure for radial corrugation and the stable corrugation mode of the nanotube surrounded by the contracting medium. Numerical calculations based on the established
theory revealed the power-law dependence of the critical pressure on the elastic modulus of the medium, which is independent of the medium thickness. Further studies are expected to shed light on
other mechanical properties of embedded nanotubes and to suggest applications based on their unique cross-sectional deformations.
This work was supported by JSPS KAKENHI (Grant nos. 24686096, 11J03636, 22760058, and 25390147). One of the authors (H. Shima) appreciates the financial support provided by the Asahi Glass
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pp. 263–278, 2008. View at Scopus | {"url":"http://www.hindawi.com/journals/jnm/2013/767249/","timestamp":"2014-04-16T20:09:07Z","content_type":null,"content_length":"441888","record_id":"<urn:uuid:fd581fdb-b46b-4c26-a7e3-52285a40d856>","cc-path":"CC-MAIN-2014-15/segments/1397609524644.38/warc/CC-MAIN-20140416005204-00074-ip-10-147-4-33.ec2.internal.warc.gz"} |
st: Sorting
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st: Sorting
From Doug Owens <owens.doug@gmail.com>
To statalist@hsphsun2.harvard.edu
Subject st: Sorting
Date Mon, 14 Mar 2005 15:38:35 -0500
How does Stata's sort algorithm work?
I was looking at the stability of the sort order. See the following 2
examples. But first note that I am not using the stable option in the
sort command.
Example 1 (changing sort order):
set obs 100
g x = _n
g herbal = _n>25 & _n<=75
sort grade
l x in 1/10, clean noo
Example 2 (stable sort order):
set obs 100
g x = _n
g herbal = _n<51
sort herbal
assert x ==_N+1-_n
Example 1 is what I've come to expect from experience. Each time it is
run the data is sorted differently.
The stability preservation of example 2 was surprising. Quoting from
the Stata manual entry on sort, "Without the stable option, the
ordering of observations with equal values of varlist is randomized."
I though that if the data was such that "herbal>=herbal[_n-1] if
_n>1", then a "sort herbal" command would not need to change the order
of the data, and thus the resulting sort order would not not vary with
multiple executions of the code. But that is not the case here. What
other conditions can lead to a (non-unique) sort command producing the
same dataset each time?
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/ | {"url":"http://www.stata.com/statalist/archive/2005-03/msg00473.html","timestamp":"2014-04-16T16:20:53Z","content_type":null,"content_length":"5640","record_id":"<urn:uuid:e6b6aacd-b16f-45e7-ba42-139feb51ce16>","cc-path":"CC-MAIN-2014-15/segments/1397609524259.30/warc/CC-MAIN-20140416005204-00104-ip-10-147-4-33.ec2.internal.warc.gz"} |
Surface Area of a Football
Date: 02/12/99 at 15:19:30
From: Brian
Subject: Surface area
What is the surface area of a football with a diameter of 22 inches
and an arc length of 14 inches?
Date: 03/19/99 at 16:10:58
From: Doctor Gene
Subject: Re: Surface area
This football has problems: it is too thick in the middle to be NFL
standard, or even to have an arc length of 14" - a radius of 11" would
require an arc length of greater than 22". Maybe you mean a
circumference of 22"? That, we can handle.
There are lots of curves that give football-like shapes when revolved
around the x-axis, and which have these dimensions; you could take
half an ellipse, a parabola, or no end of others. Most choices would
give you integrals that you cannot evaluate in closed form, but you
could use a program like Maple or Mathematica to get good
Let us try approximating the shape of a football by revolving a
parabola around the x-axis. The kind of parabola you probably want
would be of the form f(x) = b - x^2/c (if you leave off the c you get
a football that is way too fat). This has the advantage of yielding
an integral that you can compute, the disadvantage of the extra
unknown c - all shapes will involve nasty algebra.
* b f(x) = b - x^2/c
* | *
* | *
\ sqrt (bc)
When the given parabola is revolved around the x-axis, the surface has
diameter pi*b^2 = 22, so b = sqrt (22/pi), which is a little more than
2 1/2" - a bit skinny for a football. You might want to try another
curve, but I will illustrate what to do with this one.
The arc length is the integral from one x-intercept to the other of
sqrt(1 + f'(x)^2), whose closed form expression lives in the back of
most calculus books for our particular curve. Set the value to your
given arc length of 14, and solve for c, which will take a bit of
You will then know which parabola of our given form satisfies the arc
length and diameter conditions you stipulated. The area of a surface
of revolution is given by a famous integral:
2*pi* f(x)*sqrt(1 + f'(x)^2)
between the two x-intercepts. In our case it can also be managed in
closed form, but may cause you to review your integration techniques.
- Doctor Gene, The Math Forum | {"url":"http://mathforum.org/library/drmath/view/53697.html","timestamp":"2014-04-20T12:20:40Z","content_type":null,"content_length":"7327","record_id":"<urn:uuid:bfbbfc35-6bef-4d42-9e8d-52945ad6032e>","cc-path":"CC-MAIN-2014-15/segments/1397609538423.10/warc/CC-MAIN-20140416005218-00489-ip-10-147-4-33.ec2.internal.warc.gz"} |
Relationship between circumference of the circle and diametre.no nothing. - Homework Help - eNotes.com
Relationship between circumference of the circle and diametre.
no nothing.
Let's start with a few definitions. The circumference of a circle is the measurement of the whole circle, the outside of the circle. The diameter of a circle is the greatest width of the circle or
twice the radius of a circle, which is the midpoint of the circle to the edge. By knowing either the diameter of the circle of the circumference, you can determine the other. For instance, the
diameter of the circle multiplied by the pi, which is 3.1415... will get you the circumference of a circle. I am sure that there are other formulae, but this should give you a good start.
TT (often written pi) is a mathematical constant whose value is the ratio between circumference and diameter of any circle in a Euclidean space, is the same value as the ratio of area of a circle and
square of its radius.
pi=2*pi*R/2*R, where 2*pi*R=circumference of any circle and 2*R=D=diameter of any circle.
The symbol "pi" was first proposed by Welsh mathematician William Jones, in 1706. Constant value is equal to approximately 3.14159, in an ordinary decimal notation.
Pi is one of the most important constants in mathematics and physics: many formula of mathematics, engineering and other sciences involving pi.
Since pi is an irrational number, it has an infinite number of decimals that do not contain sequences that are repeated. The infinite string of digits has fascinated mathematicians and many have made
significant efforts over the past several centuries to calculate decimal places and to investigate properties of this number. Despite the analytical work and calculations performed on supercomputers
have calculated 10 thousand billion digits of π, there appeared no identifiable pattern in the digits found.
The circumference of a circle is is always pi times the diameter of the circle. Or if C is the circumference of circle and d is the diameter of the circle, then the relationship between the
circumference and the diameter is given by:
C = pi*d, where the value of pi = 3.141593654 approximately.
Also if r is the radius of thr circle, fixed distance distance from the centre of the circle to anny point on the circumference of the circle, then r = d/2. Therefore, C = pi*d = 2pir. Or the
circumference is equal to 2 pi times radius of the circle.
Circumference equals Pi (3.14) times the diameter (diameter is 2 times the radius) or simply 2 times Pi times the radius.
I always taught my math students that if you have a circle (circumference) of friends you can "Party (or Pi-D). Corny but it worked. Good luck!
Join to answer this question
Join a community of thousands of dedicated teachers and students.
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ADC Basics (Part 5): Key ADC specifications for system analysis
One can view the SAR-ADC as a one-shot converter where the output data represents a single analog sample.
Figure 3 illustrates a possible delta-sigma converter timing scenario. In this figure, the converter acquires multiple samples and internally produces intermediate conversions.
Click on image to enlarge.
Figure 3. 24-bit delta-sigma conversion timing diagram using the ADS1258.
This figure shows the intermediate, internal conversions of the delta-sigma converter with a fifth-order digital filter. Notice the “hidden conversions” are an artifact of the internal digital
filter’s order. The user never sees these hidden conversions.
Output noise
The noise magnitude produced by the SAR-ADC and delta-sigma converter, as compared to the number of bits, is dramatically different. Typically, the noise generated by the 12-bit SAR-ADC is well below
the voltage size of the converter’s LSB. For instance, a 12-bit SAR-ADC with a 4.096V full-scale input range has an LSB size of 1 mV. In contrast, a 24-bit delta-sigma converter with a 4.096V
full-scale input range has an approximate LSB size of 244 nV.
Device or converter noise is a random event, but it does follow probability theories.
Figure 4
illustrates a group of a delta-sigma converter results with a DC input. There are three points of interest.
Click on image to enlarge.
Figure 4. Continuous output data from a delta-sigma converter.
First is the mean value. The mean, or average value of the data, is a reference point needed when you calculate the data’s standard deviation. The second point of interest is the volts-RMS or
bits-RMS label. These labels are equivalent to a span from the data’s negative to the positive standard deviation. Third, if you are going to place the converter results in a display, volts p-p or
bits p-p determine how often the lower digits in your display change.
Figure 5
shows how the output data in Figure 4 translates into a histogram. The RMS value is equal to the standard deviation of this data. Between the two standard deviations or RMS lines in this graph, a
significant number of the noise occurrences are captured. The probability that an ADC produces one output value that lands between the two RMS lines is equal to about 68 percent.
Click on image to enlarge.
Figure 5. Unipolar Ideal ADC Transfer Function
With the Gaussian distribution in our histogram plot, you can see that your RMS limits exclude a lot of data. If you look at the number of converter output results between the two standard deviation
limits, you account for 68 percent of the occurrences. But if you multiply the doubled standard deviation by a constant or crest factor, you can expand the percentage of occurrences underneath the
curve. The crest factor allows you to define your peak-to-peak limits as well as determine which of the converter bits are useful in your 12-bit system.
The discussion in the article excludes a signification number of ADC specifications. These other specifications are important as you hone in on your finished solution, however, the topics mentioned
here can be used to quickly prove or disprove the appropriate direction for your system design. The 12-bit applications that we are exploring include multiplexed circuits, handheld meters, data
loggers, automotive systems, and monitoring systems.
1. “
A Glossary of Analog-to-Digital Specifications and Performance Characteristics (SBAA147B)
,” B. Baker, Texas Instruments, October 2011.
2. “
Delta-sigma ADCs in a nutshell, part 3: the digital/decimator filter
,” Baker, EDN, February 21, 2008.
For more information, see
Part 4
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JavaScript Number Format - Decimal Precision
Round a number to a certain number of places
JavaScript built-in methods toFixed and toPrecision
JavaScript has built-in methods to format a number to a certain precision. They are toFixed and toPrecision, and are part of the Number object. Any browser that supports ECMAScript version 3 should
support toFixed and toPrecision. This roughly equates to Netscape 6.0 and above and IE 5.5 and above.
Use toFixed to set precision after the decimal point. It doesn't matter how large the number is before the decimal point. For normal decimal formatting, this is your best option.
// Example: toFixed(2) when the number has no decimal places
// It will add trailing zeros
var num = 10;
var result = num.toFixed(2); // result will equal 10.00
// Example: toFixed(3) when the number has decimal places
// It will round to the thousandths place
num = 930.9805;
result = num.toFixed(3); // result will equal 930.981
Try toFixed
Use toPrecision when you're setting the overall precision. Here, it matters how large the number is before and after the decimal point. This is more useful for mathematical purposes than for
// Example: toPrecision(4) when the number has 7 digits (3 before, 4 after)
// It will round to the tenths place
num = 500.2349;
result = num.toPrecision(4); // result will equal 500.2
// Example: toPrecision(4) when the number has 8 digits (4 before, 4 after)
// It will round to the ones place
num = 5000.2349;
result = num.toPrecision(4); // result will equal 5000
// Example: toPrecision(2) when the number has 5 digits (3 before, 2 after)
// It will round to the tens place expressed as an exponential
num = 555.55;
result = num.toPrecision(2); // result will equal 5.6e+2
Floating-point errors
toFixed and toPrecision are subject to floating-point errors.
Here is a test where the starting number is 162.295. The following should show the JavaScript results:
Do they show up correctly as 162.30 in your browser? Most JavaScript implementations will display it as 162.29
Here is basically what happens when rounding 162.295 to two decimal places
num = 162.295
num *= 100 // 16229.499999999998
num = Math.round(num) // 16229
num /= 100 // 162.29
As you can tell, it's in the second step that the number changes from its actual value.
Floating-point numbers - External references
bugnet.com - JavaScript Math Errors in Netscape & Internet Explorer
wikipedia.org - Problems with floating-point | {"url":"http://www.mredkj.com/javascript/nfbasic2.html","timestamp":"2014-04-20T08:18:28Z","content_type":null,"content_length":"9168","record_id":"<urn:uuid:54e9bc22-cca1-46a2-a7ef-f302eb856159>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00252-ip-10-147-4-33.ec2.internal.warc.gz"} |
Aspects of discrete mathematics and probability in the theory of machine learning
Martin Anthony
Discrete Applied Mathematics Volume 156, Number 6, , 2008.
This paper concerns developments in the use of probabilistic and combinatorial techniques in the analysis of machine learning. Initially, probabilistic models of learning addressed binary
classification (or pattern classification). More recently, analysis has been extended to regression problems, and to classification problems in which the classification is achieved by using
real-valued functions (where the concept of a large margin has proven useful). Another development, important in obtaining more applicable models, has been the derivation of data-dependent bounds.
Here, we discuss some of the main probabilistic and combinatorial techniques and results. | {"url":"http://eprints.pascal-network.org/archive/00004604/","timestamp":"2014-04-21T15:06:28Z","content_type":null,"content_length":"6649","record_id":"<urn:uuid:2efaa95a-ae40-48b6-a93c-919f2ce73c41>","cc-path":"CC-MAIN-2014-15/segments/1397609540626.47/warc/CC-MAIN-20140416005220-00332-ip-10-147-4-33.ec2.internal.warc.gz"} |
Please Help!
March 19th 2006, 04:31 PM #1
Dec 2005
Please Help!
Let f be the function given by f(x)=2x/sqrt(x^2+x+1)
a)Find the domain for f. Justify your answer.
b)Write an equation for each horizontal asymptote of the graph f.
c)Find the range of f. Use f'(x) to justify your answer.
Note: f'(x)=(x+2)/(x^2 +x+1)^.5
Please help me with this problem. I don't really know how to do much of it.
Let f be the function given by f(x)=2x/sqrt(x^2+x+1)
a)Find the domain for f. Justify your answer.
The fraction is underfined when the denominator is zero. Also, when the radical is negative. Thus, $x^2+x+1=0$ or $x^2+x+1<0$. Thus, $x^2+x+1>0$ is the domain. But $x^2+x+1$ is always positive,
because the leading coefficient is positive and its discrimanat is negative. Thus, this function is defined for all real numbers.
b)Write an equation for each horizontal asymptote of the graph f.
The horizontal tangents is when the derivative approaches infinity. Since is derivative is,
$f'(x)=\frac{x+2}{\sqrt{x^2+x+1}}$ that happens when the denominator is zero. Thus, $\sqrt{x^2+x+1}=0$ thus, $x^2+x+1$ but that never happens because we said that $x^2+x+1$ is always a positive
number. Thus, it has no horizontal tangents.
The horizontal tangents is when the derivative approaches infinity. Since is derivative is,
$f'(x)=\frac{x+2}{\sqrt{x^2+x+1}}$ that happens when the denominator is zero. Thus, $\sqrt{x^2+x+1}=0$ thus, $x^2+x+1$ but that never happens because we said that $x^2+x+1$ is always a positive
number. Thus, it has no horizontal tangents.
No a horizontal tangent is when the derivative approaches zero.
$<br /> \frac{d}{dx}\frac{2x}{\sqrt{x^2+x+1}}=\frac{x+2}{( x^2 + x + 1)^{3/2}}$
There is a zero of this at $x=-2$, but that is not an asymtote.
Also the derivative goes to zero as $x \to \pm \infty$, so if
the limits of $f(x)$ as $x \to \pm \infty$ are finite these
will give horizontal asymtotes.
$<br /> f(x)=\frac{2x}{\sqrt{x^2+x+1}}=\frac{2\ \mbox{sign}(x)}{\sqrt{1+1/x+1/x^2}}<br />$
So as $x \to \infty\ \ f(x) \to 2$, and as $x \to -\infty\ \ f(x) \to -2$, giving horizontal asymtotes at $y=\pm 2$
Last edited by CaptainBlack; March 19th 2006 at 11:38 PM.
I see how you got the equation for the horizontal asympototes but when I graph the function, y=-2 does not quite fit the horizontal asymptote on the bottom. Why is that so?
I see how you got the equation for the horizontal asympototes but when I graph the function, y=-2 does not quite fit the horizontal asymptote on the bottom. Why is that so?
Probably because you have not extended the plot far enough to the left.
Here is a little table which shows what is going on:
$<br /> \begin{array}{cc}x&f(x)\\-1&-2\\-2&-2.31\\-3&-2.27\\-4&-2.22\\-5&-2.18\\-10&-2.10\\-100&-2.01 \end{array}<br />$
March 19th 2006, 06:43 PM #2
Global Moderator
Nov 2005
New York City
March 19th 2006, 08:47 PM #3
Grand Panjandrum
Nov 2005
March 23rd 2006, 04:52 AM #4
Dec 2005
March 23rd 2006, 05:19 AM #5
Grand Panjandrum
Nov 2005 | {"url":"http://mathhelpforum.com/calculus/2271-please-help.html","timestamp":"2014-04-19T10:01:22Z","content_type":null,"content_length":"44123","record_id":"<urn:uuid:7397545d-265c-4c50-bf59-809cb1380a13>","cc-path":"CC-MAIN-2014-15/segments/1397609537097.26/warc/CC-MAIN-20140416005217-00624-ip-10-147-4-33.ec2.internal.warc.gz"} |
set of
Here is how to determine if a set of differential equations is stiff.
The two equations below are a simplified model for a catalytic converter, where y denotes the mole fraction of CO in the gas phase and T denotes the temperature of the gas phase. The model was
simplified by excluding the reaction term.
When the constants are substituted into the above equations we get:
By solving for the eigenvalues for this system of equations, we can tell which variable, y or T, is going to its steady state solutions faster. Note.
For information regarding eigenvalues for systems of ODE's, please consult an ordinary differential equations text.
It is the ratio of the eigenvalues (larger over smaller) that tells us if the equations are stiff. A ratio of 1000 qualifies as 'stiff', while a ratio of 5000 would be 'very stiff'.
Here is the ratio of the Eigenvalues.
Page constructed by
Doug Thornhill, James Bopp, Cory Clemmons and Devin Hodgson.
For questions or suggestions, please contact us through dt@u.washington.edu
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The module implements directed acyclic word graphs (DAWGs) internaly represented as minimal acyclic deterministic finite-state automata.
In comparison to Data.DAWG module the automaton implemented here:
• Keeps all nodes in one array and therefore uses much less memory,
• When weighed, it can be used to perform static hashing with hash and unHash functions,
• Doesn't provide insert/delete family of operations.
DAWG type
newtype DAWG a b c Source
DAWG a b c constitutes an automaton with alphabet symbols of type a, node values of type Maybe b and additional transition labels of type c. Root is stored on the first position of the array.
(Eq b, Eq c, Unbox c) => Eq (DAWG a b c)
(Eq (DAWG a b c), Ord b, Ord c, Unbox c) => Ord (DAWG a b c)
(Show b, Show c, Unbox c) => Show (DAWG a b c)
(Binary b, Binary c, Unbox c) => Binary (DAWG a b c)
fromList :: (Enum a, Ord b) => [([a], b)] -> DAWG a b ()Source
Construct DAWG from the list of (word, value) pairs. First a DAWG is created and then it is frozen using the freeze function.
fromListWith :: (Enum a, Ord b) => (b -> b -> b) -> [([a], b)] -> DAWG a b ()Source
Construct DAWG from the list of (word, value) pairs with a combining function. The combining function is applied strictly. First a DAWG is created and then it is frozen using the freeze function.
fromLang :: Enum a => [[a]] -> DAWG a () ()Source
Make DAWG from the list of words. Annotate each word with the () value. First a DAWG is created and then it is frozen using the freeze function.
type Weight = IntSource
Weight of a node corresponds to the number of final states reachable from the node. Weight of an edge is a sum of weights of preceding nodes outgoing from the same parent node.
elems :: Unbox c => DAWG a b c -> [b]Source
Return all elements of the DAWG in the ascending order of their keys. | {"url":"http://hackage.haskell.org/package/dawg-0.7.1/docs/Data-DAWG-Static.html","timestamp":"2014-04-24T04:37:51Z","content_type":null,"content_length":"20081","record_id":"<urn:uuid:4b2b9f96-5613-40ac-9fda-2a62627c62e5>","cc-path":"CC-MAIN-2014-15/segments/1398223205137.4/warc/CC-MAIN-20140423032005-00462-ip-10-147-4-33.ec2.internal.warc.gz"} |
Returns number rounded up to the nearest even integer. You can use this function for processing items that come in twos. For example, a packing crate accepts rows of one or two items. The crate is
full when the number of items, rounded up to the nearest two, matches the crate's capacity.
Number is the value to round.
● If number is nonnumeric, EVEN returns the #VALUE! error value.
● Regardless of the sign of number, a value is rounded up when adjusted away from zero. If number is an even integer, no rounding occurs.
The example may be easier to understand if you copy it to a blank worksheet.
● Create a blank workbook or worksheet.
● Select the example in the Help topic.
Note Do not select the row or column headers.
Selecting an example from Help
● Press CTRL+C.
● In the worksheet, select cell A1, and press CTRL+V.
● To switch between viewing the results and viewing the formulas that return the results, press CTRL+` (grave accent), or on the Formulas tab, in the Formula Auditing group, click the Show Formulas
A B
Formula Description (Result)
=EVEN(1.5) Rounds 1.5 up to the nearest even integer (2)
=EVEN(3) Rounds 3 up to the nearest even integer (4)
=EVEN(2) Rounds 2 up to the nearest even integer (2)
=EVEN(-1) Rounds -1 up to the nearest even integer (-2) | {"url":"http://office.microsoft.com/en-us/excel-help/even-HP005209080.aspx","timestamp":"2014-04-19T18:00:29Z","content_type":null,"content_length":"22980","record_id":"<urn:uuid:8880104b-a9b3-49a8-85c0-8c1e3180a056>","cc-path":"CC-MAIN-2014-15/segments/1397609537308.32/warc/CC-MAIN-20140416005217-00589-ip-10-147-4-33.ec2.internal.warc.gz"} |
Score Higher by Avoiding Algebra
Demonstrating a contradiction is a powerful technique for data sufficiency questions on the Quant section of the GMAT.
As an example, let’s take a look at the following data sufficiency problem:
Is b < ?
(1) b < a
(2) b = -2
A) Statement 1) alone is sufficient, but statement 2) alone is not sufficient
B) Statement 2) alone is sufficient, but statement 1) alone is not sufficient
C) BOTH statements 1) and 2) TOGETHER are sufficient, but NEITHER statement ALONE is sufficient)
D) EACH statement ALONE is sufficient
E) Statements 1) and 2) TOGETHER are NOT sufficient
By looking at statement 1) alone, we can suspect that it provides insufficient data. We’re told that b < a, but that seems to tell us very little about whether b < . After all, we don’t know whether
the variable “a” is positive or negative, and the exponent in can change “a” from a negative number to a positive number. If statement 1) alone is insufficient, we can prove this without doing any
algebra simply by demonstrating a contradiction.
For a statement to provide sufficient data for a Yes/No-type DS question, it must guarantee that the answer to the question stem is either always “Yes” or always “No”.
We’ll use the technique of plugging-in to find two contradictory examples. For the first example, let’s choose a few small prime numbers, such as a = 3 and b = 2. If we plug these numbers in, we see
that b < a, so these numbers satisfy statement 1). Now, let’s answer the question stem: is b < ? We see that 2 is indeed less than = 9, so the answer is “Yes”.
We now need to find a single example of a contradictory “No” answer to the question stem. To do so, let’s switch the type of numbers used for the plug-in. We used positive, prime integers at first,
but there are no restrictions which state that the numbers must be positive values — or even integers. We should therefore consider using fractions or negative numbers.
Let’s try one set of fractional plug-ins: b=
In general, if your hunch suggests that you have insufficient data, prove you have insufficiency by demonstrating a contradiction by plugging-in a few numbers. Only use algebra if you suspect you
have sufficient data. | {"url":"https://gmat.economist.com/blog/quantitative/score-higher-avoiding-algebra-0?gsrc=GMATCLUB_TEGblogfeed","timestamp":"2014-04-17T21:34:38Z","content_type":null,"content_length":"39549","record_id":"<urn:uuid:07f135af-65e8-4ab7-ad78-5d6a34e3a65a>","cc-path":"CC-MAIN-2014-15/segments/1397609532128.44/warc/CC-MAIN-20140416005212-00280-ip-10-147-4-33.ec2.internal.warc.gz"} |
collinear points
Definition of Collinear Points
● The points lying on the same line are called Collinear Points.
More about Collinear Points
• The points that do not lie on the same line are called non-collinear points.
• Three non-collinear points form a triangle.
Examples of Collinear Points
• In the given figure, U, P, and V are the collinear points.
• Also (P, Q, R), (T, S, R), and (V, Q, S) are the other sets of collinear points.
• (U, P, T), (U, S, T), and (V, Q, R) are the sets of non-collinear points.
Solved Example on Collinear Points
Identify the three collinear points from the diagram.
A. P, Q, and R
B. R, Q, and S
C. P, Q, and S
D. none of these
Correct Answer: A
Step 1: If the points lie on the same line, then the points are called collinear points.
Step 2: In the figure, P, Q, and R lie on the same line.
Step 3: So, P, Q, and R are the three collinear points.
Related Terms for Collinear Points | {"url":"http://www.icoachmath.com/math_dictionary/collinear_points.html","timestamp":"2014-04-21T12:25:09Z","content_type":null,"content_length":"7895","record_id":"<urn:uuid:491fd60f-3eeb-465a-9b42-894466d62aea>","cc-path":"CC-MAIN-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00007-ip-10-147-4-33.ec2.internal.warc.gz"} |
Westminster, CA Trigonometry Tutor
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Statistical Analysis: an Introduction using R/R/Lists
R is very particular about what can be contained in a vector. All the elements need to be of the same type, an moreover must be either types of number
, logical values, or strings of text
If you want a collection of elements which are of different types, or not of one of the allowed vector types, you need to use a list.
1. l1 <- list(a=1, b=1:3)
2. l2 <- c(sqrt, log) #
1. ↑ There are actually 3 types of allowed numbers: "normal" numbers, complex numbers, and simple integers. This book deals almost exclusively with the first of these.
2. ↑ This is not quite true, but unless you are a computer specialist, you are unlikely to use the final type: a vectors of elements storing "raw" computer bits, see ?raw
Last modified on 23 July 2009, at 20:05 | {"url":"http://en.m.wikibooks.org/wiki/Statistical_Analysis:_an_Introduction_using_R/R/Lists","timestamp":"2014-04-20T23:29:11Z","content_type":null,"content_length":"18263","record_id":"<urn:uuid:5a0668c6-7f53-4622-9796-58b254b8201b>","cc-path":"CC-MAIN-2014-15/segments/1398223202774.3/warc/CC-MAIN-20140423032002-00405-ip-10-147-4-33.ec2.internal.warc.gz"} |
Math Forum Discussions
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12/23/12 Distinguishability of paths of the Infinite Binary tree??? Zaljohar@gmail.com
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12/28/12 Re: Distinguishability of paths of the Infinite Binary tree??? Virgil
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12/29/12 Re: Distinguishability of paths of the Infinite Binary tree??? Virgil
12/29/12 Re: Distinguishability of paths of the Infinite Binary tree??? mueckenh@rz.fh-augsburg.de
12/29/12 Re: Distinguishability of paths of the Infinite Binary tree??? Virgil
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12/29/12 Re: Distinguishability of paths of the Infinite Binary tree??? mueckenh@rz.fh-augsburg.de
12/29/12 Re: Distinguishability of paths of the Infinite Binary tree??? Virgil
12/27/12 Re: Distinguishability of paths of the Infinite Binary tree??? Virgil
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12/26/12 Re: Distinguishability of paths of the Infinite Binary tree??? mueckenh@rz.fh-augsburg.de
12/26/12 Re: Distinguishability of paths of the Infinite Binary tree??? Virgil
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12/27/12 Re: Distinguishability of paths of the Infinite Binary tree??? Tanu R.
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12/28/12 Re: Distinguishability of paths of the Infinite Binary tree??? Virgil
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Linear Algebra
From Uncyclopedia, the content-free encyclopedia
(Redirected from
Linear algebra
“Go to hell.”
“oh god please no”
Linear algebra, a branch of mathematics that describes spherical rooms (or vector room) of odd numbered dimensions (1,3,5, ...,pi) described by curved vectors, subspheres, tensors, linear
transformations , Cheese, and other completely useless abstractions of reality between spherical rooms. The term linear originates from the usage of spheres (from elementary logic).
The science was first explored around 500 B.C. when Euclid (of Greece) discovered that two spherical vectors can be used to describe a spherical volume. First believed to have no practical
applications, linear algebra was revisited in 1843 by the british monk William Rowan Hamilton, who accidentally received a ten pound century old copy of the greek scripture on his toes. Another monk,
the german Hermann Grassmann, invented the matrix, which basically is a list of numbers. Noteworthy is the contemporary usage which often involves green symbols falling downward. Despite early
difficulties, Grassmann discovered (with the help of wine) that matrices were extremely useful to describe curved vectors and spherical rooms, but the only problem with this was that to get the
proper results, one had to sacrifice a goat to God
edit Current Usage
Across the nation, many high school students are being taught algebra. Unfortunately, since not all students are on the same level academically, linear algebra is substituted for regular algebra for
students who are struggling with regular algebra. Since its linear, its much easier for students who are so stupid, they can only understand math that involves lines and nothing else. Students who
cannot understand linear algebra are often further demoted to calculus with emphasis on imaginary numbers since they can simply use their imagination instead of their analytic capabilities.
edit Elementary linear rooms
The first spherical room of study was an elliptical circular line, the fundamental example of a one-dimensional sphere. Described by a single curved vector, any point within the extended line can be
described as a linear combination of the single curved vector. That vector is then considered a base for that particular spherical room. If the vector has a length of pi, it is considered to be an
ortonormalised base.
The spherical room can be extended to include more than one dimension. Another real world example is the standard linear elliptical sphere, described by three curved vectors. Notably, it is
occasionally referred to as personal space (compare: Earth's personal space). In this particular room a base consists of two curved vectors (that are partially perpendicular in a pair). The third
vector will always be hypertransposed, moved pi units from the origin. If the base vectors are completely perpendicular, the base is considered ortonormalised. In a three-dimensional linear room, no
base can possibly be ortonormalised.
edit Subspheres
A subsphere is a linear room present in another linear room with an even number of lower dimensions. An example is an elliptical circular line present in a standard linear elliptical sphere with a
shared origin and a imaginary two-dimensional circle as a normal vector. Because of the 'hypertransposition theorem' earlier explained, a circle can only exist as a subsphere in a room of higher
dimension and never as an independent spherical room. As a result, the elliptical sphere is a subsphere of the five-dimensional hypersphere, but not to the four-dimensional hypercircle (unless both
are present in the hypersphere in the same model of calculations).
edit Linear transformations
A linear transformation is a function from one sphere to itself or a subsphere. It is also possible to have transformations from imaginary subspheres (such as the circle). The transformation is
equivalent to a matrix, a transformation matrix (compare with the unimatrix, which transforms the room into a elliptical circular line with length pi). Transformations makes it possible to perform
calculations on non-existant imaginary subspheres, by transforming them into existing, independent linear rooms. These transformations have more applications then first meets the eye, for example
Linear Algebra in nature.
edit Dangers of using Linear Algebra
Linear Algebra is known as a weapon of mass destruction in some parts of the world. The fact that it can create elliptical circular lines creates great confusion, so much that World War 2 could have
been solved by dropping a linear algebra text book into japan. This would have resulted in some one eventually drawing an elliptical circular line, which has the same effect as Dividing by zero
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Help! -Linked Lists, Memory Allocation, Time Steps, Debugg
Help! -Linked Lists, Memory Allocation, Time Steps, Debugg
I'm having trouble with effectively getting my code to do what my assignment says. Initially I can't debug it to get it to compile for what i have so far. Next, I'm not sure where and how i
should add a linked list in my code to allocate memory after a crash has occured. Then lastly i'm not sure how to make the cars go through 1000 time steps in the simulation.
Here is the Question:
Consider a square domain partitioned into 250,000 equally spaced grid cells (500 by
500). We will randomly distribute some cars into this domain such that no more than 1
car is located in any grid cell initially. Each car has a random mass between 1-50. At
each time step of the simulation, each car randomly moves one grid cell in the x or y
direction (4 possible directions). If a car touches the edges of the domain then it
“rebounds” 180o towards the interior. If two cars land on the same grid cell, then they
have had an accident, and the smaller car is removed from the simulation.
Write a program to simulate this system. Use structures to represent the cars in the
simulation. Make your program efficient by only allocating memory for active cars, and
free memory for cars that are no longer active. Hint: use a linked list to represent the
Show the initial distribution of cars and the final distribution of surviving cars (the
number of cars at each possible mass) after 1000 time steps if the simulation starts with
500 and 2000 cars (2 cases). Make a histogram of your results using excel. The bins on
the x-axis of the histogram will be the mass of each car. The value on the y-axis will be
the number of cars in each bin.
-I would greatly appreciate any help i could get, seeing how i'm stuck and just going in circles.
Thank You in advance for trying to help!
Here is my Code:
/* ************************************************************************
written by: Robert Stange (Feb09)
Student of Civil and Environmental Engineering
UC Davis ECI19
The purpose of this program is to predict the distribution of surviving cars.
*The program must generate a 250,000 equally spaced grid with 500 cars ranging in
mass from 1-50. Each car must beable to randomly move in 4 possible directions.
If there is a crash the smaller car is removed from the simulation. Then you must
allocate the memory for that crashed car and free it to the system.
-The program randomly generates the following data:
-The program will output the surviving cars after 1000 time steps.
*********************************************************************** */
//prototype definitions
//structure for x,y,and weight of an array of cars
typedef struct
float x;
float y;
float weight;
} car[];
int location()
int main()
//variable definitions
int i = 0;
int X = rand() % 500;
int Y = rand() % 500;
int Weight = rand() % 50;
float car[5]={X, Y, Weight};
//generate cars
for (i=0;i<500;i++)
car[i]={X, Y, Weight};
printf("location of car is %d,%d and weight is %d\n",X.x, Y.y, Weight.weight);
return 0;
//Function for the location of the cars
int location()
int m[5] = rand() % 4;
int g = 5;
//loop through the different randimizations
for (i=0; i<=g; i++)
car[i]=rand() % (X+1).X;
car[i]=rand() % (X-1).X;
car[i]=rand() % (Y+1).Y;
car[i]=rand() % (Y-1).Y;
return m;
//Function to check for collisions
int check()
int ix=0;
int iy=0;
//loop checking for collisions
for(i=o; i<=500; i++)
for(ix=0; ix<=500; ix++)
for(iy=0; iy<=500; iy++)
car[i]= 0.x
car[i]= 0.y
return 0;
Thanks for all you time and help!
Let's start with the following code:
int check()
int ix=0;
int iy=0;
//loop checking for collisions
for(i=o; i<=500; i++)
for(ix=0; ix<=500; ix++)
for(iy=0; iy<=500; iy++)
car[i]= 0.x
car[i]= 0.y
return 0;
Take a careful look at that function, and figure out why it won't compile. Start with counting the number of start brackets ({) vs the number of end brackets (}).
Next, take at look at your if statement.
What do you want to do if iy=ix?
car[i]= 0.x
What in the world is going on here? This doesn't even resemble valid C code. You need to do a bit more reading from your course book.
You need to start with less code. Start with a single function, and get that to compile. Then start adding more code, compiling as you go.
I think the idea is to use the linked list for the cars so you can de-allocate the memory when a car is destroyed. So you want your car typedef to reflect this:
typedef struct _car
float x;
float y;
float weight;
struct _car* next;
} car[];
Then lastly i'm not sure how to make the cars go through 1000 time steps in the simulation.
Use another for loop after you "generate cars".
ps. nice assignment | {"url":"http://cboard.cprogramming.com/c-programming/113316-help-linked-lists-memory-allocation-time-steps-debugg-printable-thread.html","timestamp":"2014-04-18T20:37:35Z","content_type":null,"content_length":"15885","record_id":"<urn:uuid:d5ee7848-5fd6-4706-94e3-d50843d3384e>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.7/warc/CC-MAIN-20140416005215-00465-ip-10-147-4-33.ec2.internal.warc.gz"} |
I need some help
$\lim_{x \to 0} \frac{sin(x) - tan^{-1}(x)}{x^2 ln(1 + x)}$ This is of the form 0/0 and so can be done by using L'Hopital's rule. However taking the derivative of this is going to be long and
susceptible to errors. What I would suggest is that we replace the sine, inverse tangent, and natural log functions with their Taylor expansions for x near 0: $sin(x) \approx x - \frac{x^3}{6}$ $tan^
{-1}(x) \approx x - \frac{x^3}{3}$ $ln(1 + x) \approx x - \frac{x^2}{2}$ So $\lim_{x \to 0} \frac{sin(x) - tan^{-1}(x)}{x^2 ln(1 + x)} \approx \frac{x - \frac{x^3}{6} - x + \frac{x^3}{3}}{x^2 \left
(x - \frac{x^2}{2} \right ) }$ = $\lim_{x \to 0} \frac{ \frac{x^3}{6} }{x^3 - \frac{x^4}{2} }$ = $\lim_{x \to 0} \frac{x^3}{6x^3 - 3x^4}$ = $\lim_{x \to 0} \frac{1}{6 - 3x}$ = $\frac{1}{6}$ -Dan
$\lim_{x \to 0} \left ( \frac{1}{sin^2(x)} - \frac{1}{x^2} \right )$ Again, I would employ the Taylor series trick here (because I just tried the L'Hopital's route and the 0/0 didn't go away after
two applications and things were again getting ugly.) First combine the fractions: $\lim_{x \to 0} \left ( \frac{1}{sin^2(x)} - \frac{1}{x^2} \right ) = \lim_{x \to 0} \frac{x^2 - sin^2(x)}{x^2sin^2
(x)}$ $sin^2(x) \approx x^2 - \frac{x^4}{3}$ $\lim_{x \to 0} \frac{x^2 - sin^2(x)}{x^2sin^2(x)} = \lim_{x \to 0} \frac{x^2 - x^2 + \frac{x^4}{3}}{x^2 \left (x^2 - \frac{x^4}{3} \right ) }$ = $\lim_{x
\to 0} \frac{\frac{x^4}{3}}{x^4 - \frac{x^6}{3}}$ = $\lim_{x \to 0} \frac{\frac{1}{3}}{1 - \frac{x^2}{3}}$ = $\frac{1}{3}$ -Dan
I loved the Tylor trick... Can you explain why we can take the Taylor expansions for x near 0? is it because the Lim is for x near 0?
Yes, that's exactly it. Normally I would consider such an approach "cheating" but it looks like the problems you were given were designed to make an application of L'Hopital's rule difficult if not
impossible. And, as I'm a Physicist and not a Mathematician I have no problems with "cheating" if it gets me the answers. -Dan | {"url":"http://mathhelpforum.com/calculus/8830-i-need-some-help.html","timestamp":"2014-04-16T04:26:49Z","content_type":null,"content_length":"45787","record_id":"<urn:uuid:663e8599-bf2c-4016-adf8-863174a99e49>","cc-path":"CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00138-ip-10-147-4-33.ec2.internal.warc.gz"} |
Combining Like Terms Examples Page 3
Combine like terms in the expression 4 + 5y + y + 2x^2 + 3x^2.
We can combine the two terms with variable part y to get
4 + 6y + 2x^2 + 3x^2,
and we can also combine the terms with variable part x^2 to get
4 + 6y + 5x^2.
Sometimes it is helpful to rearrange the terms such that like terms are next to each other. Just don't put them too close together unless they have both recently bathed. | {"url":"http://www.shmoop.com/algebraic-expressions/combining-like-terms-examples-3.html","timestamp":"2014-04-20T18:52:08Z","content_type":null,"content_length":"37151","record_id":"<urn:uuid:57b59e93-42a0-4e53-879c-f524c63b7f96>","cc-path":"CC-MAIN-2014-15/segments/1397609539066.13/warc/CC-MAIN-20140416005219-00394-ip-10-147-4-33.ec2.internal.warc.gz"} |
Re: mpz_powm crashes with modulus>2^2100000
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Re: mpz_powm crashes with modulus>2^2100000
From: Jens Kruse Andersen
Subject: Re: mpz_powm crashes with modulus>2^2100000
Date: Wed, 12 Mar 2003 20:49:43 +0100
Paul Zimmermann wrote:
> It is my fault. I'm the original author of the current mpz_powm code.
> I wrote it to optimize the computer time, without taking into account
> the memory used. For your example (exponent and modulus of 2145355 bits)
> the mpz_powm function will first create an auxiliary table containing
> x^(2i+1) for i < 2^15, i.e. 2^14 values each of 268Ko, i.e. a total of
> about 4.4Go!!!
> The workaround (which I think will be in the next gmp release) is to have
> a hard limit for the parameter 'k' in mpz_powm. For example kmax=7 seems
> reasonable, which gives a table of 2^6 elements at most, i.e. about 17Mo
> in your case. Since the cost is proportional to 1+1/(k+1), with k=7 you
> loose only about 6% wrt the "optimal" k=15.
> Below is a patch that limits k to 7, and in addition prints some information
> to show the progress in the computation. For computing 3^p mod p for
> p=3*2^2145353+1 on an old PIII-500, I estimate it will take about 120 days.
I inserted the patch including all printf. My earlier sent test program still
crashes with p=2^2100000+1 after outputting:
Calling mpz_powm_ui...Done. Calling mpz_powm...k=1
computing x^2
Note the test program only tries to compute 3^3 mod p, the exponent is tiny
and the patch does not change k.
I am running under Windows XP with 256 MB physical memory.
Before my first mail, I had already made my own workaround by writing my own
proth_powm and skipping mpz_powm completely. That worked and did not slow down
my program, so you don't have to make a patch for my sake. I just reported the
problem for others but I will be happy to test your patches if the problem is
platform dependant.
The prp test of p=3*2^2145353+1 had completed with my workaround. Just for
speed, before finding the powm bug, I had written my own proth_mod optimized
for modulus p=k*2^n+1, unsigned long k. This can be done with bitshifts and
_ui operations and is extremely fast compared to the generic mpz_mod. I
combined mpz_mul and my proth_mod and the resulting modular exponentiation
became 5.3 times faster than mpz_powm for p=54767*2^1337287+1, taking 40 hours
on my 1333 MHz Athlon XP with 133 MHz ram.
p=3*2^2145353+1 took 4.5 days for my program, which only makes a base 2^32
Fermat prp test as a side effect. My purpose is to find the _smallest_ e for
which (2^32)^e mod p = 1. This e is the order of 2^32 for modulus p. George
Marsaglia has a good and efficient rng based on huge Proth primes and he asked
me to write a program capable of computing the period which is the order of
2^32. I installed GMP for this purpose and my original program called mpz_powm
with small exponents near the end of the order computation - after saving to
file, so the 4.5 day run was not lost when it crashed. With my workaround, it
took the program a couple of minutes from the restore file to find the order
e=2^2145347 for p=3*2^2145353+1. This is probably the longest known period for
an rng - and possibly the largest modular exponentiation performed with GMP. I
would not have computed this with mpz_powm on an old PIII-500!
BTW, I see you have also worked on CYF no. 11 at
You sure were patient, I wonder if a factor of 10^999+13 will ever be found.
It is a little surprising to me that factoring efforts already fails at the
14th titanic number but I guess it is just a coincidence that the first hard
number comes so early. You better get started on gmp-ecm 6.0 after
mpz_powm :-)
Jens Kruse Andersen
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Re: [PRD] PRD Review By Example
From: Christian de Sainte Marie <csma@ilog.fr> Date: Fri, 17 Oct 2008 10:27:30 +0200 Message-ID: <48F84C72.9050207@ilog.fr> To: Gary Hallmark <gary.hallmark@oracle.com> CC: rif WG <
Gary and all,
I just browsed through your comments, yet, so this is only a quick comment re the subject: trying the semantics with examples is a very good thing, indeed. But I suggest that you try the semantics in the July FPWD, or wait a couple more days for the current wiki version to be updated.
Just as a reminder: although there is no specific warning in the current draft on the wiki, the "rules and ruleset" section is not uptodate, as I said during the PRD telecon (but maybe not clearly enough).
It is mostly the version that was published in July, with a couple ad hoc changes, but not aligned with the remainder of the document anymore. E.g., as Gary notices, the notion of an instance of a rule is still used, but it is not defined anymore, etc.
My understanding was that that part (the semantics of rulesets, including conflict resolution) would be the main subject of our discussion in Orlando, and I planned to have it tidied by next Tuesday.
Gary, actually, I think that it would be interesting if you could try your example on the smenatics as it is described in the FPWD, since I do not plan to change anything re the content, only align the form to the editorial changes in the other parts of the document.
Gary Hallmark wrote:
> This is a case-oriented review of the current PRD spec on the wiki. It
> attempts to use the spec to give meaning to
> the following simple ruleset, and identifies a number of issues
> uncovered. Overall, I'm encouraged that the spec
> generally seems to "work" -- many of the issues are largely editorial
> (save for the problems with PICK, which I address
> somewhat here and in detail in a seperate email.
> Prefix(p someIRI)
> p:A(0)
> (* r1 *)Forall ?x (p:A(?x) :- p:A(External(func:numeric-minus(?x 1))))
> Issue 1: We must state that all "initial facts" in the ruleset RS (here,
> p:A(0)) are removed from RS and become the initial working memory
> w<sub>0</sub>
> so, after this step, we have
> w = {p:A(0)}
> RS = (* r1 *)Forall ?x (p:A(?x) :- p:A(External(func:numeric-minus(?x 1))))
> c = (w, {}, nil, nil) i.e. the initial configuration is the initial
> facts and an empty set of rule instances.
> We have added a third component to the configuration: the rule instance
> that is picked.
> Now, we execute the first cycle of the transistion system.
> s = {} because instances(c) is {}
> c' = (w, INSTANTIATE(w, RS), PICK(LIFO, c))
> Issue 2: INSTANTIATE returns a set of Inst(R). This is undefined.
> I guess it is a pair (rule-id, consistent-substitution), and that
> furthermore it must return *all* such pairs.
> Issue 3: picked(c) from my new PICK specification should be changed to
> mean the rule instance picked
> from the prior configuration to arrive at the current configuration. And
> the transistion system that calls PICK must
> save the returned rule instance as picked(c').
> So,
> c' = (w, (r1, {?x->1}), nil)
> Now, the second cycle
> s = extractActions((r1, {?x->1})) because PICK is trivial - only 1 rule
> instance exists
> Issue 4: extractActions seems a bit underspecified as to how it
> "grounds" the actions in the rules,
> which may have variables. Also, its argument should be a single rule
> instance, not a list.
> I guess it is supposed to "plug in" the substitutions, so
> s = {p:A(1)}
> Now, the RIF-PRD transition system says
> ({p:A(0)}, p:A(1), w') →RIF-PRD if and only if, for every ground formula
> φ in W,
> w' |= φ if and only if And({p:A(0)} p:A(1)) |= φ
> Issue 5: And(...) is undefined. We must define the PS before the
> semantics if we use the PS in the semantics.
> Also, this And takes a set as its first argument. I'll replace the set
> with its member:
> ({p:A(0)}, p:A(1), w') →RIF-PRD if and only if, for every ground formula
> φ in W,
> w' |= φ if and only if And(p:A(0) p:A(1)) |= φ
> w' = {p:A(0) p:A(1)}
> Note: w' is not uniquely determined. E.g. w' = {And(1=1 p:A(0) p:A(1))}
> is ok, too. I guess this is fine.
> c' = (w', INSTANTIATE(w', RS), (r1, {?x->1})) = ({p:A(0) p:A(1)}, {(r1,
> {?x->1}), (r1, {?x->2})}, (r1, {?x->1}))
> Now, the third cycle. Interesting because we need to pick one of the two
> rule instances.
> Issue 6: PICK is ill-defined.
> See my email proposal. According to the new specification, assuming the
> rule is declared as repeatable, the agenda is:
> {(0, 0, (r1, {?x->2}))}
> and thus its only rule instance is returned.
> s = {p:A(2)}
> w' = {p:A(0) p:A(1) p:A(2)}
> c' = ({p:A(0) p:A(1) p:A(2)}, {(r1, {?x->1}), (r1, {?x->2}), (r1,
> {?x->3})}, (r1, {?x->2}))
> Note that if r1 is declared *not* repeatable, then the agenda would be
> empty in the third cycle and the system terminates.
Received on Friday, 17 October 2008 08:29:17 GMT | {"url":"http://lists.w3.org/Archives/Public/public-rif-wg/2008Oct/0076.html","timestamp":"2014-04-20T11:27:39Z","content_type":null,"content_length":"12886","record_id":"<urn:uuid:7f6bc496-7ad9-4a42-a12c-087c843b6d08>","cc-path":"CC-MAIN-2014-15/segments/1397609538423.10/warc/CC-MAIN-20140416005218-00205-ip-10-147-4-33.ec2.internal.warc.gz"} |
Math Power: How to Help Your Child Love
Math, Even If You Don't
Return to AWM Bibliography
AWM Newsletter
AWM Book Review:
Patricia Clark Kenschaft, Addison-Wesley, Reading, MA 1997. x+310. ISBN 0-201-77289-2 (paper), $15.00.
From: AWM Newsletter, January/February 2000.
Reviewed by: Bridget Arvold, Department of Curriculum and Instruction, College of Education, University of Illinois at Urbana/Champaign, Champaign IL 61820; email: arvold@uiuc.edu.
Many reports and articles lead Americans to believe that mathematics education in the United States is a sad state of affairs. Textbook publishers, program developers, teachers, and teacher educators
have become targets for attack. In _Math Power: How to Help Your Child Love Math, Even If You Don't_, Patricia Clark Kenschaft not only offers her perspective on a plethora of issues in mathematics
education but invites parents to become involved in their children's mathematics education. She addresses parents "who are willing to play with math for the sake of their children." Kenschaft is
motivated by her work with children and her concern for the future of mathematics. "As real mathematics struggles for survival in our culture," Kenschaft writes, "it becomes increasingly urgent that
parents share real mathematics with their children." She describes Math Power as a guide to parents who, as she noted, "may or may not possess 'mathematical minds.'"
Kenschaft shares her perspectives on the nature and value of mathematics, and on mathematical ability and development. In some cases, however, her incorporation of outside perspectives confuses
rather than clarifies matters. She emphasizes the need for parental involvement in the school community and urges parents to promote enjoyable mathematical experiences at home. She suggests that
parents help their children view failure and frustration as a natural part of learning. Although the mixed messages about the enjoyable and frustrating nature of doing mathematics might confuse some
readers, her messages speak to the complexity of learning mathematics.
Kenschaft reveals how her teaching experiences brought her to the realization that the basics of school mathematics often leave children confused. Her involvement with PRIMES, the Project for
Resourceful Instruction in Mathematics in the Elementary School, provided her with insights into early childhood education. She describes the nature of context-free mathematics terminology and shares
her feelings: "Pity the poor children trying to figure out the meaning of basic mathematical words." Her frustration spilled over into her graduate level classes and became an important topic of
discussion there. Convinced of the power of understanding the basic concepts in mathematics, she pours her convictions about developing math power into this resource for parents.
Although Kenschaft supports the organizational and relational nature of mathematics, her book focuses on coping with conventional school mathematics and downplays mathematical enjoyment. Statements
such as, "You must have a resilient ego to live with yourself while learning math" reflect little of the natural "enjoyment" she claims as necessary in parent-child activities. Remarks like, "The
U.S. system is so unreliable, we must give our children every possible advantage to endure it," reveal her feelings of frustration with the present system. Parents, especially those who were
unsuccessful with school mathematics, might trust the author and follow the suggestions to make their children "smart kids." But they are likely to be confused by messages such as, "Simple math is
harder than you thought," and "We all learn that failure is inevitable - but followed by success." While such statements might be true within given contexts, the mixed messages they offer parents may
be debilitating. In her effort to help parents thwart children's confusion, she may also incite parental confusion.
The mathematics education literature offers a host of rationales for supporting early mathematical development. Turback (1999) shares a biological research perspective that posits that although
babies may be born with neural wiring, most connections form during infancy and childhood. He adds that even while in the womb, humans begin to build connections that form the foundation of language.
Moreover, he reports that the analysis of data collected from monitoring human brain activity indicates that specific types of connections are made within small time periods. Findings suggest that if
children are not appropriately stimulated during this time, they may never be able to learn a specific concept or skill. From these findings, we might hypothesize that the foundations for the
language of mathematics are also formed in a child's earliest years and therefore parental nurturing is crucial to their mathematical development.
In The Language Instinct, Steven Pinker (1994) describes language not as a cultural artifact that is learned --- as, for example, we learn to tell time --- but as a distinct piece of our biological
makeup. From this perspective, mathematical language is integral to our very being and complements a natural penchant for organization and inquiry. Pinker adds that much of formative human experience
relies on touch, sound, and sight. Thus the construction of the language of mathematics might be enhanced by distinguishing shapes by touching, hearing the cadences of number patterns, and noting
similarities and differences. Kenschaft incorporates such approaches to language acquisition --- despite the fact that her perspective on child development is quite different from Pinker's. For
example, she believes that newborns are filled with mathematical potential at birth and that diminished potential is due to our corruptive culture. Thus Kenschaft provides parents with a quite
different basis for creating a positive learning environment for their children.
Although books of elementary mathematics activities abound, relatively few authors direct attention toward helping parents nurture the mathematical development of their children. Kenschaft approaches
this task by focusing on preparation for conventional school mathematics activities. A focus on counting and recognition of number patterns and polygonal shapes, for instance, offers children
cultural artifacts that prepare them for conventional school activities. Kenschaft's suggested activities connect to children's worlds, but few build from children's worlds or their parents' worlds.
Such building is based on fundamental mental and intuitive processes that are distinct pieces of a child's biological makeup. For example, enhancing story time with children's mathematics literature
(see Bresser 1995; Burns 1992) provides parents with subtle ways to expand children's ways of thinking. The mathematics within existing family practices provides learning opportunities for both
children and parents.
In a description of mathematics programs involving parents, Kliman (1999) suggests that adults build from natural mathematical processes that are often hidden from view. She describes how the
matching and sorting that accompany laundry chores can serve as a basis for subtle mathematical discussion. Kliman, like Kenschaft, promotes skill and concept development --- but Kliman emphasizes
relational thinking, reasoning, and problem solving as embedded within story time, outdoor activities, road trips, and household chores rather than as additions to existing activities. Yet both
approaches offer parents and children opportunities for extending themselves mathematically.
Descriptive accounts of parents' experiences with a child's mathematics --- although lengthy --- might better enable parents to foster mathematical development. For example, when my three-year old
daughter Trina voiced her wonderment about how we would share ten dinner rolls among the four at the table, we asked her to distribute the rolls for us. She passed one roll to each person and then
another to each. She broke each of the two remaining rolls into two pieces, and distributed the pieces equally among us. She smiled widely as she declared that we each received "two and a half rolls
so we each got the same." My amazement only grew when five-year old Erica piped in with, "No, we did not get the same amount. Actually, I don't have two and a half because halves must be the same
size and my half is bigger that your half." She immediately realized that the very words she had spoken, "my half is bigger than your half," contradicted her message and she broke out in a laughter
that was contagious.
The messages that parents might accord to this short account are numerous. Parents might learn a strategy for division. They might realize the power of language and children's ability to communicate
and think abstractly. Child-initiated mathematical activities are well worth the attention of parents and simple analysis of children's dialogue might help parents learn to recognize and make the
most of possible learning situations.
The belief that a young child constructs meaning has prompted the recent focus on student engagement in reasoning, problem solving, making connections and communicating mathematically (NCTM, 1989,
1999). Although Kenschaft uses a riddle and a description of a game called 7-Up to highlight these process standards, her mention of these areas of cognitive and social development is far too brief
to adequately prepare parents to ready their children for the classrooms of today. She might have relegated what she termed her "tirade" to a separate publication rather than including it in a
motivational guide to parents. Nonetheless, Math Power incites parents to become involved in their children's education and offers them insights into how they might help their children. In this
sense, it provides a much needed resource for parents.
Works cited:
Bresser, R. (1995). Math and Literature (4-6). White Plains, NY: Math Solutions Publications.
Burns, M. (1992). Math and Literature (K-3). White Plains, NY: Math Solutions Publications.
Kliman, M. (1999). "Beyond Helping with Homework: Parents and Children Doing Mathematics at Home." Teaching Children Mathematics 6, (3), 140-146.
National Council of Teachers of Mathematics (NCTM). (1999). Principles and Standards for School Mathematics, draft for 2000. Reston, VA: NCTM.
National Council of Teachers of Mathematics (NCTM). (1989). Curriculum and Evaluation Standards for School Mathematics. Reston, VA: NCTM.
Pinker, S. (1994). The Language Instinct. New York: William Morrow.
Turbak, G. (1999). "Tomorrow's Brainchild, Part One: The First Years Last Forever." Kiwanis 84 (1), 26-29, 50-51.
Copyright ©2005 Association for Women in Mathematics. All rights reserved.
Comments: awm-webmaster@awm-math.org. | {"url":"http://www.awm-math.org/bookreviews/JanFeb00.html","timestamp":"2014-04-20T08:44:14Z","content_type":null,"content_length":"13623","record_id":"<urn:uuid:ba49cc69-f0f7-4555-9d54-fbe8519d9390>","cc-path":"CC-MAIN-2014-15/segments/1397609538110.1/warc/CC-MAIN-20140416005218-00093-ip-10-147-4-33.ec2.internal.warc.gz"} |
Kimura alignment
Lloyd Allison lloyd at cs.monash.edu.au
Sun Feb 5 16:50:40 EST 1995
histone at acpub.duke.edu (Ronald DeBry) writes:
>article <D39Dv5.Ax8 at zoo.toronto.edu> mes at zoo.toronto.edu (Mark Siddall) writes:
>>Given a multiple hit site wherein roman numerals represent taxa and letters
>>are bases and the following distribution of states:
>>Taxon I II III IV V VI
>> A A G G T T
>>There are two possible explanations:
>>a) one transition and one transversion.
>>b) two transversions:
>>The decision as to which, requires reference to some tree. But we do not
>>yet have a tree, in fact, the tree is the end point of what we are
>>attempting to do.
>alignment should really be conditional on the evolutionary relationships
>(whether you are using a Kimura substitution model or not). Two kinds
>of solutions have been proposed. One can use an iterative reciprocal
>process, where a trial alignment is used to give a trial phylogeny,
>which is used to generate a new alignment then a new phylogeny and so on
>until it converges on a stable alignment/phylogeny. This approach has
>been implemented as a computer program by Jotun Hein. The on;y
>alternative that I know of is a series of papers by Rich Thorne. In his
>method (essentially) all possible alignments are generated. Each
>alignment is given a likelihood based on a model incorporating both
>insertion/deletion and substitution, and a distance matrix is calculated
>by weighting each alignment based on its likelihood. The phylogeny is
>then inferred from that weighted distance matrix.
> As someone who subscribes to the idea that phylogenetic inference is a
>statistical problem and that maximum likelihood in some form is the best
>approach, ...
>Ron DeBry Department of Medicine Duke Univ Medical Center
>histone at acpub.duke.edu
In the following we do Gibbs sampling over many, but not all
(that is infeasible), multiple alignments to get a good approx'
to the tree's true edge lengths.
Gibbs sampling is simulated annealing at a constant temperature;
if you cool it you also get a simulated annealing search for an
optimal alignment. This is for a fixed given tree, but it can be used to
compare a "few" competing evolutionary trees. (In principle you could
let the tree vary too, but then I don't know that I would like to wait up
for it to finish.)
%A L. Allison
%A C. S. Wallace
%T The posterior probability distribution of alignments and its application to
parameter estimation of evolutionary trees and to optimization of multiple
%J J. Mol. Evol.
%V 39
%P 418-430
%D 1994
Lloyd ALLISON
Central Inductive Agency,
Dept. of Computer Science, Monash University, Clayton, Victoria 3168, AUSTRALIA
tel: 61 3 905 5205 fax: 61 3 905 5146 email: lloyd at cs.monash.edu.au
<A HREF="http://www.cs.monash.edu.au/~lloyd/tildeStrings">Molecular Biology</A>
More information about the Mol-evol mailing list | {"url":"http://www.bio.net/bionet/mm/mol-evol/1995-February/002378.html","timestamp":"2014-04-18T05:34:34Z","content_type":null,"content_length":"5682","record_id":"<urn:uuid:52ee6184-4b38-4aa3-a774-59ba056f19c4>","cc-path":"CC-MAIN-2014-15/segments/1397609538787.31/warc/CC-MAIN-20140416005218-00026-ip-10-147-4-33.ec2.internal.warc.gz"} |
Learning Math Process Download - Free Download Learning Math Process
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CCSS.Math.Content.HSA-REI.C.6 - Wolfram Demonstrations Project
US Common Core State Standard Math HSA-REI.C.6
Demonstrations 1 - 11 of 11
Description of Standard: Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables.
Description of Standard: Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables. | {"url":"http://demonstrations.wolfram.com/education.html?edutag=CCSS.Math.Content.HSA-REI.C.6","timestamp":"2014-04-19T04:28:56Z","content_type":null,"content_length":"27561","record_id":"<urn:uuid:b9fa06d3-c224-4ad2-972b-8fef76ab9536>","cc-path":"CC-MAIN-2014-15/segments/1397609535775.35/warc/CC-MAIN-20140416005215-00643-ip-10-147-4-33.ec2.internal.warc.gz"} |
Minimal Surfaces, Differential Geometry, and Partial Differential Equations
Last night in a lecture my professor explained that some partial differential equations are used to observe events on minimal surface (e.g. membranes).
A former advisor, someone that studied differential geometry, gave a brief summary of minimal surfaces but in a diffy G perspective.
1.) Are there any connections between the two studies and their perspectives on minimal surfaces?
2.) Are there any papers that use minimal surfaces, PDE's, and Diffy G together in some manner?
I'm curious to know what do PDE's and Diffy G have in 'common' via minimal surfaces. Any input is appreciated. | {"url":"http://www.physicsforums.com/showthread.php?t=441921","timestamp":"2014-04-18T03:03:36Z","content_type":null,"content_length":"20063","record_id":"<urn:uuid:09a0ad84-20ac-4165-8034-3f7532000c21>","cc-path":"CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00099-ip-10-147-4-33.ec2.internal.warc.gz"} |
Dealing with Multivariate Limits?
April 13th 2009, 10:31 PM #1
Apr 2009
Dealing with Multivariate Limits?
Just revising, and came across a few limits I couldn't remember how to prove.
1. lim(x,y)-->(0,0) (cosx-1+x^2/2)/(x^4-y^4)
2. lim(x,y)-->(0,0) (y-2x+sin2x)/(x^3+y)
And these, which are supposed to use sandwich rule
1. 7x^2*y^2/(x^2+2y^4)
2. 3y*x^2/(x^2+y^2)
The most promising way to find such types of limits is working in polar coordinates with the substitution...
$x= \rho \cdot \cos \theta$
$y = \rho \cdot \sin \theta$ (1)
... and searching [if it exists...] the limit with $\rho \rightarrow 0$ that is independent from $\theta$.
Let's start with your example nr. 2. With substitutions (1) we have...
$\lim_{[x,y]\rightarrow [0,0]} \frac{y- 2\cdot x + \sin x}{x^{3}+y} = \lim_{\rho \rightarrow 0} \frac {\rho\cdot (\sin \theta - 2\cdot \cos \theta) + \sin (2\cdot \rho \cdot \cos \theta)}{\rho \
cdot (\rho^{2}\cdot \cos^{3}\theta+ \sin \theta)}$ (2)
For 'small' value of $\rho$ is...
$\sin (2\cdot \rho \cdot \cos \theta) \approx 2\cdot \rho \cdot \cos \theta$
$\rho^{2}\cdot \cos^{3}\theta+ \sin \theta \approx \sin \theta$ (3)
... so that...
$\lim_{\rho \rightarrow 0} \frac {\rho\cdot (\sin \theta - 2\cdot \cos \theta) + \sin (2\cdot \rho \cdot \cos \theta)}{\rho \cdot (\rho^{2}\cdot \cos^{3}\theta+ \sin \theta)} = \lim_{\rho \
rightarrow 0} \frac {\rho \cdot (\sin \theta - 2\cdot \cos \theta + 2\cdot \cos \theta)}{\rho \cdot \sin \theta} = 1$ (4)
And now your example nr. 1. With substitutions (1) we have...
$\lim_{[x,y]\rightarrow [0,0]} \frac{\cos x - 1 + \frac{x^{2}}{2}}{x^{4}- y^{4}} = \lim_{\rho \rightarrow 0} \frac{\cos(\rho\cdot \cos \theta) -1 + \frac{\rho^{2}\cdot \cos^{2}\theta}{2}}{\rho^
{4}\cdot (\cos^{4} \theta - \sin^{4} \theta)}$ (5)
For 'small' value of $\rho$ is...
$\cos (\rho\cdot \cos \theta) \approx 1 - \frac {\rho^{2}\cdot \cos^{2} \theta}{2} + \frac {\rho^{4}\cdot \cos^{4} \theta}{24}$ (6)
... so that...
$\lim_{\rho \rightarrow 0} \frac{\cos(\rho\cdot \cos \theta) -1 + \frac{\rho^{2}\cdot \cos^{2}\theta}{2}}{\rho^{4}\cdot (\cos^{4} \theta - \sin^{4} \theta)} = \frac{1}{24}\cdot \frac{\cos^{4} \
theta}{\cos^{4} \theta - \sin^{4} \theta}$ (7)
Now we observe that the limit (7) does depend from $\theta$ so that we must conclude that limit (5) doesn't exist...
Kind regards
April 14th 2009, 12:44 AM #2 | {"url":"http://mathhelpforum.com/calculus/83641-dealing-multivariate-limits.html","timestamp":"2014-04-20T08:39:49Z","content_type":null,"content_length":"38315","record_id":"<urn:uuid:5fcc6733-8e6b-4db3-bb1a-14f683da8bd3>","cc-path":"CC-MAIN-2014-15/segments/1397609538110.1/warc/CC-MAIN-20140416005218-00439-ip-10-147-4-33.ec2.internal.warc.gz"} |
10.2 Higher Approximations and Taylor Series
Home | 18.013A | Chapter 10 Tools Glossary Index Up Previous Next
10.2 Higher Approximations and Taylor Series
We address the following questions:
What are these higher, non-linear approximations to f in terms of its derivatives?
Why do we do these things?
How accurate are these approximations?
What happens when f is a function of several variables?
The linear approximation to f at x[0] is the linear function with value f(x[0]) and first derivative f '(x[0]) there.
The quadratic approximation is the quadratic function whose value and first two derivatives agree with those of f at argument x[0]. Being quadratic it can be written as f(x[0]) + a(x - x[0]) + b(x -
We determine a and b by applying the condition that its derivatives are those of f at argument
We can extend this argument to create the cubic approximation, etc, when f is suitably differentiable by applying the same steps with still higher derivatives. If we do this on forever, we get the
"Taylor series expansion of f at argument x[0]."
10.1 Write down the Taylor series expansion about x[0] for a general infinitely differential function f.
10.2 Write down the approximation formula of degree 5 for a general function that is 5 times differentiable, and apply it explicitly for the sine function at x[0] = 0. Give the cubic approximation to
the sine, formed at x[0] = 1.
10.3 The exponential function, being its own derivative, can be factored out of its Taylor series expansion. Apply that expansion around x[0], to deduce the relation between exp(x) and exp(x[0]).
The following applet allows you to enter a standard function and look at what the first three of these approximations look like, as defined over a domain of your choosing. | {"url":"http://ocw.mit.edu/ans7870/18/18.013a/textbook/HTML/chapter10/section02.html","timestamp":"2014-04-17T09:42:31Z","content_type":null,"content_length":"4520","record_id":"<urn:uuid:71ecbf2e-c2dd-4797-b9c8-cba5598cad00>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00199-ip-10-147-4-33.ec2.internal.warc.gz"} |
Basic fractional equation
Definitely No! $\frac{x^2+x-12}{x^2-9x+18} = \frac{(x+4)(x-3)}{(x-6)(x-3)}$ 1. Cancel the common factor. 2. Determine the domain of the term at the LHS. 3. Set the numerator equal to zero, solve for
x. By the way: The simplification isn't necessary here: Multiply both sides of the equation by the denominator after you've determined the domain of the complete term.
Sorry, but this is completely wrong. By your reasoning, you would simplify something like $\frac{x^2 - 4}{x^2 - 1}$ as $\frac{-4}{-1} = 4$. You are expected to factorise the numerator and the
denominator and then cancel the common factor. Can you factorise $x^2 + x - 12$? Can you factorise $x^2 - 9x + 18$?
Thanks for the answer earboth ^^ worked it out now
The domain is which input values give real output values. Essentially this boils down to a few simple rules of which you should use the first The denominator can never equal 0 Values inside
logarithms must be greater than 0 Values inside even roots (such as square root) must be greater than 0 In your case $\frac{(x+4)}{(x-6)}$ which means any value that makes $x-6 = 0$ is not part of
the domain
So... it's safe to assume the following; $x+4=0$ , $x+6eq{0}$ ; $x=-4$? then how does this work out? Quote: Originally Posted by earboth $\frac{x^2+x-12}{x^2-9x+18}$ $x^2-9x+18eq{0}$ ; $x^2+x-12=0$
does this not grant two answers for $x$? Edit: O wait, I think I got it now. The domain also includes $x-3eq{0}$ which in the end means you do have to factorise both nominator and denominator of the
original equation.
I actually missed out something in my answer and that is that 3 is not in the domain either. | {"url":"http://mathhelpforum.com/algebra/122374-basic-fractional-equation-print.html","timestamp":"2014-04-20T11:53:00Z","content_type":null,"content_length":"14170","record_id":"<urn:uuid:556ca0aa-381c-4941-9027-532f59ed79cd>","cc-path":"CC-MAIN-2014-15/segments/1397609538423.10/warc/CC-MAIN-20140416005218-00198-ip-10-147-4-33.ec2.internal.warc.gz"} |
Vector Direction
A vector contains two types of information: a magnitude and a direction. The magnitude is the length of the vector while the direction tells us which way the vector points. Vector direction can be
given in various forms, but is most commonly denoted in degrees. Acceleration and velocity are examples of vectors.
Vectors, Vectors are any unit that have both magnitude and direction. Well magnitude that's just an amount that could be anything mass, density, volume et cetera but direction that's pretty specific
right you've got to be doing you know east or north or west or someone else could talk about angles 90 degrees, 0 degrees 7, 270 degrees.
So let's look at some examples of vectors. Both acceleration and velocity are often used as vectors, so here's a vector and let's say this vector is 20 meters per second okay? Well again that's the
magnitude but the direction is east okay? Or I might have the same vector going in the north direction 20 meters per second north or 20 meters per second west okay or south. Another way we can depict
this is if we're not talking about direction we might be talking about angles, so if we're heading along the x axis we're going to call that 0 degrees okay and then going clockwi- I'm sorry counter
clockwise, we're going to be going up to where, what would be the y axis would be 20 meters per second at 90 degrees okay, where it would be going in the westerly direction where we would say it's
180 okay and then going in the downward direction would be 270 degrees.
So these are some of the ways that we depict vectors, they have both magnitude and direction.
vector direction magnitude velocity | {"url":"https://www.brightstorm.com/science/physics/linear-and-projectile-motion/vector-direction/","timestamp":"2014-04-16T10:22:15Z","content_type":null,"content_length":"62137","record_id":"<urn:uuid:18c89bf7-faf3-418c-93d0-5fc45d74fc8d>","cc-path":"CC-MAIN-2014-15/segments/1398223206672.15/warc/CC-MAIN-20140423032006-00294-ip-10-147-4-33.ec2.internal.warc.gz"} |
normal distribution
Author Message
normal distribution [#permalink] 07 Dec 2009, 17:08
adam15 E
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Joined: 23 Nov 5% (low)
Question Stats:
Posts: 5
Followers: 0
(03:00) correct
Kudos [?]: 0 [0]
, given: 5 0% (00:00)
based on 2 sessions
Two different groups of test-takers received scores on the GXYZ standardized test. Group A's scores had a normal distribution with a mean of 460 and a standard deviation of 20. Group
B's scores had a normal distribution with a mean of 520 and a standard deviation of 40. If each group has the same number of test-takers, what fraction of the test-takers who scored
below 440 belonged to Group B?
Re: normal distribution [#permalink] 07 Dec 2009, 20:46
This post received
Joined: 29 Jul KUDOS
1/9 I think. This could be kinda tricky even if you know the normal distribution, bc of the estimation. I did it this way. 1 std deviation away to the left would be all the people
Posts: 226 who got less than 440 in group A, which means 14% (1 std below)+2% (2 std below)=16% in the Group A. Now for Group B, it's two std deviations to the left. So about 2% in group B.
Assume there are 100 people in each group. 16% of Group A would give you 16 people below 440, and 2% of group B would give you 2 people in Group B who scored below 440.
Followers: 2
To find the fraction of students below 440 from B ====> Group B/(Total)
Kudos [?]: 33 [1
] , given: 6 ===> 2/(2+16)
Reduces to 1/9
Re: normal distribution [#permalink] 09 Dec 2009, 02:30
GMAT Instructor
This post received
Joined: 24 Jun KUDOS
There is no way to answer this question without consulting a stats table. Since that's impossible during a GMAT, it's not a realistic question. I know that the source gives an OA of
Posts: 967 B for this question, but that is the wrong answer. While you don't need to know this for the GMAT, when data is normally distributed, approximately 68.3% of values are within one
standard deviation of the mean, and approximately 95.4% are within two standard deviations of the mean. If you round these values off to 68% and 95%, you'll get the answer given by
Location: the source. This rounding affects the answer, however; if you use exact values, you'll find that the correct answer to this question is closer to 1/8 than to 1/9. It's a very poorly
Toronto designed question, and completely out of the scope of GMAT statistics, so it's best to ignore this question (and similar questions from the same source) and move on to better
Followers: 236
Kudos [?]: 576 [
1] , given: 3 Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.
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amit2k9 Re: normal distribution [#permalink] 17 May 2011, 23:39
VP got the concept now.
Status: There is in A between 440-460 34% in b between 480-520 34%
always something
new !! in A between 420-440 14% in B between 440-480 14%
Affiliations: in A less than 420 2% in B less than 440 2%
Global,eXampleCG hence (14+2) in A the percentage of people
Joined: 08 May and 2 in B the percentage of people having scores <440
thus ratio = 2/ (16+2) = 1/9 B
Posts: 1368
Followers: 9
Visit -- http://www.sustainable-sphere.com/
Kudos [?]: 121 [ Promote Green Business,Sustainable Living and Green Earth !!
0], given: 10
gmatclubot Re: normal distribution [#permalink] 17 May 2011, 23:39 | {"url":"http://gmatclub.com/forum/normal-distribution-87727.html?fl=similar","timestamp":"2014-04-19T07:15:23Z","content_type":null,"content_length":"154366","record_id":"<urn:uuid:2b2310c6-7b8b-4c8b-a547-9a33e417fa11>","cc-path":"CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00314-ip-10-147-4-33.ec2.internal.warc.gz"} |
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complement help
September 23rd 2009, 10:08 PM
complement help
I'm confused about these problems. How would I prove?
$(A \cup B \cup C)^c = A^c \cap B^c \cap C^c$
The complement of the complement of A = A.
I drew a venn diagram, and I can see from the diagram that they are equal. I'm just not sure how to approach this from a proof standpoint.
September 24th 2009, 12:53 AM
For both problems, show containment in both directions. I'll help get you started.
$<br /> x\in (A\cup B\cup C)^c<br />$
$<br /> \Rightarrow xotin A\cup B\cup C<br />$
$<br /> \Rightarrow xotin A\ \text{and}\ xotin B\ \text{and}\ xotin C<br />$
$<br /> \Rightarrow x\in A^c\ \text{and}\ x\in B^c\ \text{and}\ x\in C^c<br />$
$<br /> \Rightarrow x\in A^c\cap B^c\cap C^c<br />$
$<br /> \Rightarrow (A\cup B\cup C)^c\subset A^c\cap B^c\cap C^c<br />$
September 24th 2009, 03:08 AM
Thanks. :) So, would the other side be...
$x \in A^c \cap B^c \cap C^c$
$\Leftarrow x otin (A \cap B \cap C)$
$\Leftarrow x otin A$ and $x otin B$ and $x otin C$
$\Leftarrow x \in A^c$ and $x \in B^c$ and $x \in C^c$
$\Leftarrow x \in A^c \cup B^c \cup C^c$
$\Leftarrow x \in (A \cup B \cup C)^c$
September 24th 2009, 10:49 AM
Not quite.
Remove your second statement, $xotin (A\cap B\cap C)$, as the third follows directly from the first. If you'd like to be precise, maybe replace $xotin (A\cap B\cap C)$ with $x\in A^c\ \text{and}\
x\in B^c\ \text{and}\ x\in C^c$, your fourth statement.
Think about $xotin A\ \text{and}\ xotin B\ \text{and}\ xotin C$. Since $x$ isn't in a single of the three sets, then it cannot be in their union. So your fourth statement should be $xotin (A\cup
B\cup C)$. This then implies $x$ is on the set you want.
I highly suggest writing out a sentence or two justification for each step; because x is in this set..., by this definition..., by this theorem..., ect. These ideas will keep reoccurring
throughout math, but there will not always symbolic logical connections from each step to the next, it your job to convince not only the reader of your proof, but more importantly yourself, that
what you have written down is right.
Hope this helps.
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[SciPy-user] Chirp Z transform
Stéfan van der Walt stefan@sun.ac...
Thu Jan 22 09:05:09 CST 2009
Hi Nadav
2009/1/22 Nadav Horesh <nadavh@visionsense.com>:
> Chirp Z transform is a generalization of the Fourier transform.
> Attached here a module for chirp z transform written by Paul Kienzle and I. We tried to follow scipy's coding-style directions. Is it possible (and how) to make it a part of the scipy project?
Thanks for working on this; I, for one, would like to see it in SciPy.
Recently I referred you to another implementation at
Your version is much more complete, but the following struck me as
slightly strange:
data = np.random.random(10000)
a = czt.czt(data, w=np.exp(-2*1j*np.pi/float(len(data)))
b = chirpz_s.chirpz(data, 1, np.exp(-2*1j*np.pi/float(len(data))), len(data))
target = np.fft.fft(data)
err_a = np.sum(np.abs(a - target))
err_b = np.sum(np.abs(b - target))
In [152]: err_a / err_b
Out[152]: 1.6138562461610748
The only reason I mention this is because you speak about the
inaccuracy in the docstring. The errors are, on average, in the
vicinity of 1e-10 vs. 5e-11 respectively, so I'm probably on a wild
goose chase.
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Polynomials - Microsoft Research
Understanding Polynomials
A catalog of the types of shapes generated by polynomials of various orders in one, two and three dimensions.
Computer graphicists model shapes with polynomials. Simple linear polynomials generate flat polygons; higher order polynomials generate curved lines and surfaces. In order to develop algorithms for
modeling and rendering such shapes we must have an intimate understanding of how the coefficients of the polynomials relate to the geometric shapes they generate. The answers to various geometric
questions are expressed in terms of new polynomials built out of the coefficients of the original polynomial. The problem is that, for fairly simple shapes, these new polynomials can get incredibly
complicated. For example, the condition that a cubic curve has a double point results in a test expression that is a 12^th order polynomial in the cubic’s coefficients, and this test expression has
over 10000 terms! Fortunately there is a better way to deal with this problem. The research we are doing here builds on a graphical notation technique that was originally proposed by Sylvester and
Clifford over 100 years ago. We are now translating many of the results of classical invariant theory into this notation. This has led to a much better visualization of the relation between a
polynomial’s shape and its coefficients and has produced new algorithms for solving and factoring polynomials. Our ultimate goal is a new catalog of all the interesting algebraic relations between
polynomial coefficients and their shape, describing curves and surfaces of orders up to 4 or 5 in a way that their processing can be accurately computed by parallel processors such as modern GPU’s. | {"url":"http://research.microsoft.com/en-us/projects/polynomials/default.aspx","timestamp":"2014-04-20T18:59:15Z","content_type":null,"content_length":"11926","record_id":"<urn:uuid:38fb0ca1-298d-426a-9d36-0e883dd9ea88>","cc-path":"CC-MAIN-2014-15/segments/1397609539066.13/warc/CC-MAIN-20140416005219-00591-ip-10-147-4-33.ec2.internal.warc.gz"} |
Reference for explicit calculation of blowups (of ideals) and strict transforms
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.
Can one suggest some references where explicit calculations for blow up technique(along an ideal) and strict transformation is done in different examples?
up vote 1 down vote favorite ag.algebraic-geometry birational-geometry
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Can one suggest some references where explicit calculations for blow up technique(along an ideal) and strict transformation is done in different examples?
Both Hartshorne and Fulton's Intersection theory contain many examples. Look for the terms "Blow up" and "Monoidal transform". There are also some "hands on" examples in Dolgachev's
up vote 4 classical algebraic geometry online book . The standard example is describing a cubic surface as a plane with 6 blown up points; see Hartshorne V.4
down vote
add comment
Both Hartshorne and Fulton's Intersection theory contain many examples. Look for the terms "Blow up" and "Monoidal transform". There are also some "hands on" examples in Dolgachev's classical
algebraic geometry online book . The standard example is describing a cubic surface as a plane with 6 blown up points; see Hartshorne V.4
I think, you'd like to see explicit equations and if possible, a singularity completely solved by blowups. One can take a look at such thinks in Lectures on resolution of singularities by
Janos Kollar (http://press.princeton.edu/titles/8449.html). The book starts with more than ten different methods of how to solve singularities on curves and then continues on resolution of
up vote 3 singularities on surfaces. There you will see specific examples and explicit equations.
down vote
add comment
I think, you'd like to see explicit equations and if possible, a singularity completely solved by blowups. One can take a look at such thinks in Lectures on resolution of singularities by Janos
Kollar (http://press.princeton.edu/titles/8449.html). The book starts with more than ten different methods of how to solve singularities on curves and then continues on resolution of singularities on
surfaces. There you will see specific examples and explicit equations.
I suggest Beauville's Complex Algebraic Surfaces.
up vote 1 down vote
add comment
There are several explicit calculations given in Chapter IV.2 of Eisenbud and Harris' Geometry of Schemes
up vote 1 down vote
add comment
There are several explicit calculations given in Chapter IV.2 of Eisenbud and Harris' Geometry of Schemes
You can try out Hauser's seven short stories (with lots of examples and exercises!) http://homepage.univie.ac.at/herwig.hauser/Publications/
seven%20short%20stories%20on%20blowups%20and%20resolution/seven%20short%20stories%20on%20blowups%20and%20resolution.pdf some more examples are in http://homepage.univie.ac.at/
up vote 0 herwig.hauser/Publications/seventeen-obstacles-97.pdf
down vote
add comment
You can try out Hauser's seven short stories (with lots of examples and exercises!) http://homepage.univie.ac.at/herwig.hauser/Publications/seven%20short%20stories%20on%20blowups%20and%20resolution/
seven%20short%20stories%20on%20blowups%20and%20resolution.pdf some more examples are in http://homepage.univie.ac.at/herwig.hauser/Publications/seventeen-obstacles-97.pdf | {"url":"http://mathoverflow.net/questions/5012/reference-for-explicit-calculation-of-blowups-of-ideals-and-strict-transforms/5657","timestamp":"2014-04-16T14:06:11Z","content_type":null,"content_length":"64180","record_id":"<urn:uuid:6cc21a47-2186-453c-9c62-732911484edb>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00224-ip-10-147-4-33.ec2.internal.warc.gz"} |
Finding Pythagorean Triples
May 21st, 2010, 11:22 AM #1
Junior Member
Join Date
May 2010
Finding Pythagorean Triples
Given a set 'S' of 'N' numbers, find all x, y and z such that x^2 + y^2 = z^2.
One method:
1. I can store the z^2's in a hashmap (put every number squared ^2 into a hashmap). I can do this in time=N with additional space (the hashmap) of N elements.
2. I can then create a nested loop and multiple every number 'i' by every other number 'j' - looking for the result in the hashmap. That'd be N^2 time.
Am I overlooking a "better" way to do this?
And one more question, what particular class or category of algorithms does this problem fall into?
I can leverage HashMap properties to lookup answers but looking for other structures that would help improve this case.
Thanks in advance
Re: Finding Pythagorean Triples
In this specific case you can't do better than N^2, because at the least you need to sum N^2 different pairs:
Number of different pairs out of N numbers = N over 2 = N!/(2!*(N-2)!) = (N-1)*N/2 = O(N^2)
But it is possible to dismiss the hash table and remain with O(N^2).
Re: Finding Pythagorean Triples
Ah - I see. Indeed, this is a Combination. The math helps. No better than N^2.
So, if I were to sort the array first, I'd add n lg n but that would still be < O(N^2). That means that on average, I would only have to look at value larger than the largest value I multiplied -
but it doesn't seem right to me. Although it'd be shrinking as I moved up the indexes, I'm still walking some variable steps to the right on every iteration .... which, while it may likely avg
out to be n/2 steps on each iteration which ... would still bring us to O(N^3)
The hashmap gives me O(1) lookup ... I'm not sure how else I can get that. I must be thinking of it wrong. I'm thinking that I must find a needle in a haystack and either I read all the needles
first and save them into a hashmap or I iterate "over the rest of the needles" to find the correct value on each iteration.
Without actually giving it away, can you nudge me with an analogy or something? A hint at some property I am ignoring maybe? What other kind of problem is this like?
Re: Finding Pythagorean Triples
Try to keep it simple.
- Can you iterate over the pair sums in increasing order ?
- Can you iterate over the original number squares in increasing order ?
- Can you "combine" both iterations described above ?
Re: Finding Pythagorean Triples
Ok - I think I've got it now.
Sort the integers into an array A and then initialize indexes i, j and k with the notion that A[i]^2 + A[j]^2 = A[k]^2.
Furthermore, i represents the outer loop while j represents the inner loop. For each outer loop increment, the inner loop starts j over at i+1 and k over at j+1.
As I multiple A[i] and A[j], I search for the hypotenuse with A[k]. It should be noted that k never moves backward. The integers at i and j get progressively larger such that, at a minimum, k
will move forward one position for each inner loop iteration. In the slowest case, k always ends up at j+1 where A[j+1]^2 > A[i]^2 + A[j]^2.
Another way to say this ... since the array is sorted, A[i]^2 + A[j]^2 are only getting larger and so A[k] is either on the correct integer or must move forward some number of steps until it
finds the integer or steps one position greater than it.
I believe that the cost of actually moving 'k' the max of N-j steps on each round averages out to a constant with respect to N.
So, search = O(nlgn) and then iterate with nested loops O(n^2) including incrementing i, j and k indexes that all move forward only (except for when the outer loop increments in which case they
start over at just above the ith index). Indeed, it turns out to be O(n^2).
This was actually my very very first idea - but I didn't appreciate that k didn't have to keep starting over at j+1 on each check for hypotenuse. I originally posited a third nested loop to walk
through all of the integers looking for the value at k and since said approach is O(n^3) ... I just went the opposite way looking for a better solution - thus, the HashMap.
Hopefully ... this is what you were thinking. If not ... I need to continue searching. Many many thanks for your assistance.
Last edited by MacInTosh; May 26th, 2010 at 12:16 AM.
May 23rd, 2010, 03:26 AM #2
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George Hart :: Geometric Sculptures
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George Hart :: Geometric Sculptures
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"The Susurrus of the Sea," by George W. Hart "Star Corona," by George W. Hart "Eights," by George W. Hart
(www.georgehart.com) (www.georgehart.com) (www.georgehart.com)
"Frabjous," by George W. Hart "72 Pencils," by George W. Hart "Gyrangle," by George W. Hart
(www.georgehart.com) (www.georgehart.com) (www.georgehart.com) | {"url":"http://ams.org/mathimagery/thumbnails.php?album=15","timestamp":"2014-04-17T23:04:18Z","content_type":null,"content_length":"33817","record_id":"<urn:uuid:fb1f8522-3835-46de-acfc-43c90c56abe3>","cc-path":"CC-MAIN-2014-15/segments/1397609532128.44/warc/CC-MAIN-20140416005212-00192-ip-10-147-4-33.ec2.internal.warc.gz"} |
Time Space Matter Characteristics
Hey All,
Maybe it is a dumb question, I donno, but do time, space and matter (the building block of the known universe) SHARE properties or are they COMPLETELY independent but interwoven with each other? For
example, like the Union of three sets in Set Theory. Three independent entities combined to each other.
Tank you, | {"url":"http://www.physicsforums.com/showthread.php?s=16db9a4951cb69f98818fb5b80c8bf5d&p=4313694","timestamp":"2014-04-25T08:20:33Z","content_type":null,"content_length":"27137","record_id":"<urn:uuid:a3d97e09-5784-4694-9431-a18fcc7b6b49>","cc-path":"CC-MAIN-2014-15/segments/1398223211700.16/warc/CC-MAIN-20140423032011-00203-ip-10-147-4-33.ec2.internal.warc.gz"} |
Topic 3: Fractions
Try the problem below. After you try the problem, you can watch a video to see a correct solution.
Xu Li and Elsa received a box of chocolates as a gift. Xu Li ate \(\frac{2}{7}\) of the box of chocolates and Elsa ate \(\frac{3}{5}\) of the same box of chocolates. What fraction of the box of
chocolates is left to serve to guests?
After watching the video, try this problem to practice more.
James and Gary bought a pizza pie for dinner. James ate \(\frac{1}{6}\) of the pie and Gary ate \(\frac{2}{9}\) of the same pizza pie. How much of the pizza pie is left for the rest of the family? | {"url":"http://areyoureadyformath.com/help-videos/fractions/","timestamp":"2014-04-16T19:09:10Z","content_type":null,"content_length":"37775","record_id":"<urn:uuid:03ed3a75-1cee-4661-9c47-e2826eafaebe>","cc-path":"CC-MAIN-2014-15/segments/1397609524644.38/warc/CC-MAIN-20140416005204-00275-ip-10-147-4-33.ec2.internal.warc.gz"} |
Trapezium problem
September 12th 2005, 07:09 AM
Trapezium problem
Look at the drawing and declare the x and y. The intersection is the point we get if we lenghten the c and d as far as they intersect.
The final task is to calculate the area of the formed triangle.
September 12th 2005, 07:56 AM
Ok, there is an easy way : mathematically the area of the triangle CDE is the same in two figure (figure 1 and figure 2) because you see the two blue areas (figure 3) are the same so it does not
change anything to the area of CDE if we replace DE without changing its length.
So I'll work with the second figure : we have two triangles ABC and CDE which have a common angle C and having DE parallel to AB, the angle CDE is the same as CAB so the two triangle have
proportional sides. (the relation between the two triangles has a name but I just know it in french. sorry)
So all homologous sides have the same proportion k = m(AB)/m(DE) = m(CB)/m(CE) = m(CA)/m(CD) and we have one of them which is m(AB)/m(DE) = 1.8/1.2 = 1.5 so k=1.5= m(CB)/m(CE) = m(CA)/m(CD).
We replace what we know : k=1.5= 1.5/m(CE) = 1.2/m(CD). So m(CE) = 1 and m(CD) = 1.2/1.5=0.8 So we have all our sides.
Now you can use either the Heron formula or draw a height and calculate A(triangle CDE) = h*b/2
By Heron : the semiperimeter is p = (0.8+1+1.2)/2 = 1.5 and A(triangle CDE) = sqrt(1.5 (1.5-1) (1.5-0.8) (1.5-1.2)) = 0.397 units^2.
September 12th 2005, 09:42 AM
CD=2,4 CE=3 or contrary. You have a mistake on your solution.
September 12th 2005, 11:12 AM
Sorry, I misunderstood your figure. I thought m(AC) = 1.2 when it was m(AD)=1.2.
"Ok, there is an easy way : mathematically the area of the triangle CDE is the same in two figure (figure 1 and figure 2) because you see the two blue areas (figure 3) are the same so it does not
change anything to the area of CDE if we replace DE without changing its length.
So I'll work with the second figure : we have two triangles ABC and CDE which have a common angle C and having DE parallel to AB, the angle CDE is the same as CAB so the two triangle have
proportional sides. (the relation between the two triangles has a name but I just know it in french. sorry)
So all homologous sides have the same proportion k = m(AB)/m(DE) = m(CB)/m(CE) = m(CA)/m(CD) and we have one of them which is m(AB)/m(DE) = 1.8/1.2 = 1.5 so k=1.5= m(CB)/m(CE) = m(CA)/m(CD).
We replace what we know : k=1.5= (1.5+m(CE))/m(CE) = (1.2+m(CD))/m(CD). So 1.5m(CE) = 1.5+m(CE) so m(CE) =3 and 1.5m(CD) = 1.2+m(CD) so m(CD) = 2.4. So we have all our sides.
Now you can use either the Heron formula or draw a height and calculate A(triangle CDE) = h*b/2
By Heron : the semiperimeter is p = (1.2 + 3 + 2.4)/2 = 3.3 and A(triangle CDE) = sqrt(3.3 (3.3-1.2) (3.3-2.4) (3.3-3)) = 1.378 units^2. | {"url":"http://mathhelpforum.com/geometry/879-trapezium-problem-print.html","timestamp":"2014-04-18T20:43:24Z","content_type":null,"content_length":"6666","record_id":"<urn:uuid:204a5b7b-839a-4784-96de-1e427aab0f02>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.9/warc/CC-MAIN-20140416005215-00043-ip-10-147-4-33.ec2.internal.warc.gz"} |
Homework Help
Posted by Aj on Tuesday, January 16, 2007 at 4:55pm.
Oregon is about 400 miles from west to east, and 300 miles from north to south. If a map of Oregon is 15 inches tall (from north to south), about how wide is the map?
Set up a proportion:
15/300 (the ratio of inches to miles, north to south)
and x/400 (the ratio of inches to miles, west to east)
15/300 = x/400
Can you solve from there?
not really
i'm confused
What kind of math do you need help with like decimals or what.
Are you needed with most tipes of math or just some tipes of math
• math - a dude on earth who u will never know, Tuesday, November 30, 2010 at 8:46pm
when you set up a proportion 15/300 it can be reduced so 15/300 will become 1/20. 20x20= 400 so 1x20=20.
the answer is 20
your welcome Aj and people of earth
btw im a 7th grader
• math - Brittany, Tuesday, November 30, 2010 at 9:00pm
if your wondering who a dude on earth who u will never know is its me i got bored but your welcome for the anwser Aj now your not confused anymore your welcome =) =) =)
im a random person who likes cheese, fat bunnys,animals,Julies S. dog coco,my best buddy ever= Annabelle W.,Some people,my dogs,dogs,cookies,basketball=),OHIO!,TEXAS!,CALIFORNIA!,and the 2
hottest celebs Jared Leto!!, and Taylor Launter!!
That will be all =)
THE END
GO NINJA PENGUINS!
IM SOO WEIRD
NINJA PENGUINS=AWSOMEST THING EVER!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!K BYE BYE
PERSON WHO I DONT KNOW OR MAYBE KNOW I DONT KNOW B/C IM WEIRD!!
PIP PIP CHEERIO TA TA PEACE OUT!!!!
• math - noone, Tuesday, February 8, 2011 at 10:15pm
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Co-ordinate systems and velocities
If you have any vector in Cartesian coordinates and you want to express it in Cylindrical coordinates then you use:
r = sqrt(x²+y²)
theta = atan(y/x)
z = z
If you have any vector in Cylindrical coordinates and you want to express it in Cartesian coordinates then you use:
x = r cos(theta)
y = r sin(theta)
z = z
It makes no difference whatsoever what the units are nor whether the vector represents position, acceleration, velocity, momentum, etc.
The only trick comes if you have a position as a function of time in Cylindrical coordinates and you want to find the velocity or acceleration as a function of time in the Cylindrical coordinates
without converting to and from Cartesian. Then you use the formulas I linked to above.
So radial velocity would be vr = sqrt(vx^2+vy^2). However, for an orbiting body, radial velocity is largely zero, which with that equation would imply that vx and vy are both largely zero, which is
not the case.
Why are there sign functions in your expression?
To find velocities you need only differentiate withe respect to time:
[tex]v_x\equiv\frac{dx}{dt}=\frac{d}{dt}(r\cos\theta)=\frac{dr}{dt}\cos\thet a-r\sin\theta\frac{d\theta}{dt}=v_r\cos\theta- v_{\theta}\sin\theta[/tex]
To get [itex]v_y[/itex] and [itex]v_z[/itex] you just do the same thing to your expressions for [itex]y[/itex] and [itex]z[/itex]
To get [itex]v_r[/itex] and [itex]v_{\theta}[/itex] in terms of Cartesians you just differentiate the expressions for [itex]r[/itex] and [itex]\theta[/itex] with respect to time:
[tex]v_{\theta}\equiv r\frac{d\theta}{dt}=r\frac{d}{dt} \tan^{-1}\left(\frac{y}{x}\right)=\sqrt{x^2+y^2}\frac{d}{dt} \tan^{-1}\left(\frac{y}{x}\right)[/tex]
OR you solve the 3 equations you got for v_x, v_y and v_z for v_r, v_theta and v_z the same way you would solve any system of 3 equations for 3 unknowns.
The signs are because x and y become negative and the signs on the velocities need to be right.
Sorry guys if I'm missing the point or I'm not explaining it well enough. Like I said, I really suck at angle conversions and co-ordinate systems etc.
To try to explain things better, I'm attaching a sketch of what I'm trying to do:
I basically start with information in the form x,y,z,vx,vy,vz. I then have to calculate a force for which the equations work in azimuthal and radial values and depend on the azimuthal and radial
velocities (in the same units as vx,vy,vz). So I have to first convert vx and vy into rotational velocity and radial velocity. Then the resulting force has to be converted back to the force in x,y,z.
So, vx,vy,vz,v_azimuthal,v_radial all have the same units (i.e. m/s). Theta is calculated by using y/x but since y/x changes sign depending on which quadrant it is in, I've added those sign modifiers
to try to make things work.
: Right, I think I've worked out the formulae that I need so I'll post them up on here in case anyone else finds this via search and is as similarly lacking as I am in the angles and conversions
theta = atan(y/x)
v_x = abs(v_radial*cos(theta))*sign(x) + abs(v_rotational*sin(theta))*sign(Cy)
v_y = abs(v_radial*sin(theta))*sign(y) - abs(v_rotational*cos(theta))*sign(Cx)
v_radial = abs(v_x*cos(theta))*sign(x) + abs(v_y*sin(theta))*sign(y)
v_rotational = abs(v_x*sin(theta))*sign(Cy) - abs(v_y*cos(theta))*sign(Cx)
where abs() denotes the absolute value (or modulus) of a number (i.e. -1.5623 -> 1.5623 and +1.2479 -> 1.2479) and sign() gives the sign of a number (i.e. 1.2435 -> 1 and -1.7829 -> -1).
C is a sign factor depending on the direction of rotation. For anticlockwise motion, C is -1, for clockwise motion, C is 1. This is for a system where a body is rotating around another, which is at x
= y = 0.
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I’m liveblogging from the Taj Mahal
No particular news to report — it’s about the same as it was 400 years ago, I guess. I just wanted to liveblog from the Taj Mahal, is all. (Jonathan Walgate is the one who suggested it.). Now I’ll go
back to looking at it.
Greg Egan Says:
Comment #1 December 22nd, 2007 at 7:29 am
Come on Scott, we’re not going to believe you’re at the Taj Mahal from four lines of text, with not so much as a single descriptive phrase. Where are the pictures?
Or better yet, a short YouTube video of you in a dance routine with 1,000 Bollywood Extras.
huh Says:
Comment #2 December 22nd, 2007 at 10:36 am
they have wireless there?
Scott Says:
Comment #3 December 22nd, 2007 at 11:24 am
My Blackberry worked there. (Which was surprising, since it didn’t work in downtown Agra or Dayalbagh just a few miles away. That Shah Jahan thought of everything.)
Sarang K Says:
Comment #4 December 22nd, 2007 at 12:01 pm
Hey Scott,
Are you in India? Any plans to visit Indian SiO2 valley a.k.a. B’lore?
Please reply me at the above email, I would be grateful to have a chance to talk to you.
Scott Says:
Comment #5 December 22nd, 2007 at 2:18 pm
Sarang: Yes, I’m in India (as you know, that’s where the Taj Mahal is
Bilal Shaw Says:
Comment #6 December 22nd, 2007 at 5:48 pm
What’s a trip to the Taj without a photograph? Talk about love! That Shah Jahan really knew how to leave his mark.
Scott Says:
Comment #7 December 22nd, 2007 at 10:55 pm
Bilal: Don’t worry, photos of one of the world’s most-photographed landmarks are a-comin’ — as soon as I’m able to access the Internet from something more than three inches across.
milkshake Says:
Comment #8 December 23rd, 2007 at 2:20 am
I think Taj needs a livecam. Or a radio beacon (like they do with migratory animals) to follow its progress on the map in real time
Cheshire Cat Says:
Comment #9 December 23rd, 2007 at 9:19 am
The Taj is ineffable. But for want of something better, Andre Weil’s description will have to serve:
“bastardized offspring of Italian Baroque grafted on to the ostentatious whims of a Mughal despot”
D'Andray While Says:
Comment #10 December 23rd, 2007 at 5:09 pm
Forget the landmarks. Does P=NP?
Job (UG) Says:
Comment #11 December 23rd, 2007 at 7:27 pm
It does, but it’s not provable – hence the landmarks.
harrison Says:
Comment #12 December 23rd, 2007 at 9:51 pm
I disagree. P != NP, but it’s not provable. If Scott could prove it, he’d have the money to build his own Taj Mahal. (Or at least a 1/1000 scale model or something).
Job (UG) Says:
Comment #13 December 24th, 2007 at 3:11 am
Scott would have have to be really selfish to intentionally prove that P != NP. His fame and fortune in exchange of everyone’s happy hours of thoughts entertaining P=NP. And all of that to no one’s
Job (UG) Says:
Comment #14 December 24th, 2007 at 3:20 am
God won’t allow it.
Scott Says:
Comment #15 December 24th, 2007 at 4:25 am
Does P=NP?
(To answer the inevitable next question: no, I can’t prove it. I also can’t prove that crap doesn’t smell like violets. But had you asked me if it does, I would also have answered no.)
John Sidles Says:
Comment #16 December 24th, 2007 at 10:49 am
Scott says: Does P=NP? No.
Scott, since you are traveling in the east, perhaps you might (temporarily) adopt the far-east POV that “Does P=NP? ” is not properly a question, but a koan?
That is, a question whose best answer embodies an understanding of why it is not quite the best question to ask.
In which case, what is (in your view) the best question to ask?
Might the right question be as concise as “Does P=PPAD?” (or something similar)?
A thread of “Questions that are at least as much fun as ‘Does P=NP?’” would IMHO be very enjoyable.
cody Says:
Comment #17 December 24th, 2007 at 12:12 pm
i recently heard (on audiobook), Ian Stewart’s essay the mathematics of 2050, from the next fifty years essays, and he guessed that P vs NP will be proven formally undecidable by then (which i had
never even considered). but i got to thinking about it more, and i realized i dont really know if he means undecidable like the halting problem, or undecidable like CH (which i now see might more
properly be called independent). and that got me wondering if computational complexity is built on axioms, and if so what are they. CC seems somehow “more bound” to reality than pure mathematics, and
so it seems as if P vs NP were independent, then an additional axiom would be in order. at the same time, i cant really picture it being undecidable like the halting problem, though i cant say why.
which now has me wondering: do we ever examine the “space complexity” or anything similar of proofs? maybe these are silly questions, im really more of a low-level physics guy than mathematician or
computer scientist.
John Sidles Says:
Comment #18 December 24th, 2007 at 12:29 pm
Cody, being a “physics guy” like you, may I say that the literature on the complexity of search problems seems more natural—or at least, more physically motivated—than the literature on the
complexity of decision problems?
The two categories of problems are of course closely related, but they are not the same. That’s my limited understanding anyway … perhaps others have helpful comments?
cody Says:
Comment #19 December 24th, 2007 at 12:47 pm
ah, yes, i think i see what you mean. would the relationship between search problems and decision problems be sort of the relationships between FP and P? (or any other functional version of a
decision problem).
John Sidles Says:
Comment #20 December 24th, 2007 at 12:57 pm
Cody asks: Would the relationship between search problems and decision problems be sort of the relationships between FP and P? (or any other functional version of a decision problem?
Speaking personally, questions like yours make me feel like the farmer in that old Iowa saying: “I feel like a hog being shown a wristwatch.”
You’ve asked a great question, so maybe a lurking complexity expert will tackle it?
Scott Says:
Comment #21 December 24th, 2007 at 1:12 pm
John: P vs. NP is of course just the “canonical representative” of a very large set of questions (P vs. BQP, P vs. PSPACE, NP vs. P/poly, FP vs. PPAD, NP vs. DTIME(n^polylog), existence of OWF’s,
etc.) that we’d like to answer but can’t. Personally, I don’t necessarily see it as more important than the rest; I think people just focus on it for concreteness (similarly to how mathematicians
talk about the Riemann Hypothesis, even though many of the implications actually require stronger versions like ERH).
In the case of P vs. NP, what justifies focusing on decision problems is their well-known equivalence to search problems. (That is, P=NP iff FP=FNP.) For the TFNP classes (PPAD, PPP, etc.), one has
to talk about search problems, so that’s exactly what people do.
roland Says:
Comment #22 December 24th, 2007 at 1:48 pm
CH being independent means that number theory does not imply CH or not CH. In contrast every TM either halts or goes on forever but no TM decides this question.
The definition of P and NP does not need any special axioms.
P=NP being independent of number theory would mean that either
a) There is an algorithm that solves SAT in polynomial time but its correctness and/or running time has no proof in the number theory framework.
b) There is no such algorithm and no number theoretic proof for this fact.
I don’t exactly understand what you mean by space complexity of proofs.
John Sidles Says:
Comment #23 December 24th, 2007 at 1:48 pm
Scott says: P=NP iff FP=FNP
Thanks Scott, not only for your thoughtful response to my specific question, but especially, for all the trouble you take to run a forum that is humorous, good-natured, and thought-provoking.
There aren’t too many on-line forums that have all of these features, which IMHO are very valuable to the community (and I am sure that many people feel this way).
WIth regard to “P=NP iff FP=FNP”, that very assertion is the subject of a thought-provoking (to me) standalone appendix to Jiri Hanika’s thesis Search Problems and Bounded Arithmetic.
The title of Hanika’s appendix is “Three Myths About Function Problems”. The three myths are called (by Hanika) “Reducibility to Decision”, “Totality Makes [a?] Difference”, and “Decisions are
When I say the appendix is “thought-provoking” what I mean is “it reassured me that I was not the only person to whom these issues were non-obvious”.
That doesn’t mean I understand these issues … far from it. But am I right in guessing that they would be good things to understand better?
Job Says:
Comment #24 December 24th, 2007 at 2:23 pm
Here’s something i don’t understand entirely, if a function f(x) = {0,1} alternates infinitely between chunks (of varying length) that are computable with polynomial lower bounds (of varying degrees)
and chunks that are computable with exponential lower bounds, then what is the lower bound of an algorithm computing f(x), polynomial or exponential?
cody Says:
Comment #25 December 24th, 2007 at 5:13 pm
roland, do you think that Ian Stewart meant P vs NP would be proven independent of number theory then? and also, is computational complexity theory founded on number theory? or just ZFC (is there a
difference)? and if that were the case, couldnt we “choose” whatever answer we would like, for P vs NP and make it a new axiom? as i understand it, most mathematicians work under the assumption that
CH is false, though not all, and that it is up to them.
the idea of “space complexity” of proofs isnt necessarily a coherent one; it is a curiosity of what might result from looking at the number of steps and amount of information required to prove
something, (though i do understand that proofs are highly arbitrary).
i guess really im wondering if anything useful could result in applying computability and complexity sort of ideas to proof sort of problems, though it feels like a dumb question because proofs seem
like such radically different problems than algorithms. clearly im beyond my comprehension.
Scott, i just noticed i can resize this comment window, thats a neat little touch.
cody Says:
Comment #26 December 24th, 2007 at 5:20 pm
oh, i meant to say that i am comfortable with arbitrarily rejecting CH or not, because in pure mathematics you can choose what sort of system you are studying. my view is that in math, you define
your axioms and see what sort of world results, as where in physics you are handed a set of unknown axioms and the resulting world, and we are trying to work backwards. P/NP independent of ZFC would
seem weird because P and NP seem somehow more grounded in reality, as if they were handed to us along with electric charge and gravity, so it seems silly to say we could choose to work with either
Scott Says:
Comment #27 December 25th, 2007 at 2:37 am
John, I took a look at Hanika’s thesis. He’s correct that the reduction from search to decision problems is necessarily a Turing reduction. But we can perform Turing reductions in P — so if all we
care about is the P vs. NP question, then we don’t have to talk about function problems.
But as I said before, sometimes you do have to talk about function problems — and when you have to, you do. This is not a deep ideological issue, and there’s really not much to understand.
Scott Says:
Comment #28 December 25th, 2007 at 2:57 am
Cody, my intuition about CH vs. P≠NP accords with yours. When people throw around the idea that P≠NP is independent, they don’t mention that in the whole history of mathematics, we’ve never once seen
a “natural” question phrasable in terms of Turing machines and whether they halt proven independent of set theory.
(The Gödel sentence is not “natural,” in the sense that it’s specifically constructed to be independent. The Continuum Hypothesis is not phrasable in terms of Turing machines and whether they halt.
The Paris-Harrington example is only independent of arithmetic, not of set theory.)
On the other hand, in the history of mathematics we’ve seen many, many examples of questions that were answerable, but that were asked centuries or millennia before the tools existed to answer them.
That’s an extremely well-known phenomenon.
So I’d think the “default” conjecture should be that P≠NP, that there’s indeed a proof, and that mathematics simply hasn’t yet advanced to the point where we can find it.
(If you want to know more about independence and P vs. NP, see this survey paper I wrote a while back. Unfortunately, nowhere in the paper did I bother to state my own opinion, and some people
misinterpreted me to mean I actually think independence is a serious possibility.)
cody Says:
Comment #29 December 26th, 2007 at 1:48 am
thanks for the point to the paper. i find these things very interesting, but am not really comfortable
cody Says:
Comment #30 December 26th, 2007 at 1:49 am
oops, i mean to say, but im not really comfortable speculating on these things myself.
Peter Sheldrick Says:
Comment #31 December 27th, 2007 at 9:40 am
cody says: i guess really im wondering if anything useful could result in applying computability and complexity sort of ideas to proof sort of problems, though it feels like a dumb question because
proofs seem like such radically different problems than algorithms.
The way i understand it, is that an algorithm is basically constructing a mathematical object. So in constructive mathematics you always prove that a mathematical obejct exists by constructing it –
which gives you a way to “make” that object “by hand” (e.g. an algorithm). In other words
algorithm = proof of existence of something (in con. math)
John Sidles Says:
Comment #32 December 27th, 2007 at 11:02 am
cody says: i guess really i’m wondering if anything useful could result in applying computability and complexity sort of ideas to proof sort of problems?
That is a fine topic for an end-of-year “quantum confection”.
The old-school answer is “no” and the reason is aesthetic. As Chandrasekhar phrased it “The simple is the seal of the true” and “Beauty is the splendour of truth”. The implication being, that
proof-related results originating in complexity theory may well be formally true, but results derived by this path will in general lack the simplicity and beauty that Chandra prized.
The new-school answer is “yes”, and Kurt Gödel’s incompleteness proofs are of course the shining example. But this doesn’t mean the old-school thinking is wrong. As Scott noted, Gödel sentences lack
Chandra’s “splendour of simple truth.”
The post-modern view might be “yes-and-no”, with the aesthetic the point being that in the real world, truths are always embedded within computational ecosystems, such that the full meaning of a
truth is illuminated only in light of its informatic embedding.
Let’s construct an example of a post-modern kind of question. We suppose that Alice is running an error-corrected quantum computation in a finite-dimensional Hilbert space and Bob is doing the same.
But without knowing it—and this is the eco-informatic “catch”—Alice and Bob are operating in the same physical Hilbert space.
This sharing is not immediately apparent to either Alice or Bob, because Bob’s qubit basis basis is random relative to Alice’s. In consequence, Bob’s computation looks like noise relative to Alice,
and vice versa.
As a toy problem, we ask “Is quantum error-correction possible within this reciprocal computational ecosystem?” Just for fun, I coded up a toy computation, and obtained the surprising (to me) result
that Alice and Bob’ respective computations are fratricidal … each destroys the other.
For example, if Bob implements an computation as simple as a single pi-pulse on a single Bob qubit, this one operation appears in Alice’s frame as a randomizing POVM that completely “mixmasters” her
entire computation, beyond any hope of error correction.
Does this toy problem have any interest at all? Well, the converse is interesting: if Alice’s computation succeeds, this is convincing evidence for Alice that Bob does not exist in her universe (or
if he does, is not carrying out computations).
So this toy problem helps us (well, me at least) understand better why we perceive only one classical universe … it’s because the shared universes all committed quantum fratricide!
Now, where is Chandra’s “splendour of simple truth” in all this?
Dallas Fortworth Says:
Comment #33 December 27th, 2007 at 11:23 am
Why is the attribute “simple” considered to be the complete equivalent of the attribute “beautiful”? They are two totally different characteristics. Is a beautiful woman a simple woman? In addition,
the predicate “elegant” is different from the properties “simple” and “beautiful.” Is such careless talk accepted and condoned in mathematics? If so, it should be consciously recognized as such.
John Sidles Says:
Comment #34 December 27th, 2007 at 1:48 pm
Dallas Fortworth said: Why is the attribute “simple” considered to be the complete equivalent of the attribute “beautiful”? … is such careless talk accepted and condoned in mathematics?
Subrahmanyan Chandrasekhar (for physicists there is only one “Chandra”) wrote a book on this precise topic, entitled Truth and Beauty: Aesthetics and Motivations in Science.
Without offering an opinion on whether Chandra’s book is “true”, it is surely true that its philosophy motivated and guided at least one person—namely Chandra himself—to do some outstanding physics
and mathematics.
Dallas Fortworth Says:
Comment #35 December 27th, 2007 at 11:21 pm
That is a good pragmatic definition of beauty. Whatever results in valuable work is true and beautiful, then. Maybe it’s even elegant (von Neumann), pretty (Oppenheimer), tasteful, fine, and nice.
Chandrasekhar’s authority must be overwhelming. Did he have as much success in his expensive book on Newton? In it, he tried to convert Newton’s idiosyncratic, pseudo-mathematics into differential
John Sidles Says:
Comment #36 December 28th, 2007 at 9:29 am
Dallas Fortworth sez … Whatever results in valuable work is true and beautiful, then.
Well, I dunno. Assembly language code is beautiful? Manure shoveling is beautiful? More provocative IMHO is the idea that new forms and ideals of beauty become apparent in every century.
Supposed we adopt for the moment Chandra’s categories of “basic science” and “derived science” (while admitting that the exact boundary between basic and derived science is somewhat indistinct).
Then last century saw a lot of beautiful “basic science” that (for example) united variational principles, path integrals, symmetries, and symmetry-breaking; these ideas added up to the Standard
But as Feynman foresaw, this flowering of basic science was a one-time event:
“We are very lucky to live in an age in which we are still making discoveries. It is like the discovery of America—you only discover it once. The age in which we live is the one in which we are
discovering the fundamental laws of nature, and that day will never come again. It is very exciting, it is marvelous, but this excitement will have to go. Of course, in the future there will be
other interests. There will be the interest in the connection of one level of phenomena with another—phenomena in biology and so on, or, if you are talking about exploration, exploring other
planets, but there will not still be the same things that we are doing now.”
What comes next? Well, it would be cool—which is to say “beautiful”—if in the twenty-first century, one percent of humanity (say) could find employment as scientists or mathematicians.
This will require creating on the order of 100 million new scientific jobs in this century, which is about 3,000 new scientific jobs per day … so we had better get busy!
There definitely are scientific enterprises that potentially “scale” to this immense size: the sky surveys and the genome surveys come to mind. These are examples of what Chandra called “derived”
It is uniquely characteristic of our new century that humanity has begun to undertake these derived enterprises on an global scale … a necessity that John von Neumann foresaw.
And who is better equipped, by training and talent, to think creatively about these global enterprises, and to provide the fundamental tools for undertaking them, than complexity theorists?
That concludes my optimistic New Years’ Essay … whose main point is that the scientific community has in-hand both urgent challenges and interesting questions. Perfect!
Coturnix Says:
Comment #37 January 1st, 2008 at 9:51 pm
Scott, please e-mail me ASAP, about the science blogging anthology
Bora aka Coturnix
A Blog Around The Clock
Zelah Says:
Comment #38 January 2nd, 2008 at 8:08 am
Hi Everyone,
First a Happy New year!
Secondly, a couple of questions:
1. If RP = NP does this imply P=NP?
2. If RP = NP does this imply that RP has non zero p measure?
(I believe that there is an Zero One theroem that states that if RP has non zero p-measure then ZPP=EXP, but that is for another day!)
Thanks in advanced
Scott Says:
Comment #39 January 2nd, 2008 at 9:38 am
1. Not known. (But probably true, since we believe P=RP.)
2. Not known either. Since P has p-measure 0, that implication would imply (RP=NP implies P≠NP). Which not only is unknown, but goes in the opposite direction from what we’d expect!
B S Gupta Says:
Comment #40 January 19th, 2008 at 5:28 pm
I saw Mr Kohli comments somewhere else -Mr Kohli is right that you were lucky to have Dr Prem Saran Satsangi during your lecture at Dayalbagh-what you try to compute ,probate and prove- A PARTICLE OR
A STRING HERE AS WELL THERE AND PROBABLY NOWHERE AND POSSIBLY EVERYWHERE -He sees it directly-he has described the phenomenon of `Para’( Science of Other world) and Apara( of this Material world)-It
would be worthwhile to read his following publication ,to know something about His Superhuman mind and a most wonderful humanbeing:-
British Library Direct: Systems movement:Autobiographical …Systems movement:Autobiographical Retrospectives. Author. Satsangi, PS. Journal title. INTERNATIONAL JOURNAL OF GENERAL SYSTEMS …
direct.bl.uk/research/16/3C/RN190205690.html – Similar pages – Note this
Scott Says:
Comment #41 January 19th, 2008 at 6:17 pm
To the people repeatedly leaving comments about how lucky I was to be graced by the Guru’s “superhuman mind”: do you realize just how ridiculous this sort of talk sounds to outsiders? | {"url":"http://www.scottaaronson.com/blog/?p=301","timestamp":"2014-04-19T04:19:11Z","content_type":null,"content_length":"39687","record_id":"<urn:uuid:e500a0c5-5b62-4bf9-a929-3746c28cca9b>","cc-path":"CC-MAIN-2014-15/segments/1397609535775.35/warc/CC-MAIN-20140416005215-00209-ip-10-147-4-33.ec2.internal.warc.gz"} |
doing a research on aberrations in order to
Hello everyone,
I am doing a research on aberrations in order to prepare a presentation. I looked at Optics, Hecht and Introduction to Optics, Pedrotti; however, I think they use different conventions for
aberrations. Is there an established standard for aberrations(I heard the name Seidel)? I also wonder what the third order aberration means; is it sine expanded to include angle cubed or the
aberrations are written up to 3rd order of the aperture diameter for example. I am pretty confused about this topic and in need of a good resource which will guide me and possibly derive some of the
aberration formulas explaining each one in detail. Any suggestion is appreciated.
The Seidel aberrations are referred to as 'third order' because the expansion sin(q) = q + q^3/3! + q^5/5!+... has a third-order polynomial after the paraxial sin(q) ~ q approximation. There are 7
third order aberrations (piston, tilt, spherical, coma, petvzal, distortion, astigmatism), a bunch of 5th order, 7th order, etc.
Alternatively, the wavefront aberration is written in terms of Zernike polynomials- mapping a Zernike coefficient with a Seidel aberration is not possible, but there are ways to convert one to the
A good resource (free, etc.) is the MIL-HDBK 141 (Optical Design)
3rd order aberrations are in Chapter 8. | {"url":"http://www.physicsforums.com/showthread.php?p=3811024","timestamp":"2014-04-20T00:52:33Z","content_type":null,"content_length":"34985","record_id":"<urn:uuid:4060aef2-8b21-4763-bbf5-c1bc93869ec6>","cc-path":"CC-MAIN-2014-15/segments/1397609537804.4/warc/CC-MAIN-20140416005217-00325-ip-10-147-4-33.ec2.internal.warc.gz"} |
Some Tips
November 7th 2007, 07:47 AM #16
About black triangles, if someone wants to use $\blacktriangleright$ or $\blacktriangleleft$ just click on the image
Now watch
$\frac{2}{6} = \frac{2}{{2 \cdot 3}} = \frac{{\mathop {\rlap/2}\limits^1 }}{{\mathop {\rlap/2}\limits_1 \cdot 3}} = \frac{1}{3}.$
Really cool, isn't it? I made it with MathType. To generate the cross line, you gotta put {\rlap/}. See the code above.
Krizalid you so good, are you sure you are not Japanese?
You actually had written this:
Krizalid you so good at LaTeX that even those crazy Japanese people would be jealous of your skill
I'm chilean!
I like LaTeX, and I like help on it.
I had made up this way to cancel fractions . . .
. . $\frac{ot4^2}{ot6_3} =\frac{2}{3}$ . . . \frac {\not4 ^2} {\not6 _3} = \frac{2}{3}
But I really like your "rlap" code . . . it's looks much better!
. . $\frac{\rlap/4 ^2}{\rlap/6 _3}$ . . \frac{\rlap/4 ^2} { \rlap/6 _3}
Let me try something: . frac{1{\rlap/9}} {{\rlap/9}5} = \frac{1}{5}
. . $\frac{1{\rlap/9}}{{\rlap/9}5} =\frac{1}{5}$ . . . . . Ha! It works!
the thing is, i could not make the strike through go through two symbols, for instance, if you wanted to strike through (cancel) 19, you'd have to do one strike for the 1 and one for the 9. does
anyone know how to go through two or more symbols at a time, or am i just being too picky?
I was just browsing around and found some interesting information on this. I saw an example where x - x was canceled out by a diagonal line and had a 0 in the upper right corner.
I couldn't find code for it, but the article mentioned using the "cancel package." So.......
@MathGuru: Do we have the cancel package loaded into the site?
November 7th 2007, 10:03 AM #17
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New York City
November 7th 2007, 10:06 AM #18
November 7th 2007, 10:49 AM #19
Super Member
May 2006
Lexington, MA (USA)
November 7th 2007, 12:40 PM #20
November 7th 2007, 05:20 PM #21
November 8th 2007, 06:13 AM #22 | {"url":"http://mathhelpforum.com/latex-help/19082-some-tips-2.html","timestamp":"2014-04-20T10:32:13Z","content_type":null,"content_length":"59679","record_id":"<urn:uuid:fe524570-00f0-4283-9c5d-d950d0eb0ff8>","cc-path":"CC-MAIN-2014-15/segments/1397609538110.1/warc/CC-MAIN-20140416005218-00548-ip-10-147-4-33.ec2.internal.warc.gz"} |
[Numpy-discussion] speed of numpy vs matlab on dot product
Paulo Jose da Silva e Silva pjssilva at ime.usp.br
Sun Jun 11 18:03:18 CDT 2006
Em Sáb, 2006-06-10 às 15:15 -0700, JJ escreveu:
> python
> import numpy
> import scipy
> a = scipy.random.normal(0,1,[10000,2000])
> b = scipy.random.normal(0,1,[10000,2000])
> c = scipy.dot(a,scipy.transpose(b))
Interesting enough, I may have found "the reason". I am using only numpy
(as I don't have scipy compiled and it is not necessary to the code
The problem is probably memory consumption. Let me explain.
After creating a, ipython reports 160Mb of memory usage. After creating
b, 330Mb. But when I run the last line, the memory footprint jumps to
1.2gb! This is four times the original memory consumption. In my
computer the result is swapping and the calculation would take forever.
Why is the memory usage getting so high?
Obs: As a side not. If you decrease the matrix sizes (like for example
2000x2000), numpy and matlab spend basically the same time. If the
transpose imposes some penalty for numpy, it imposes the same penalty
for matlab (version 6.5, R13).
More information about the Numpy-discussion mailing list | {"url":"http://mail.scipy.org/pipermail/numpy-discussion/2006-June/020870.html","timestamp":"2014-04-16T04:32:06Z","content_type":null,"content_length":"3797","record_id":"<urn:uuid:34cf7a63-4443-4543-b552-0b48303e5887>","cc-path":"CC-MAIN-2014-15/segments/1397609530136.5/warc/CC-MAIN-20140416005210-00538-ip-10-147-4-33.ec2.internal.warc.gz"} |
order of vertices in 2d polygon... [Archive] - OpenGL Discussion and Help Forums
i've got some vertices and i want to draw them as a convex polygon, in other words, i need to determine correct order of vertices in convex polygon. it it possible?
my vertices have only x,y coordinate, there are no duplicates or other evil stuff and always can be arranged to polygon | {"url":"https://www.opengl.org/discussion_boards/archive/index.php/t-159596.html?s=b5957fed27b8b929a42b3db1e35ae205","timestamp":"2014-04-17T10:13:58Z","content_type":null,"content_length":"6859","record_id":"<urn:uuid:07873013-29b5-4a2b-8a18-c6f6b492d0d9>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00089-ip-10-147-4-33.ec2.internal.warc.gz"} |