content stringlengths 86 994k | meta stringlengths 288 619 |
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Linear Algebra/Vector Spaces
The first chapter began by introducing Gauss' method and finished with a fair understanding, keyed on the Linear Combination Lemma, of how it finds the solution set of a linear system. Gauss' method
systematically takes linear combinations of the rows. With that insight, we now move to a general study of linear combinations.
We need a setting for this study. At times in the first chapter, we've combined vectors from $\mathbb{R}^2$, at other times vectors from $\mathbb{R}^3$, and at other times vectors from even
higher-dimensional spaces. Thus, our first impulse might be to work in $\mathbb{R}^n$, leaving $n$ unspecified. This would have the advantage that any of the results would hold for $\mathbb{R}^2$ and
for $\mathbb{R}^3$ and for many other spaces, simultaneously.
But, if having the results apply to many spaces at once is advantageous then sticking only to $\mathbb{R}^n$'s is overly restrictive. We'd like the results to also apply to combinations of row
vectors, as in the final section of the first chapter. We've even seen some spaces that are not just a collection of all of the same-sized column vectors or row vectors. For instance, we've seen a
solution set of a homogeneous system that is a plane, inside of $\mathbb{R}^3$. This solution set is a closed system in the sense that a linear combination of these solutions is also a solution. But
it is not just a collection of all of the three-tall column vectors; only some of them are in this solution set.
We want the results about linear combinations to apply anywhere that linear combinations are sensible. We shall call any such set a vector space. Our results, instead of being phrased as "Whenever we
have a collection in which we can sensibly take linear combinations ...", will be stated as "In any vector space ...".
Such a statement describes at once what happens in many spaces. The step up in abstraction from studying a single space at a time to studying a class of spaces can be hard to make. To understand its
advantages, consider this analogy. Imagine that the government made laws one person at a time: "Leslie Jones can't jay walk." That would be a bad idea; statements have the virtue of economy when they
apply to many cases at once. Or, suppose that they ruled, "Kim Ke must stop when passing the scene of an accident." Contrast that with, "Any doctor must stop when passing the scene of an accident."
More general statements, in some ways, are clearer.
Last modified on 12 November 2010, at 22:49 | {"url":"https://en.m.wikibooks.org/wiki/Linear_Algebra/Vector_Spaces","timestamp":"2014-04-19T04:30:58Z","content_type":null,"content_length":"17989","record_id":"<urn:uuid:585f769d-18c2-4acc-a7d7-849fdb66dbb6>","cc-path":"CC-MAIN-2014-15/segments/1397609535775.35/warc/CC-MAIN-20140416005215-00562-ip-10-147-4-33.ec2.internal.warc.gz"} |
Distance between two points on the surface of a el
Dear all,
I want to calculate the ' great circle distance' between two points on the surface of an ellipsoid. Though because of the extra radius, the great circle distance is not what I am looking for. Also I
have not a center of the ellipsoid on [0,0,0], which makes the calculation more difficult, besides this I only have the cartesian coordinates of the two points and the center.
This is what I have now, anyone knows how to calculate this distance?
% construct of center ellipsoid
a= rand(1,3)*10
% construct the three different radii
b= rand(1,3)*10
% construct ellipsoid
[x,y,z] = ellipsoid(cx,cy,cz,rx,ry,rz,25)
hold on
% point 1 on the surface of the ellipsoid
N1=ceil(length(x)*rand(1,1)) % Random number for surface point 1
Point1 = [x(N1,M1) y(N1,M1) z(N1,M1)]
Point2 = [x(N2,M2) y(N2,M2) z(N2,M2)] | {"url":"http://www.mathworks.com/matlabcentral/newsreader/view_thread/281983","timestamp":"2014-04-24T02:00:17Z","content_type":null,"content_length":"42900","record_id":"<urn:uuid:e94b0cf1-d834-47d9-bb65-32f3200e0269>","cc-path":"CC-MAIN-2014-15/segments/1398223204388.12/warc/CC-MAIN-20140423032004-00548-ip-10-147-4-33.ec2.internal.warc.gz"} |
year 4 optional sats mental maths paper
• In the Year 7 and 8 Optional Tests more ...Condensed courses offered in: Maths Yes English ...[Of those who took] maths SATs [Key Stage3 ...
• This pdf remains the intellectual property of satspapers.org satspapers.org Year 4 Optional Test. Mental Mathematics Questions. Practice question. What is eight ...
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• Year 5 Maths Optional SATs 2003 Teacher Guidance Introduction Since the introduction of optional tests for years 3, 4 and 5 from 1997, there has been
• The years 3, 4 and 5 optional tests in mathematics offer schools a means of monitoring and measuring children's ... Each child will sit two tests: a written mathematics test and a mental
mathematics test. It is ... photocopying onto coloured paper;.
• SATs The BSN no longer holds SATs examinations ...In-class maths support is provided for students ...As the course is optional these students are ...
• 1 1. Introduction Since the introduction of optional tests for years 3, 4 and 5 from 1997, there has been much development in the teaching of mathematics.
• SATs, or Standard Assessment Tests, are the former, but still known as, name ... For example, a 'one mark' answer in a reading paper only requires one word, ... may be selected to sit the
optional Level 6 paper in the afternoon of the same day .
• Year 10 Curriculum Core Subjects Optional Subjects ... Drama Environmental Science French Geography...SATS results at Level 4 in Maths, Science & ...
• optional tests in the forthcoming year Year 3 ...was not able to use them in 'the SATs week'...maths so they have no grasp of the basic ... | {"url":"http://www.doc-live.com/year-4-optional-sats-mental-maths-paper","timestamp":"2014-04-18T18:11:00Z","content_type":null,"content_length":"21604","record_id":"<urn:uuid:f13d38fd-8c7f-4452-835f-73493c1baedc>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00055-ip-10-147-4-33.ec2.internal.warc.gz"} |
Complexity of random knot with vertices on sphere
up vote 18 down vote favorite
Connect $n$ random points on a sphere in a cycle of segments between succesive points:
I would like to know the growth rate, with respect to $n$, of the crossing number (the minimal number of crossings of any diagram of the knot) $c(n)$ of such a knot. I only know that $c(n)$ is $O(n^
2)$, because it is known that the crossing number is upper-bounded by the stick number $s(n)$: $$\frac{1}{2}(7+\sqrt{ 8 c(K) + 1}) \le s(K)$$ for any knot $K$. And $s(n) \le n$ is immediate.
I feel certain this has been explored but I am not finding it in the literature. Thanks for pointers!
knot-theory stick-knots reference-request gt.geometric-topology
What is the definition of the crossing number? – Igor Rivin Feb 5 '11 at 16:30
I presume he means the minimal crossing number (on all plane diagrams) – Bruno Martelli Feb 5 '11 at 16:50
Yes, as Bruno says. Sorry for not being clear. Edited now. – Joseph O'Rourke Feb 5 '11 at 16:58
What do you mean by growth rate? This could be interpreted in different ways: the maximal crossing number for a given n, the average (with respect to the spherical measure), or the average (take
all knot types constructed this way, and take the average crossing number). – Ian Agol Feb 5 '11 at 17:23
@Agol: Good point! I meant the expected crossing number of the knot determined by $n$ random points on the sphere, which is I guess your 2nd option. – Joseph O'Rourke Feb 5 '11 at 18:13
show 2 more comments
4 Answers
active oldest votes
This is not really an answer but an over-long comment following up suggestions of Ian Agol and Bill Thurston.
Experiment suggests (with 97% confidence) that the crossing probability (in a specified or random projection, for two line segments with the four endpoints chosen independently uniformly at
random with respect to Haar measure) is greater than $.2499$ and less than $.2501$. I have a motto never to compute by integrating what can be computed by symmetry, so the hope would be,
for some integer $S$, to write a total of $4S$ symmetry-related expressions for the probability whose sum is identically $S$. So far any such trick eludes me; does anyone else see a way?
It would surprise me, for very large $n$, if even a constant factor improvement over a randomly chosen obvious projection is possible. I would wager, then (although I would hate to have to
prove it) that $$\lim\limits_{n\rightarrow\infty} \frac{\bar{c}(n)}{n^2} = \frac18,$$ where $\bar{c}(n)$ is the expected value of crossing number for $n$ random points on the sphere.
up vote The suggestion of a Voronoi-like spine whose dual would triangulate the knot complement is an interesting one. The individual faces are sections of hyperbolic paraboloid. It seems
14 down reasonable that their number would be strictly between linear and quadratic, although I don't yet see a good heuristic for guessing the correct order.
EDIT: I was able to satisfy myself that the crossing probability is exactly $\frac14$, through a fairly (and probably unnecessarily) involved process. At most stages one can reduce to
simpler calculations using symmetries or nice facts such as the probability-preserving projection map from a $(d-1)$-sphere in $\mathbb{R}^d$ onto the ball of its first $d-2$ components.
For the final step I did have to compute an integral, though—the same one that tells you the angular momentum of a spinning coin (whose axis of rotation is in the plane of the coin).
I have started to suspect that in a certain precise sense an integral is unavoidable—that the boundary of crossing configurations is curved in ways that prevent abutment, just as the region
north of $30^\circ$ latitude carries exactly $\frac14$ of the surface area of a perfect globe, but no finite collection of $4S$ rotated copies of it can be an exact $S$-fold cover. (The
transcendental answer to the usual Sylvester's four-point problem in the disc also discourages the finite cover approach.)
2 What a wonderfully specific conjecture! And the subquestion of proving that the probability of two random segments crossing in projection is $\frac{1}{4}$ is very pretty. – Joseph
O'Rourke Feb 9 '11 at 11:59
3 If the projections of the 4 points are not in convex position then the segments do not intersect. If the projections of the 4 points are in convex position, then 1 of the 3 of the ways
of pairing the points causes the line segments to cross. So the $1/4$ conjecture is saying that the answer to Sylvester's problem for the projection of the uniform measure on the surface
of the sphere is $3/4$. – Douglas Zare Feb 9 '11 at 23:01
@Douglas Zare: Yes, in fact this is exactly how the experiment was run: the proportion of convex positions, divided by 3. The name of Sylvester's problem for the same question (with
uniform measure) in a convex region of the plane is helpful. Apparently the problem is usually stated as the probability that the four points are $\textit{not}$ in convex position, so
the crossing probability would be $\frac{1-p}3$ where $p$ is the Sylvester's problem for the sphere projection measure in a circle. (Conjecturally $p=\frac14$, with non-convex as likely
as each of the 3 ways of crossing.) – Tracy Hall Feb 10 '11 at 1:30
add comment
Edited 2/9 after discussion with Dylan Thurston
It seems unlikely that the obvious knot projections can be simplified by more than a constant factor, so a quadratic lower bound for the expected minimum crossing number seems likely.
Crossing number by itself though is a strange measure of complexity, and it is hard to compute. However, it's bounded below by hyperbolic volume of the knot complement.
It would be possible to get some experimental evidence by feeding output of your random process through snappea, and looking at the distribution of hyperbolic volume. However, I think
hyperbolic volume probably grows at a less than quadratic rate. You can imagine thickening the knot into a growing solid torus, pushing outward until every part of the boundary has bumped
up vote into other boundary --- similarly to a Voronoi subdivision. With tubes of diameter some constant times $n^{-.5})$, the total volume of tubes is on the order of the volume of the ball, so
12 down typical tube spacing should be $O(n^{-.5})$. This suggests the number of faces in this subdivision should be $O(n^{3/2})$, which would give a triangulation having $O(n^{3/2})$ tetrahedron
vote where the knot is in the 1-skeleton, implying that the typical Gromov norm or hyperbolic volume probably grows as $O(n^{3/2})$. This would only imply $n^{3/2}$ crossings.
Marc Lackenby, in SPECTRAL GEOMETRY, LINK COMPLEMENTS AND SURGERY DIAGRAMS, developed a beautiful method to give lower bounds for crossing numbers for knots. His method possibly could be
applicable to improve this situation, provided the Cheeger constants for these manifolds can be shown to be not too small.
It's also possible that one could estimate the degree of the Alexander polynomial, to get an estimate of the crossing number.
There's a strange formatting issue in the rendering of this on my computer, where $n^2$ is displaced from where it is in the text. If anyone knows how to fix this, feel free. – Bill
Thurston Feb 5 '11 at 17:16
1 I appreciate your guess that a quadratic lower bound seems likely. And I like the idea of gathering some experimental evidence. Thanks! – Joseph O'Rourke Feb 5 '11 at 18:01
add comment
I think you should look at this paper "The average crossing number of equilateral random polygons" by Y Diao, A Dobay, R B Kusner, K Millett and A Stasiak. http://iopscience.iop.org/
It can't seem to get the full text right now, so I'm not sure exactly what their random knot model is. It certainly isn't quite what you wanted, as they require their polygons to be
up vote 7 down equilateral. In their model the crossing number grows as $n \; \ln n$.
If this doesn't answer your question, consider looking through the rest of Ken Millett's work. He is one of the main people working on random knots.
Thanks for these useful references! I will look into Millett's work. – Joseph O'Rourke Feb 5 '11 at 18:13
add comment
You can find estimates nd other relevant information in the following two papers:
Arsuaga, J., Blackstone, T.*, Diao,Y., Karadayi, E. , Saito, Y. Sampling Large Random knots in Confined Spaces. J. Phys. A: Math Gen. (2007) 40; 11697-11711
up vote 4 down vote Arsuaga, J., Blackstone, T.*, Diao,Y., Karadayi, E. , Saito, Y. Linking of Uniform Random Polygons in Confined Spaces. J. Phys. A: Math. Gen. (2007) 40; 1925-1936
and The linking number and the writhe of uniform random walks and polygons in confined spaces E Panagiotou et al 2010 J. Phys. A: Math. Theor.
add comment
Not the answer you're looking for? Browse other questions tagged knot-theory stick-knots reference-request gt.geometric-topology or ask your own question. | {"url":"http://mathoverflow.net/questions/54412/complexity-of-random-knot-with-vertices-on-sphere/54417","timestamp":"2014-04-19T15:11:28Z","content_type":null,"content_length":"81074","record_id":"<urn:uuid:c77e2ee9-e378-435b-8939-ccfe71cdb6a8>","cc-path":"CC-MAIN-2014-15/segments/1398223207046.13/warc/CC-MAIN-20140423032007-00096-ip-10-147-4-33.ec2.internal.warc.gz"} |
st: re: Efficient, foolproof calculation of matrix quadratic form with
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st: re: Efficient, foolproof calculation of matrix quadratic form with
From Kit Baum <baum@bc.edu>
To statalist@hsphsun2.harvard.edu
Subject st: re: Efficient, foolproof calculation of matrix quadratic form with
Date Fri, 25 Sep 2009 11:16:17 -0400
Venable said
Alternatively, I know that I could do the sum of the N terms
Xn'*Om_Inv_Block*Xn but I am worried that I would somehow mess this
up. So, another solution, I suppose, would be some idiot-proof way to
do this sum.
This is very similar to the problem of calculating the -sureg- estimator (with the difference that in -sureg- there is no constraint that the X matrices have the same number of columns). The
-suregub- routine codes -sureg- in Mata and avoids creating the monster matrix by doing the above sum. It really isn't that hard to set up that loop, and if you think about it, it is incredibly
wasteful to be calculating the full matrix (even if memory is not an issue) given the block- diagonal structure implied by the Kronecker. You may find the - suregub- code in the downloadable
materials related to ITSP below. That routine implements -sureg- for the special case where T is not necessarily equal, but it also works fine if all equations' T values are the same.
Kit Baum | Boston College Economics & DIW Berlin | http://ideas.repec.org/e/pba1.html
An Introduction to Stata Programming | http://www.stata-press.com/books/isp.html
An Introduction to Modern Econometrics Using Stata | http://www.stata-press.com/books/imeus.html
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/ | {"url":"http://www.stata.com/statalist/archive/2009-09/msg01049.html","timestamp":"2014-04-19T17:40:26Z","content_type":null,"content_length":"7380","record_id":"<urn:uuid:37b6560b-3535-4ddb-ab7f-ea3ab20adbe1>","cc-path":"CC-MAIN-2014-15/segments/1397609537308.32/warc/CC-MAIN-20140416005217-00406-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
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Slopes of Tangents and Secants 2
Sorry I'm pretty sure it said post a new thread for a different question....well here it is:
lim f(a + h) - f(a) / h
Use that forumla to find the slope of the tangent line of the cubic funciotn y = x^3 at the point (1,1)
This formula keeps messing me up. | {"url":"http://mathhelpforum.com/calculus/73394-slopes-tangents-secants-2-a.html","timestamp":"2014-04-19T00:37:34Z","content_type":null,"content_length":"40532","record_id":"<urn:uuid:3dd4c501-4483-4865-974d-46d038d7ffcf>","cc-path":"CC-MAIN-2014-15/segments/1397609537097.26/warc/CC-MAIN-20140416005217-00360-ip-10-147-4-33.ec2.internal.warc.gz"} |
What math comes after calculus and differential equations?
I know there is basic math, Algebra I, Intermediate Algebra, Geometry, Trigonometry, College Algebra, Statistics, Calculus I,II,III and then differential equations. What comes after differential
What are the different levels of math? There must be more math after differential equations, right?
Suggestion by shadowsage702
I think it’s multivariable calculus
Give your answer to this question below!
How is math involved in Civil engineering?
Im really good in math and im somewhat interested in this career. I was just wondering how much of civil engineering in involved in math and what kind of math (algebra, geometry, etc.). Im thinking
more of a construction, structural or mechanical engineering. And what are the main subjects in getting a degree for it?(math, science?)
Also can anyone recommend a math related career?
Suggestion by Steve
I am a civil engineer and took 4 semesters of calculus, differential equations, and another math class in the engineering department. Almost all of engineering is based off physics. You like physics,
then you will probably like engineering. I am happy. If not engineering, a lot of engineering drop outs enter into computer science. You can always be a math professor. Good luck with life.
Suggestion by Prathamesh
hey dear….
the following link can give you a brief of kind of work with civil…
so you wont have troubles thinking about what ull go thru in tht courses..
also i post a link for a few courses and subjects ull hav to do in various engg departments and careers
go thru that too it might help
Suggestion by I whip my bread back’n'furrth
I just finished my Ba in Civil Engineering and I can tell you that at uni, you will focus a lot more on calculus than algebra, although in first year all engineering students learn the basics of many
fields of math weather they are relevant for civil or not.
With geometry, yes it is involved, however most likely the majority of your geometry work will be done with computer programs such as CAD or maybe even a modelling program which uses numerical
methods such as finite elements.
In summary, yes there is quite a bit of math, however most of the tricky questions you will be asked in exam conditions will not be about understanding the math but about understanding the physical/
engineering concepts which have mathematical relationships.
Give your answer to this question below!
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Ackerman function
04-26-2011, 10:29 AM
Ackerman function
I'm working on reading through the first chapter of SICP, I know this is not java, but I'm hoping it's fine to post this here, I know at least Jos has been through the book.
I'm having a bit of trouble tracing how it works. I get that it's a linear recursive process but I am at a bit of a sticking point. I omitted the first few evaluations, however; this is what I
have so far
(A -4(A -3(A -2(A -1(A 0(A 1 5))))))
(A -5(A -4(A -3(A -2(A -1(A 0(A 1 4)))))))
(A -6(A -5(A -4(A -3(A -2(A -1(A 0(A 1 3))))))))
(A -7(A -6(A -5(A -4(A -3(A -2(A -1(A 0(A 1 2))))))))
(A -8(A -7(A -6(A -5(A -4(A -3(A -2(A -1(A 0(A 1 1)))))))))
(A -8(A -7(A -6(A -5(A -4(A -3(A -2(A -1(A 0 2)))))))))
(A -8(A -7(A -6(A -5(A -4(A -3(A -2(A -1 4))))))))
I'd like to finish it and then see if it works as I am starting to think, however; I am not sure where to go from here. I get that the next call is
but would the A(x -1 be put on the back, like this
(A -2(A -8(A -7(A -6(A -5(A -4(A -3(A -2(A -1 3))))))))
or should it be added onto the newly being worked on portion and look like this
(A -8(A -7(A -6(A -5(A -4(A -3(A -2(A -1(A -1 3)))))))))
Also, please let me know if I am even doing it correctly.
BTW: The original call is (A 1 10)
I am starting to think I did it very incorrect and instead it should look something like this
(A 1 10)
(A 0(A 1 9))
(A 0(A 0(A 1 8)))
(A 0(A 0(A 0(A 1 7))))
(A 0(A 0(A 0(A 0(A 1 6)))))
and so on.
edit: I figured out how to trace (A 1 10), the last bit was the correct way to go, the next step is to evaluate(A 2 4), and (A 3 3). I get fairly small numbers when doing it by hand,
(A 3 3) = 8
(A 2 4) = 16
However, a quick write up of the code in java, and in scheme, returns 65535. I am unable to understand why, it also looks like it overflows the stack for stuff like (A 2 10).
05-01-2011, 09:39 AM
Maybe this dumping program helps a bit:
import java.util.Stack;
public class T {
private static void print(Stack<Integer> s) {
for (int i= 0; i < s.size()-1; i++)
for (int i= 0; i < s.size()-1; i++)
public static void main(String[] args) {
Stack<Integer> s= new Stack<Integer>();
while (s.size() > 1) {
int y= s.pop();
int x= s.pop();
if (x == 0)
else if (y == 0) {
else {
kind regards,
05-01-2011, 10:08 AM
Thanks for that jos, I'll have to run it when I get off work. Boring night on these forums today. Feel like helping me out with turning a recursive process into an iterative one? The process
seems natural to do as a recursive process but I am just blocked when thinking of the iterative approach.
(define(f n)
(if(< n 3)
(+(f(- n 1))
(f(- n 2))
(f(- n 3)))))
is the recursive process I came up with. Thanks for the help, I amgetting the book soon, gonna have to read the first chapter quite a few times(and all the others as well)
Btw, I got and ran your game of life rpl code, the code was interesting, not sure what all of it meant however.
05-01-2011, 10:48 AM
Thanks for that jos, I'll have to run it when I get off work. Boring night on these forums today. Feel like helping me out with turning a recursive process into an iterative one? The process
seems natural to do as a recursive process but I am just blocked when thinking of the iterative approach.
(define(f n)
(if(< n 3)
(+(f(- n 1))
(f(- n 2))
(f(- n 3)))))
is the recursive process I came up with. Thanks for the help, I amgetting the book soon, gonna have to read the first chapter quite a few times(and all the others as well)
Btw, I got and ran your game of life rpl code, the code was interesting, not sure what all of it meant however.
w.r.t. the example above: think about a Fibonacci series with three numbers; the first three numbers are 0, 1 and 2.
w.r.t. RPL, yep, I'm afraid I don't have to just finish the reference manual but I also have to write a tutorial ;-)
kind regards,
ps. I'll email you an RPL version of Ackermann's function including memoization.
05-01-2011, 10:58 AM
Awesome thanks :)
I should be getting sicp within a few days, may have to bug you a bit when I get stuck on stuff.
Rpl seems pretty well done, seems like it can handle quite a lot, it's pretty impressive from my quick look without much experience. I can't wait to try and understand all the source code.
05-01-2011, 11:15 AM
Awesome thanks :)
I should be getting sicp within a few days, may have to bug you a bit when I get stuck on stuff.
Rpl seems pretty well done, seems like it can handle quite a lot, it's pretty impressive from my quick look without much experience. I can't wait to try and understand all the source code.
For fun, compare the speed of the RPL version (with memoization turned on as in the example I emailed you) with a properly written recursive Java version. RPL beats Java by miles here ;-)
kind regards,
ps. I understand the comparison isn't fair.
05-01-2011, 08:09 PM
The ackerman function shown in the book looks to be a bit different then yours:
(define (A x y)
(cond ((= y 0) 0)
((= x 0) (* 2 y))
((= y 1) 2)
(else (A (- x 1)
(A x (- y 1))))))
I modified the dumping program to handle everything exactly as the above. It showed me that my original hand evaluation for (A 1 10) was good, equals 1024, but I am still unsure not sure why (A 2
4) and (A 3 3) produce 65536. Anything much larger than 3 3 overflows java. Am I mis-interpreting the above function?
05-02-2011, 06:46 AM
The ackerman function shown in the book looks to be a bit different then yours:
(define (A x y)
(cond ((= y 0) 0)
((= x 0) (* 2 y))
((= y 1) 2)
(else (A (- x 1)
(A x (- y 1))))))
I modified the dumping program to handle everything exactly as the above. It showed me that my original hand evaluation for (A 1 10) was good, equals 1024, but I am still unsure not sure why (A 2
4) and (A 3 3) produce 65536. Anything much larger than 3 3 overflows java. Am I mis-interpreting the above function?
Nope, your interpretation is correct; my Ackermann function is the 'classical' one; the version above is just as bad (big-Oh wise speaking). In Java, your version looks like this:
public static long a(long x, long y) {
if (y == 0) return 0;
if (x == 0) return 2*y;
if (y == 1) return 2;
return a(x-1, a(x, y-1));
(type int works just as well ...)
kind regards,
05-02-2011, 06:53 AM
I tried playing around with it to use BigIntegers instead of ints, it still caused stack overflows as well; I think I understand this function a bit more, however; the book asks to describe what
functions of
A 0 n
A 1 n
A 2 n
can be thought of as mathematical equations. For 1, it seems like it is binary; A 1 10 produces 1024.
I'd like to thank you for the suggestion to think of fibbonaci for the function i described above. I was able to solve it with
(define(f n)
(define(f-iter a b c count)
(if(= count 0)
((+ a b c)a b(- count 1))))
(f-iter(2 1 0 n)))
I really appreciate you helping me out with this stuff; and sending me your rpl IDE jos. I got the reference material you have sent me but I haven't gotten to read much of it yet.
I hope you won't mind if I ask for your help on some of the problems on SICP. Is a lot of the stuff in the book very heavily math based? My math skills aren't as good as I'd like, but I am
willing to learn more as well.
05-02-2011, 07:29 AM
I tried playing around with it to use BigIntegers instead of ints, it still caused stack overflows as well; I think I understand this function a bit more, however; the book asks to describe what
functions of
A 0 n
A 1 n
A 2 n
can be thought of as mathematical equations. For 1, it seems like it is binary; A 1 10 produces 1024.
I'd like to thank you for the suggestion to think of fibbonaci for the function i described above. I was able to solve it with
(define(f n)
(define(f-iter a b c count)
(if(= count 0)
((+ a b c)a b(- count 1))))
(f-iter(2 1 0 n)))
I really appreciate you helping me out with this stuff; and sending me your rpl IDE jos. I got the reference material you have sent me but I haven't gotten to read much of it yet.
I hope you won't mind if I ask for your help on some of the problems on SICP. Is a lot of the stuff in the book very heavily math based? My math skills aren't as good as I'd like, but I am
willing to learn more as well.
You're welcome of course and I don't mind; I like that theoretical fiddling; the math is just simple discrete math (i.e. no nasty differential equations or such ;-)
About A(1, n); it equals A(0, A(1, n-1)) which is 2*A(1, n-1); So the recurrent relation is A(1, n) == 2*A(1, n-1); the sentinel value A(x, 1) == 1, so A(1, n) == 2^n. The value A(2, n) can be
dissected similarly.
kind regards,
05-02-2011, 07:39 AM
At least this book will get me ready for discrete math when I go back to university.
05-02-2011, 01:15 PM
If you want some discrete mathematics, try to find the three volume "The Art Of Computer Programming" by Donald Knuth; they are a must read: Donald Knuth proves everything in those books; very
fundamental stuff.
kind regards,
ps. I think nowadays those three bibles can more easily be found in antique shops; "XXX for dummies in 21 days" seems to be more populair nowadays (which is a pity imho) | {"url":"http://www.java-forums.org/forum-lobby/43053-ackerman-function-print.html","timestamp":"2014-04-19T06:58:01Z","content_type":null,"content_length":"24346","record_id":"<urn:uuid:fa56f416-cb52-408a-9813-8e800610c403>","cc-path":"CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00423-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Do partitions of unity exist if we impose additional conditions on the derivatives?
up vote 9 down vote favorite
Let $ ~~\cup_{k=-1}^{\infty} U_k = \mathbb{R} $ be an open covering of $\mathbb{R}$. It is a well known fact that partitions of unity subbordinate to the cover exists, i.e. there exists smooth
functions $ \varphi_{k} : U_k \rightarrow \mathbb{R} $ with compact support such that $$ \sum_{k=-1}^{\infty} \varphi_k(x)^2 \equiv 1.$$
My first question is vague: Do there exist partitions of unity subbordinate to the cover if we impose some additional conditions on the derivatives (and the nature of the conditions are in terms of
equalities, not inequalities.) ?
A more precise question is as follows: Given a smooth function $f: \mathbb{R} \rightarrow \mathbb{R}$, do there exist functions
$ \varphi_{k} : U_k \rightarrow \mathbb{R} $ with compact support such that in addition to being a partition of unity subbordinate to the cover, it also satisfies $$ \sum_{k=-1}^{\infty} \varphi_{k}^
{\prime}(x)^2 \equiv f(x)$$ ? Here prime denotes derivative with respect to $x$. I assume the answer should depend on what $f(x)$ is. Ideally $\varphi_k$ should be smooth functions, but at the very
least they ought to be $C^1$.
I would prefer if we do not assume $f$ is nowhere vanishing, but if there is an answer assuming that, I would still like to see it.
Remark: The answer probably also depends on the open covering.
$\textbf{Very specific question:}$ Choose some number $\tau \in (\sqrt{2}, 2) $. Say $\tau = 1.5$. Now define $$ U_k = \{ x \in \mathbb{R}: \frac{2^k}{\tau} < |x| < 2^k \tau \} \qquad k=0,1,2, \ldots
$$ U_{-1} = (-1,1).$$
The collection $\{U_k\}_{k=-1}^{\infty} $ is an open covering of $\mathbb{R}$. I want smooth functions $\varphi_k: U_k \rightarrow \mathbb{R}$ and numbers $n_k$ such that $$ \sum_{k= -1}^{\infty} \
varphi_k(x)^2 \equiv 1 $$ and $$ \sum_{k=-1}^{\infty} n_k^2 (\varphi_k(x)^2 + \varphi_k^{\prime}(x)^2) \equiv e^x$$
It is easy to see that if I set $n_k =1$ and $f(x) = e^x-1$, then it is the previous question I had asked. This question is "easier", because one is allowed to choose the numbers $n_k$ (in other
words there is a better chance that the answer here might be yes, because of the freedom in choosing $n_k$).
real-analysis differential-topology smooth-manifolds partition-of-unity
@Pietro I am very sorry, I made a mistake. I have corrected the question now. Now if you derive the equation you get $\sum_{k} 2 \varphi_k(x) \varphi^{\prime}(x) \equiv 0.$ – Ritwik Sep 5 '13 at
1 Yet I think some assumption on the cover is in order. What if all the $U_k$ but one are empty, or more generally, what if all the $U_k$ but one are disjoint from the interval $[0,1]$? – Pietro
Majer Sep 5 '13 at 11:30
I agree, the answer probably depends on the open covering. In my specific case, the second condition you have suggested is actually true, i.e. all but one of the $U_k$ is disjoint from $[0,1]$. In
fact any point in $\mathbb{R}$ belongs to at most two $U_k$ in the specif case I have. – Ritwik Sep 5 '13 at 11:45
It is pretty hard for a function supported on a short interval and having not too large derivative to rise to $1$ and, as you said itself, in your case you have only two shots at each point. Since
I see no point in obtaining the full description to cover just one particular case, I suggest you just tell what exactly your "specific case" is. – fedja Sep 5 '13 at 12:03
1 @Ritwik As posed, the answer is trivially "No" because on $(-0.1,0.1)$, the function $\varphi=\varphi_{-1}$ should satisfy $\varphi^2=1$ and $n(\varphi^2+(\varphi')^2)=e^x$ simultaneously. However
it makes more or less clear what you might really be after. It beats me how you could hope to get a sum of squares equal to $e^x-1$ for $x<0$ though ;). – fedja Sep 5 '13 at 23:43
show 4 more comments
1 Answer
active oldest votes
I hope the following construction will give you what you really need. If not, you'll have to explain why.
Take any nice locally finite covering $\mathbb R\subset\cup_j U_j$ and take any smooth partition of unity $1=\sum_j\psi_j^2$ subordinated to this covering. Take any smooth positive
function $F$ on $\mathbb R$. Choose the numbers $n_j>0$ so that $\sum_j n_j^2(\psi_j^2+(\psi_j')^2)\le F/2$. Define the smooth function $\theta$ by $$ (\theta^{\,\prime})^2\sum_j n_j^
up vote 5 down 2\psi_j^2=F-\sum_j n_j^2(\psi_j^2+(\psi_j')^2) $$
vote accepted Finally, associate with each $U_j$ two functions $\varphi_{j,0}=\psi_j\cos\theta$ and $\varphi_{j,1}=\psi_j\sin\theta$ and enjoy the identities $$ \sum_{j,k}\varphi_{j,k}^2=1, \quad \
sum_{j,k}n_j^2(\varphi_{j,k}^2+(\varphi_{j,k}')^2)=F\,. $$
When you are defining $\theta^{\prime}$, did you intend to put the $\psi_j^2$ on the rhs? If the $\psi_j^2$ is not there, then every thing works out as you have said, unless I am
making some calculation mistake. – Ritwik Sep 6 '13 at 12:20
@Ritwik Let's see: $\varphi_0^2+\varphi_1^2=\psi^2$, $(\varphi_0')^2+(\varphi_1'^2)=\psi'^2+\psi^2(\theta')^2$, so, yes, $\psi_j^2$ should be there (unless the condition is that the
pure sum of squared derivatives with coefficients $n_j^2$ is prescribed instead, in which case it shouldn't). – fedja Sep 6 '13 at 12:30
I am sorry, you were absolutely right; there is no mistake in what you said. I was making a simple error. – Ritwik Sep 6 '13 at 13:13
In any case, this is a great answer. This is exactly what I was after. – Ritwik Sep 6 '13 at 13:14
@Ritwik Glad to hear it :-). The moral is: ask exactly what you need and let other people generalize or consider partial cases. Unless you can solve the problem yourself, you never
know what the first step or the first reduction should be. – fedja Sep 6 '13 at 13:43
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Not the answer you're looking for? Browse other questions tagged real-analysis differential-topology smooth-manifolds partition-of-unity or ask your own question. | {"url":"http://mathoverflow.net/questions/141317/do-partitions-of-unity-exist-if-we-impose-additional-conditions-on-the-derivativ/141395","timestamp":"2014-04-20T03:47:32Z","content_type":null,"content_length":"67886","record_id":"<urn:uuid:f0e98c87-d24c-45d8-b280-0d05b63e91ce>","cc-path":"CC-MAIN-2014-15/segments/1397609537864.21/warc/CC-MAIN-20140416005217-00193-ip-10-147-4-33.ec2.internal.warc.gz"} |
Re: GR or QED?
David Bowman (dbowman@tiger.gtc.georgetown.ky.us)
Wed, 29 Apr 1998 13:21:11 EDT
Concerning Glenn's comments about his source (H. Ross) for the claim that
GR is the most precisely confirmed theory:
> I was corrected by both George Murphy and David Bowman when I said that GR
> was the best confirmed theory. I have thought about where I got that idea
> all day and figured out that it was from Hugh Ross. I found something that
> is close to that statement. he Says, " A recent experiment has confirmed
> the accuracy of Einstein's equation to at least twenty-one places of the
> decimal (within 0.0000000000000000001 percent.)" Creation and Time, p. 99.
> He cites S. K. Lamoreaux et al, "New limits on Spatial Anisotrophy from
> Optically Pumped 201Hg and 199Hg", Physical Review Letters 57(1986) pp
> 3125-3128.
George Murphy then surmised:
> I don't have this book of Hugh's but will check the PR article
>today. From the title it sounds like a special relativity confirmation
>- in GRT space _isn't_ strictly isotropic.
I don't have Hugh Ross' book either, but I did get a chance to check out
the PRL paper to which he refers. It seems that George's surmise from
the title is quite correct. The paper *does* only report on a particular
test of one aspect of local Lorentz invariance, (i.e. *special*
relativity, SR). That aspect is the local isotropy of space regarding
the possibility of anisotropic quadrupolar electromagnetic couplings to
the nuclear energy levels in Hg arising from a local anisotropy of space.
The paper claims an upper bound of about 10^(-21) for the fractional size
of the magnitude of such couplings. The results also provide a de facto
confirmation for the Equivalence Principle, EP, since it is via the EP
that we are allowed to ignore gravitational (curvature) effects when
considering phenomena in freely falling frames at the most local scale
sizes in spacetime. Crudely speaking, the EP (the strong form of it)
states that for all physical phenomena observed in a sufficiently small-
scale region in spacetime there exist local inertial frames, i.e. the
freely falling ones, such that *special* relativity holds in them. Thus,
if a laboratory-scale experiment demonstrated an invalidity of an aspect
of SR it would automatically be a demonstration of the invalidity of the
EP. It is true that if the EP is invalidated then GR is invalidated as
well since the EP is the basis for how GR embeds the local Lorentz
structure of spacetime into a theory that treats gravitational effects in
terms of spacetime curvature.
The main problem with Ross's claim regarding this paper is that it is not
a confirmation of the *specific distinctive* features of the predictions
that GR makes to the exclusion of all other theories of gravitation.
Bare SR (along with GR, and all other gravitation theories that
incorporate the EP) predict that space is locally isotropic (in a freely
falling frame). Even Newton's theory holds that space is isotropic
(if one sticks to mechanics and doesn't try to embed electromagnetism
into it via a luminiferous ether moving relative to the observer).
Since QED also incorporates SR in its formulation, an anisotropy of space
would would invalidate it as well. This paper could also be taken as a
confirmation of QED as much as a confirmation of GR. Thus the results of
this paper do nothing to discriminate among a host of potential theories.
The only theories that could possibly be disconfirmed/falsified by these
results are theories that specifically predicted that space *was* locally
*an*isotropic by a relative amount significantly greater than 1 part in
10^21. I don't know of any serious physicists seriously entertaining any
such theories (although this does not mean, by any means, there are
none--just that I don't know of them.)
Glenn asked:
> My question: Is Hugh correct about the applicability of this level of
> accuracy?
It seems so. But all that the results indicate is that, to this
high precision, the energy levels (or at least their differences) of Hg
nuclei (as seen by the precession of their spins) in a magnetic field
remain the same for all orientations (in the plane perpendicular to the
Earth's spin axis) of the magnetic field w.r.t. the fixed stars.
> Is the referenced article relevant to the issue?
It is relevant to the question of the an/isotropy of space. Any
relevance it has to *specifically* GR is misleading.
> Has any part of QED been verified to that many decimal places?
Since this result confirms QED just as much as it confirms GR, (since
both presume the validity of SR) then, I guess, the answer is yes. But
such confirmations *are* misleading as to their significance or their
David Bowman | {"url":"http://www2.asa3.org/archive/asa/199804/0456.html","timestamp":"2014-04-18T21:32:54Z","content_type":null,"content_length":"6963","record_id":"<urn:uuid:fa1d3879-463e-4d11-9938-2e660b88c1fd>","cc-path":"CC-MAIN-2014-15/segments/1398223202457.0/warc/CC-MAIN-20140423032002-00208-ip-10-147-4-33.ec2.internal.warc.gz"} |
How do I find an orthogonality condition for the Laguerre polynomials weighed by some particular n and x dependent phase factors? (self.math)
submitted ago by x3nodox
sorry, this has been archived and can no longer be voted on
That is to say, how do I find some [; w(x) ;] such that [; \int dx \; w(x)\; L_n(x)\; L_m(x)\; e^{i\;(n+m)\;(\frac{c_1}{\sqrt{x}}-1)}=f(n)\delta_{mn} ;] where [;L_n(x);] is the n-th Laguerre
polynomial? I know this is oddly specific, but it's getting to be kind of a roadblock on an optics project I'm working on. Thanks in advance for the help!
Edit: to clarify based on some of the responses, I'm looking for an inner product under which [; L_n(x)\;e^{i\;n\;(\frac{c_1}{\sqrt{x}}-1)} ;] are orthogonal. Just using the inner product that makes
the Laguerre polynomials orthogonal doesn't work, and I'm not sure how to tweek it to make it work. Left the f(n) on the right side just for generality sake. Sorry if that lead to any confusion; I
don't care about f(n), just that for mismatched n and m the inner product goes to 0.
all 3 comments
[–]ssrg1615Number Theory0 points1 point2 points ago
sorry, this has been archived and can no longer be voted on
[–]pomadecellphone0 points1 point2 points ago
sorry, this has been archived and can no longer be voted on
[–]pomadecellphone0 points1 point2 points ago
sorry, this has been archived and can no longer be voted on
Could you tell us some more about this problem? Laguerre polynomials are orthonormal when given the inner-product <L_n,L_m> = int L_n L_m e^-x from 0 to infinity, so getting a delta on the right-hand
side is reasonable; but the f(n) is somewhat concerning, since you could switch the two polynomials in the integrand with no bearing on the evaluation of the integral (unless w(x) has some other
dependence on n,m we don't know about). | {"url":"http://www.reddit.com/r/math/comments/1akufc/how_do_i_find_an_orthogonality_condition_for_the/","timestamp":"2014-04-21T00:02:14Z","content_type":null,"content_length":"57730","record_id":"<urn:uuid:14781e86-5d20-46ce-95bd-79bbe778ce55>","cc-path":"CC-MAIN-2014-15/segments/1397609539337.22/warc/CC-MAIN-20140416005219-00400-ip-10-147-4-33.ec2.internal.warc.gz"} |
i want help in problem
Whats the problem? What parts do you need help with in particular? The descriptive you are, tje better we can help.
First, if this problem is really of interest to you, it would have looked better, and perhaps encouraged more people to respond, if you had taken the trouble to type the problem in rather than just
copying it to a file and requiring others to open the file. That's always a nuisance and always has the possiblity of opening a virus. You are told that, on a certain island, everyone either always
tells the truth or always lies. Also you are told that if you ask people at the university on this island, they will always tell you the following two things, no matter which employee you ask: 1.
There are fewer than N employees who work more than me. 2. At least M employees of the university have a larger salary than me You are asked to write a computer program that given the two values, N
and M, will return the number of employees at the University. They they give the example of N and M both equaling 1 in which case, we are told, the program should return "2". That is, any employee at
the University must respond either: 1. There is less than 1 employee who works more than me. (I.e. there are NO employees who work more than me.) 2. At least 1 employee of the university has a larger
salary than me But remember that some employees always tell the truth and some always lie. So you would have to consider a number of possiblities. It is possible that all those who tell the truth
will answer (1) and all those who lie will answer (2). Or the other way around. But you should also consider the possibility that some people who tell always tell the truth could answer (1) and some
who always lie could answer (1). What would that mean? | {"url":"http://mathhelpforum.com/algebra/155135-i-want-help-problem.html","timestamp":"2014-04-17T03:01:03Z","content_type":null,"content_length":"39214","record_id":"<urn:uuid:017b27e8-cc71-4870-857e-ec5b325a8bc5>","cc-path":"CC-MAIN-2014-15/segments/1397609526102.3/warc/CC-MAIN-20140416005206-00084-ip-10-147-4-33.ec2.internal.warc.gz"} |
Weil reciprocity law
Weil reciprocity law
Weil reciprocity is a theorem of André Weil about the quotient field $K(C)$ of an algebraic curve $C$ over an algebraically closed field $K$.
The following 2 articles make parallel between some notions of QFT and of number theory and in particular about the analogy between the Weil reciprocity law for function fields and the Takahashi-Ward
identities of quantum field theory:
• Leon Takhtajan, Quantum field theories on algebraic curves and A. Weil reciprocity law, arxiv/0812.0169; Quantum field theories on an algebraic curve, pdf, 2000
Created on July 25, 2011 20:11:50 by
Zoran Škoda | {"url":"http://ncatlab.org/nlab/show/Weil+reciprocity+law","timestamp":"2014-04-17T07:30:17Z","content_type":null,"content_length":"12973","record_id":"<urn:uuid:dc9fce90-5a9b-437a-b0fa-613a17db13a2>","cc-path":"CC-MAIN-2014-15/segments/1397609526311.33/warc/CC-MAIN-20140416005206-00491-ip-10-147-4-33.ec2.internal.warc.gz"} |
What do these new symbols mean...?I can't start this without knowing
Ok, good. So you see how k isn't a variable you can really mess with, right? It takes on the values 0 to n just to generate the terms in the sum. In fact, when you expand the sum out, there is no k
appearing anymore because it was just a dummy variable. So you can't do stuff like assume restrict k to just one value to try get the result you want.
Similarly, you're asked to evaluate the sum for any value of n, so you can't set n to any one value.
Now, are you familiar with the binomial theorem? That is, what is the expansion of (a+b)^n? | {"url":"http://www.physicsforums.com/showthread.php?t=531143&page=5","timestamp":"2014-04-18T08:28:25Z","content_type":null,"content_length":"49160","record_id":"<urn:uuid:81604768-8db5-4f4e-b5ea-8824faf5df45>","cc-path":"CC-MAIN-2014-15/segments/1397609533121.28/warc/CC-MAIN-20140416005213-00636-ip-10-147-4-33.ec2.internal.warc.gz"} |
Physics..=> drwls - I need your help
Posted by ~christina~ on Tuesday, February 26, 2008 at 10:26pm.
I need help seeing if my thoughts are correct and how to do some things.
Block 1 of mass 0.200 kg is sliding to the right over a frictionless elevated surface at a speed of 8.00 m / s. The block undergoes an elastic collision with stationary block 2, which is attached to
a spring of spring constant 1208.5 N / m. (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in SHM with a period of 0.140 s, and block 1 slides off the
opposite end of the elevated surface landing a distance d from the base of that surface after falling height h = 4.90 m.
(a) Write an expression that gives the displacement of block 2 as a function of time. This expression must include the values of the amplitude of vibration and the angular frequency.
I came up with
x(t)= Acos(omega*t + pi/2)
not sure about +/- for the angle though and how you know which sign to have.(need help on this determination)
I found omega and m2 through: T= 2pi/omega= 2pi sqrt(m/k)
m1v1i + m2v2i = m1v1f + m2v2f
and found v2f and v1f
I was thinking that the v1f I found was the same velocity that the block 1 leaves with and travels off the table with IS THIS CORRECT?
Then I was thinking of using the v1f and v2f in energy equation to find the distance that the spring compresses (Amplitude) so I can plug it into the equation for cos
1/2mv1f + 1/2m2vf = 1/2kx^2
Is this alright?
b) Use differential calculus to obtain expressions for the velocity and acceleration of block 2 as functions of time.
once again I'm not sure if phi's angle or even if phi is correct.
x(t)= A cos (omega*t + pi/2)
v(t)= -omega A sin( omega*t + pi/2)
a(t)= -omega^2 cos (omega*t + pi/2)
c) What are the displacement, velocity, and acceleration of block 2 at t = 0.520 s?
I think I'd just plug into the equation after I find the values from a
(d) What is the value of d?
I know this is projectile motion problem with I think v in x direction...but if it is then would an angle be included? I think yes but I haven't worked with many problems with a object falling after
sliding off a level surface.
how would I approach this?
Thank you drwls =)
• Physics..=> drwls - I need your help - drwls, Wednesday, February 27, 2008 at 12:36am
a) The displacement of block 2 will be 0 at t=0, and it will vibrate about this position. They already tell you the period is P = 0.140 s. Figure out tne mass m2 from the relation
P = 2 pi sqrt (m2/k) = 0.140 s
m2/k = [P/(2 pi)]^2 = 4.965*10^-4.
I get m2 = 0.600 kg, 3 times the mass of m1.
You need to compute the amplitude of vibration to complete this part. You can get that by using energy and momentum conservation to compute the velocity of m2 right after collision. I believe you
will find this to be 4.00 m/s. Vmax of mass2 equals omega*A, where A is the amplitude. In your case,
omega = 2 pi/P = 44.88 rad/s
Therefore A = (4.00 m/s)/44.88 rad/s) = 8.91*10^-2 m
Another way to get the amplitude is to set (1/2) k A^2 equal to the kinetic energy of m2 right after the collision, which you suggested. It should give the same answer. Try it and see.
The displacement equation for mass 2 is, if I'm right,
X = 8.91*10^-2 m * sin(2 pi t/P)
= 8.91*10^-2 sin (44.88 t)
2) This step is straightforward since you know know X(t) -- assuming I did the calculations correctly. This you need to verify.
3) Yes, just plug in the numbers. Since that is exactly 3 periods later, you should get the same values you had at t=0
4) In doing the elastic collision problem, you should find that mass 1 bpunces back with a velocity of 4.0 m/s. The time it takes to fall a veritcal height of 4.90 m is
t = sqrt (2H/g) = sqrt 1 = 1.00 s
The distance d will therefore be 4.0 m from the base
• Physics..=> drwls - I need your help - ~christina~, Wednesday, February 27, 2008 at 1:05am
X = 8.91*10^-2 m * sin(2 pi t/P)
= 8.91*10^-2 sin (44.88 t)
I don't get this...why did you use sin? and how did you get 2pi t/P ??
I used these formulas below but are they incorrect?
x(t)= A cos (omega*t + pi/2)
v(t)= -omega A sin( omega*t + pi/2)
a(t)= -omega^2 cos (omega*t + pi/2)
P.S.- I was also wondering if there is a good site on the web that can explain the relation of the position of a spring to the sin/cos function since I have problems visualizing the two and which
one I should use for which situation.
• Physics..=> drwls - I need your help - drwls, Wednesday, February 27, 2008 at 3:15am
I used sin because the displacement at time =0, and then it becomes positive at first.
cos (wt + pi/2) is the same thing as -sin wt, anyway.
Your formula is OK if you define positive motion to be opposite to the direction of m1 before impact.
However, in the first part I think they want you to provide the actual value of A.
2 pi t/P is the same thing as w t, since 1/P is the frequency f and
2 pi f = w
• Physics..=> drwls - I need your help - ~christina~, Wednesday, February 27, 2008 at 12:15pm
Thanks drwls
I was wondering though if I used sin then wouldn't the
v(t)= omega A cos (omega* t)?
a(t)= -omega^2 A sin(omega* t)?
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FTC2 - Not sure how to deal with this one
March 31st 2011, 07:13 PM
FTC2 - Not sure how to deal with this one
Find g'(x) by using Part 2 of the Fundamental Theorem of Calculus.
$g(x) = \int^x_0 (1+\sqrt{t})dt$
Here is what I have done so far:
$g(x) = t + \frac{1}{2\sqrt{t}}]^x_0$
$g(x) = x + \frac{1}{2\sqrt{x}} - (0 + \frac{1}{2\sqrt{0}})$
Assuming that these two steps are correct (are they?), how do handle the fact that $\frac{1}{2\sqrt{0}}$ is undefined?
March 31st 2011, 07:23 PM
Those steps aren't quite right, consider
$\displaystyle \int t^{\frac{1}{2}} ~dt= \frac{2}{3}\times t^{\frac{3}{2}}+C$
March 31st 2011, 07:28 PM
Obviously, you are able to compute the integral and then take the derivative. This will not always be the case, and eventually (!) you will become familiar with the following:
$g(x) = \int^x_a f(t)dt = f(x)$
Oh... it's on wikipedia!
Fundamental theorem of calculus - Wikipedia, the free encyclopedia
Just replace the "t" inside the integral with x.
If your upper limit has an "interesting" derivative, you'll need to use the chain rule.
March 31st 2011, 08:18 PM
I took the derivative instead of the antiderivative. Duh. Thanks, pickslides, for pointing that out. Chaz, the question specifically requires that I use part 2 of the FTC. Your post uses part 1
(at least that's what our book calls them). The question actually makes us do it both ways. I got the answer easily using part 1, but when I did it with part 2, I got hung up. Now I know why.
Thanks to both of you. | {"url":"http://mathhelpforum.com/calculus/176484-ftc2-not-sure-how-deal-one-print.html","timestamp":"2014-04-16T14:02:48Z","content_type":null,"content_length":"6559","record_id":"<urn:uuid:b7c5631e-2ebe-492a-b139-c9a184952824>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00300-ip-10-147-4-33.ec2.internal.warc.gz"} |
Clock Class HELP
11-04-2006 #1
Is Trying to Learn
Join Date
Mar 2006
Hutton, Preston
Clock Class HELP
hi i have a problem with another class problem.
i have this code so far but i am getting a few problems
1] it comes up with 2 times and i dont want that
2] i would like it so output seconds
3]T3 = T1.Add(T2); T2 is added to T1 and the result returned is the sum of the two times which is assigned to T3.
4]Will initialise the object T1 to 15:20:35.
at the moment i have not managed to get it into a function or add the 2 times to to output T3.
can anyone help?
here is my code
#include <iostream>
using namespace std;
class Time
Time(int hours2, int minutes2);
int getHours() const;
int getMinutes() const;
void setHours(const int hours2);
void setMinutes(const int minutes2);
Time operator+(const Time &t1) const;
void show() const;
int hours, minutes;
hours = minutes=0;
Time::Time(int hours2, int minutes2)
hours = hours2;
minutes = minutes2;
int Time::getHours() const
return hours;
int Time::getMinutes() const
return minutes;
void Time::setHours(const int hours2)
hours = hours2;
void Time::setMinutes(const int minutes2)
minutes = minutes2;
Time Time::operator+(const Time &t1) const
Time sum;
sum.minutes = minutes + t1.minutes;
sum.hours = hours + t1.hours + sum.minutes/60;
sum.minutes %=60;
return sum;
void Time::show() const
int main()
Time t1;
Time t2;
Time t3;
t3 = t1 + t2;
I don't really understand your question . . . .
Right now I can see a few things:
□ system() is in <cstdlib>. There are better ways to keep the console window open: http://faq.cprogramming.com/cgi-bin/...&id=1043284385
□ You call t3.show() before you set t3 to anything, which will output 0:0. Is that what you wanted?
□ How is Time:
Seek and ye shall find. quaere et invenies.
"Simplicity does not precede complexity, but follows it." -- Alan Perlis
"Testing can only prove the presence of bugs, not their absence." -- Edsger Dijkstra
"The only real mistake is the one from which we learn nothing." -- John Powell
Other boards: DaniWeb, TPS
Unofficial Wiki FAQ: cpwiki.sf.net
My website: http://dwks.theprogrammingsite.com/
Projects: codeform, xuni, atlantis, nort, etc.
Did you write this code yourself? If so, what's the problem?
have this code so far but i am getting a few problems
1] it comes up with 2 times and i dont want that
Time::show is called twice in main(), do you mean that?
2] i would like it so output seconds
Well, add int seconds to the class members and change all the methods...
3]T3 = T1.Add(T2); T2 is added to T1 and the result returned is the sum of the two times which is assigned to T3.
I see an overloaded + operator. Time::Add would be very similar.
4]Will initialise the object T1 to 15:20:35.
Change the constructor...
at the moment i have not managed to get it into a function or add the 2 times to to output T3.
After you assign t3 = t1 + t2, it shows 2:25, as it should???
11-04-2006 #2
11-04-2006 #3
The larch
Join Date
May 2006 | {"url":"http://cboard.cprogramming.com/cplusplus-programming/84965-clock-class-help.html","timestamp":"2014-04-19T18:02:01Z","content_type":null,"content_length":"50906","record_id":"<urn:uuid:1c52edba-3af9-40e0-b17f-2e82011e273a>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00187-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Composing two-term sums from the same primes
up vote 7 down vote favorite
The following is an old result of Erdős and Turán (American Mathematical Monthly, 1934):
Given a set of $2^n + 1$ distinct positive integers, all of its two-term sums cannot be composed of the same $n$ primes. (An integer $m$ is said to be composed of primes $p_1, p_2, \dots$ is every
prime factor of $m$ is one of those primes. And in two-term sums we exclude sums of two copies of the same integer.)
Let $f(n)$ be the minimum number such that given $f(n)$ distinct positive integers, all its two-term sums cannot be composed of the same $n$ primes. Erdős and Turán conjecture that the upper bound $f
(n) \le 2^n + 1$ can be greatly improved. For example, they conjecture that $f(n) = O(n^{1+ \epsilon})$ for every fixed $\epsilon > 0$.
Does anyone know if any progress has been made on this problem since? In particular is there any polynomial upper bound known?
Also I would be interested to know about lower bounds on $f(n)$.
nt.number-theory prime-numbers
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1 Answer
active oldest votes
This problem and generalizations of it are discussed in the following papers:
1. P. Erdős, A. Sárközy, C. Stewart, On prime factors of subset sums. J. London Math. Soc. (2) 49 (1994), no. 2, 209–218.
2. C. Stewart, On prime factors of integers which are sums or shifted products. Anatomy of integers, 275–287, CRM Proc. Lecture Notes, 46, Amer. Math. Soc., Providence, RI,
3. C. Stewart, R. Tijdeman, On prime factors of sums of integers. II. Diophantine analysis (Kensington, 1985), 83–98, London Math. Soc. Lecture Note Ser., 109, Cambridge Univ.
Press, Cambridge, 1986.
I am fairly sure that no improvement to the Erdos-Turan result is known. However, there is a very closely related problem where one considers two sets. First note we can
reformulate the Erdos-Turan result as the estimate
up vote 5 down vote $$w(\prod_{a,b \in S} (a+b)) \gg \log(|S|)$$
where $w$ denotes the number of prime factors of an integer. Now a natural generalization of the above is the inequality
$$w(\prod_{a \in A, b \in B, |A|=|B|=k} (a+b)) \gg \log(k)$$
This was conjectured by Erdos and Turan and proved by Gyory, Steward, and Tijdeman in 1986. In this more general setting it was proved by Erdos, Steward and Tijdeman in 1994
that this estimate is nearly optimal. More specifically,
$$w(\prod_{a \in A, b \in B, |A|=|B|=k} (a+b)) \geq (1/8+\epsilon) \log(k)^2 \log\log(k)$$ can't hold for any $\epsilon >0$.
Thanks very much for the references. – Matthew Kahle May 24 '13 at 16:29
add comment
Not the answer you're looking for? Browse other questions tagged nt.number-theory prime-numbers or ask your own question. | {"url":"http://mathoverflow.net/questions/129645/composing-two-term-sums-from-the-same-primes","timestamp":"2014-04-20T03:35:55Z","content_type":null,"content_length":"53063","record_id":"<urn:uuid:43e6a98e-49bf-4ba5-af29-f07977c80756>","cc-path":"CC-MAIN-2014-15/segments/1397609537864.21/warc/CC-MAIN-20140416005217-00232-ip-10-147-4-33.ec2.internal.warc.gz"} |
Math 3rd grade | Digital clock reading: How much later? 10 to and 5 past, etc., max. difference 6 hours
3rd Grade: Digital clock reading: How much later? 10 to and 5 past, etc., max. difference 6 hours
You need to enable javascript for this exercise to work.
The answer you've given is incorrect:
The correct answer is: | {"url":"http://www.mathabc.com/math-3rd-grade/time/digital-clock-reading-how-much-later-10-to-and-5-past-etc-max-difference-6-hours/id/216","timestamp":"2014-04-18T11:36:22Z","content_type":null,"content_length":"26115","record_id":"<urn:uuid:49e586f2-1367-4e8d-b99f-72086787bd06>","cc-path":"CC-MAIN-2014-15/segments/1398223206672.15/warc/CC-MAIN-20140423032006-00383-ip-10-147-4-33.ec2.internal.warc.gz"} |
Level set examples
Example 1
Let $f(x,y) = x^2-y^2$. We will study the level curves $c=x^2-y^2$.
First, look at the case $c=0$. The level curve equation $x^2-y^2=0$ factors to $(x-y)(x+y)=0$. This equation is satisfied if either $y=x$ or $y=-x$. Both these are equations for lines, so the level
curve for $c=0$ is two lines.
If $c \ne 0$, then we can rewrite the level curve equation $c=x^2-y^2$ as \begin{align*} 1 = \frac{x^2}{c} - \frac{y^2}{c}. \end{align*} If you remember you conic sections, you'll recognize this as
the equation for a hyperbola. If $c$ is positive, the hyperbolas open to the left and right. If $c$ is negative, the hyperbolas open up and down.
For example, if $c=1$, the equation is $x^2-y^2=1$. If $c=-1$, the equation is $y^2-x^2=1$. A number of level curves are plotted below.
We can “stack” these level curves on top of one another to form the graph of the function. Below, the level curves are the thick blue lines. Drag the green point from 0 to 1. When the point is at 1,
each level curve given by $c=f(x,y)$ will be at the height $z=c$.
Level curves of a hyperbolic paraboloid. When the green point is at 0, as it is in the default view, the figure shows a standard level curve plot of $f(x,y)=x^2-y^2$, though it is floating in a three
dimensional space. When you drag the green point to 1, each level curve $f(x,y)=c$ moves to the height $z=c$, so that they are in the same position as in the graph of $z=f(x,y)$. In this way, the
figure demonstrates the correspondence between the level curve plot and the graph of the function.
This is yet another way to visualize the relationship between the level curves and the graph of $z=f(x,y)$ shown below, which is a hyperbolic paraboloid.
Graph of a hyperbolic paraboloid. The graph of the function $f(x,y)=x^2-y^2$.
Example 2
Let $f(x,y,z) = x^2+y^2+z^2$. Although we cannot plot the graph of this function, we can graph some of its level surfaces. The equation for a level surface, $x^2+y^2+z^2=c$, is the equation for a
sphere of radius $\sqrt{c}$.
Spherical level surfaces. The level surfaces $f(x,y,z) = x^2+y^2+z^2=c$ are spheres of radius $\sqrt{c}$. The level surface with $c=1$ is the sphere of radius 1 drawn in dark red. The level surface
with $c=4$ is the sphere of radius 2 drawn in light green. | {"url":"http://mathinsight.org/level_set_examples","timestamp":"2014-04-17T06:43:42Z","content_type":null,"content_length":"22328","record_id":"<urn:uuid:5b6a9afb-791c-49e4-9a93-d32417e21436>","cc-path":"CC-MAIN-2014-15/segments/1398223206770.7/warc/CC-MAIN-20140423032006-00171-ip-10-147-4-33.ec2.internal.warc.gz"} |
The HoTT book is finished!
Since spring, and even before that, I have participated in a great collaborative effort on writing a book on Homotopy Type Theory. It is finally finished and ready for public consumption. You can get
the book freely at http://homotopytypetheory.org/book/. Mike Shulman has written about the contents of the book, so I am not going to repeat that here. Instead, I would like to comment on the
socio-technological aspects of making the book, and in particular about what we learned from open-source community about collaborative research.
We are a group of two dozen mathematicians who wrote a 600 page book in less than half a year. This is quite amazing, since mathematicians do not normally work together in large groups. In a small
group they can get away with using obsolete technology, such as sending each other source LaTeX files by email, but with two dozen people even Dropbox or any other file synchronization system would
have failed miserably. Luckily, many of us are computer scientists disguised as mathematicians, so we knew how to tackle the logistics. We used git and github.com. In the beginning it took some
convincing and getting used to, although it was not too bad. In the end the repository served not only as an archive for our files, but also as a central hub for planning and discussions. For several
months I checked github more often than email and Facbook. Github was my Facebook (without the cute kittens). If you do not know about tools like git but you write scientific papers (or you create
any kind of digital content) you really, really should learn about revision control systems. Even as a sole author of a paper you will profit from learning how to use one, not to mention that you can
make pretty videos of how you wrote your paper.
But more importantly, the spirit of collaboration that pervaded our group at the Institute for Advanced Study was truly amazing. We did not fragment. We talked, shared ideas, explained things to each
other, and completely forgot who did what (so much in fact that we had to put some effort into reconstruction of history lest it be forgotten forever). The result was a substantial increase in
productivity. There is a lesson to be learned here (other than the fact that the Institute for Advanced Study is the world’s premier research institution), namely that mathematicians benefit from
being a little less possessive about their ideas and results. I know, I know, academic careers depend on proper credit being given and so on, but really those are just the idiosyncrasies of our time.
If we can get mathematicians to share half-baked ideas, not to worry who contributed what to a paper, or even who the authors are, then we will reach a new and unimagined level of productivity.
Progress is made by those who dare break the rules.
Truly open research habitats cannot be obstructed by copyright, profit-grabbing publishers, patents, commercial secrets, and funding schemes that are based on faulty achievement metrics.
Unfortunately we are all caught up in a system which suffers from all of these evils. But we made a small step in the right direction by making the book source code freely available under a
permissive Creative Commons license. Anyone can take the book and modify it, send us improvements and corrections, translate it, or even sell it without giving us any money. (If you twitched a little
bit when you read that sentence then the system has gotten to you.)
We decided not to publish the book with an academic publisher at present because we wanted to make it available to everyone fast and at no cost. The book can be freely downloaded, as well as bought
cheaply in hardcover and paperback versions from lulu.com. (When was the last time you paid under $30 for a 600 page hardcover academic monograph?) Again, I can feel some people thinking “oh but a
real academic publisher bestows quality”. This sort of thinking is reminiscent of Wikipedia vs. Britannica arguments, and we all know how that story ended. Yes, good quality of research must be
ensured. But once we accept the fact that anyone can publish anything on the Internet for the whole world to see, and make a cheap professionally looking book out of it, we quickly realize that
censure is not effective anymore. Instead we need a decentralized system of endorsments which cannot be manipulated by special interest groups. Things are moving in this direction with the recently
established Selected Papers Network and similar efforts. I hope these will catch on.
However, there is something else we can do. It is more radical, but also more useful. Rather than letting people only evaluate papers, why not give them a chance to participate and improve them as
well? Put all your papers on github and let others discuss them, open issues, fork them, improve them, and send you corrections. Does it sound crazy? Of course it does, open source also sounded crazy
when Richard Stallman announced his manifesto. Let us be honest, who is going to steal your LaTeX source code? There are much more valuable things to be stolen. If you are tenured professor you can
afford to lead the way. Have your grad student teach you git and put your stuff somewhere publicly. Do not be afraid, they tenured you to do such things.
So we are inviting everyone to help us improve the book by participating on github. You can leave comments, point out errors, or even better, make corrections yourself! We are not going to worry who
you are, how much you are contributing, and who shall take credit. The only thing that matters is whether your contributions are any good.
My last observation is about formalization of mathematics. Mathematicians like to imagine that their papers could in principle be formalized in set theory. This gives them a feeling of security, not
unlike the one experienced by a devout man entering a venerable cathedral. It is a form of faith professed by logicians. Homotopy Type Theory is an alternative foundation to set theory. We too claim
that ordinary mathematics can in principle be formalized in homotopy type theory. But guess what, you do not have to take our word for it! We have formalized the hardest parts of the HoTT book and
verified the proofs with computer proof assistants. Not once but twice. And we formalized first, then we wrote the book because it was easier to formalize. We win on all counts (if there is a race).
I hope you like the book, it contains an amazing amount of new mathematics. | {"url":"http://audrey.fmf.uni-lj.si/hott.html","timestamp":"2014-04-19T14:29:34Z","content_type":null,"content_length":"8213","record_id":"<urn:uuid:924ec3b4-b7da-42f6-81ca-6eea8af8b98e>","cc-path":"CC-MAIN-2014-15/segments/1397609537271.8/warc/CC-MAIN-20140416005217-00310-ip-10-147-4-33.ec2.internal.warc.gz"} |
Finding unknown integer-valued polynomials using inequalities
up vote 5 down vote favorite
I've come across this interesting inequalities problem recently, which seemed straight-forward at first glance but has proven interesting enough to ask about it here.
Suppose you are given the degree of an unknown polynomial, and are told that all the coefficients are integers and are all within $min \le a\_k \le max$. Also, you have access to an oracle who will
evaluate p(q), the unknown polynomial, and compare it to your guess $g\_q$, where q is the number of previous guesses you have made, and determine if your guess was less, greater than, or exactly
equal to the unknown polynomial.
What is the optimal method of choosing guesses, given previous results, in order to make a correct guess as soon as possible in the worst case?
If the degree is d and the coefficients are bounded by [min, max] then it would seem the best possible method would require slightly less than $ \frac {d \log (max - min + 1)}{\log 2}$ by binary
search. Its slightly less because with an answer of > from the oracle, you exclude both the values less than and those equal to the guess, which could be more than 50% of possibilities.
If the median of the evaluated values at q of all of the remaining possible polynomials can be found, than guessing that value would be guaranteed to eliminate at least half of the possibilities. But
is there any efficient way to find the median of a function of a set of polynomials that are only identified by inequalities?
For the first guess, q is zero and therefore all but the constant term of the polynomial are irrelevant. It makes sense then to make g(0) to be $\frac {min + max}{2}$. But after that, the best way of
finding the median quickly seems elusive.
As an example, make min = -100, max = 100 and d = 1. $\frac {-100 + 100}{2}$ = 0, so g(0) should be 0. If the oracle returns > then we know $0 < a\_0 \le 100$ or $1 \le a\_0 \le 100$ since we are
dealing with integer coefficients.
An ideal method would include an efficient way to find the median and characterize the set of possibilities given the previous answers from the oracle. But medians can be hard to calculate so an
efficient method to calculate the mean of p(q) for the set of possibilities would be close enough if an efficient method doesn't exist for medians.
algorithms computer-science polynomials
What are p and g? – Qiaochu Yuan Feb 1 '10 at 16:16
1 p is (obviously) the unknown polynomial. g is (as stated) your guess. – TonyK Feb 1 '10 at 17:10
I can replace the g(q) notation with $g\_q$ if that makes more sense to use subscript notation. I was just trying to describe the set of guesses made in a compact way. Each guess of course depends
on the answer from the previous one, so I can see why describing at as a function is confusing. – Anthony Peterson Feb 1 '10 at 20:55
1) The best method does not have to use medians because it is conceivable that what is lost at a given step is regained at the next step. 2) A sensible approach might be to try proving that using
means is a good algorithm, but though looking simpler, even finding a mean is a hard computational problem since it amounts to integration of a polynomial on a polyhedron. This is a
computationally hard problem: see for example math.ucdavis.edu/~deloera/RECENT_WORK/integrationsimplex.pdf – Boris Bukh Feb 1 '10 at 21:14
Thanks for the link, I will look into it. – Anthony Peterson Feb 2 '10 at 2:26
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1 Answer
active oldest votes
Just determine the coefficient at the highest power first by plugging in huge numbers and doing binary search (comparing to what you'd get for the half-integer coefficients). Once you know
it, you can easily figure out the coefficient at the next power in the same way and so on.
Now, if you also have a bound on $q$ you can plug in, it becomes interesting.
Oops, sorry for misunderstanding.
up vote OK, you can easily do $Cd^M\log N$ then. The trick is that no matter how you split the d-dimensional simplex of volume 1 by a hyperplane through its center and no matter which piece you'll
2 down take, you'll be able to find a simplex of volume $1-d^{-m}$ conatining this piece where $m$ is some fixed number, which I will need some time to compute precisely (my educated guess would be
vote m=4). Now just use this fact to obtain a simplex of either volume less than $N^{-2d}$ or with one vertex outside the ball of radius $N^{3d}$ containing the set of remaining polynomials after
just $O(d^M\log N)$ steps. In both cases, you'll be able to find a linear dependence between the coefficients that is precise up to $N^{-2}$, which means that you can eliminate one
coefficient from the polynomial entirely (the $d$-th power of the variable is still well below $N$, so the precision is enough to distinguish the integer values and each exactly attained
equality which will not allow you to tell for sure which half you are in gives you food for interpolation).
Sorry if it is too sketchy. I'll try to edit it into something more reasonable later unless somebody gives a better solution.
The question was that one can ask question about q(0), then about q(1), then about q(2), etc. One cannot ask question about arbitrary values in arbitrary order. – Boris Bukh Feb 1 '10 at
Yeah, if you can compare your guess to arbitrary k in p(k) then your technique works. You can go backwards, from the lowest power as well. But my curiosity was based on the idea that you
can't pick what x is in p(x), you have to use all different values of x, and therefore its the inequalities or some other representation of them that has to be worked with. – Anthony
Peterson Feb 2 '10 at 2:30
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Braingle: 'Find the Sum' Brain Teaser
Find the Sum
Math brain teasers require computations to solve.
Puzzle ID: #39976
Category: Math
Submitted By: shenqiang
Find the sum of all prime numbers between 1 and 100 that are simultaneously 1 greater than a multiple of 4 and 1 less than a multiple of 5.
Numbers that are 1 greater than a multiple of 4 are 5, 9, 13, 17, 21... where 9 is 1 less than a multiple of 5. lcm(4, 5)=20, therefore the pattern repeats in each 20 numbers.
9 (composite) 29 (prime) 49 (composite) 69 (composite) 89 (prime)
Therefore, the sum of all such numbers is 29+89=118. Hide
What Next?
javaguru I did it a little different. The numbers on either side must be even. The number below is 2 x an even number, so the number above is 2 x 5 x an odd number. That gives the odd multiples of
Jan 13, 10 as the number above the prime, leaving just the 5 numbers to check. | {"url":"http://www.braingle.com/brainteasers/teaser.php?id=39976&op=2&comm=1","timestamp":"2014-04-21T12:25:17Z","content_type":null,"content_length":"24679","record_id":"<urn:uuid:0d62febe-424d-4a3b-ad1d-4b352150d84d>","cc-path":"CC-MAIN-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00039-ip-10-147-4-33.ec2.internal.warc.gz"} |
The 10 gallon hat.
November 19th 2010, 04:40 PM #1
Jul 2010
The 10 gallon hat.
anyway i was telling my friend a riddle about a dude who just bought a 10 gallon hat but its really 6 gallons and all this determining of its real volume but he interrupted me and claimed that no
one can have a 10 gallon hate because no one can have a 10 gallon head. I found one gallon is 3785.4 cubic centimeters, figuring a head is a sphere i did the cubed root and divided by 4/3 to get
25, i then multiplyed by 2 to get a diameter of around 50 centimeters. is this right? check me on everything please
Last edited by mr fantastic; November 19th 2010 at 08:17 PM. Reason: Title.
ps: i did 3785.4 times 10 before i did cubed root
anyway i was telling my friend a riddle about a dude who just bought a 10 gallon hat but its really 6 gallons and all this determining of its real volume but he interrupted me and claimed that no
one can have a 10 gallon hate because no one can have a 10 gallon head. I found one gallon is 3785.4 cubic centimeters, figuring a head is a sphere i did the cubed root and divided by 4/3 to get
25, i then multiplyed by 2 to get a diameter of around 50 centimeters. is this right? check me on everything please
$10 \, gal = 37,854 \, cm^3$
$\displaystyle \frac{4}{3}\pi r^3 = V$
$\displaystyle r^3 = \frac{3V}{4\pi}$
$\displaystyle r = \sqrt[3]{\frac{3V}{4\pi}}<br />$
try again ...
oh so that means the radius is 20.8 cm?
November 19th 2010, 04:44 PM #2
Jul 2010
November 19th 2010, 04:53 PM #3
November 19th 2010, 05:03 PM #4
Jul 2010
November 19th 2010, 05:09 PM #5
November 21st 2010, 04:19 AM #6 | {"url":"http://mathhelpforum.com/math-topics/163787-10-gallon-hat.html","timestamp":"2014-04-17T11:21:54Z","content_type":null,"content_length":"43931","record_id":"<urn:uuid:0ef5a0bc-0268-4026-a898-53927bfc4669>","cc-path":"CC-MAIN-2014-15/segments/1397609527423.39/warc/CC-MAIN-20140416005207-00182-ip-10-147-4-33.ec2.internal.warc.gz"} |
About Sectional Curvature
up vote 5 down vote favorite
In a paper by Yann Ollivier:
Let $x$ be a point in $X$, $v$ a small tangent vector at $x$, $y$ the endpoint of $v$, $w_x$ a small tangent vector at $x$, and $w_y$ the parallel transport of $w_x$ from $x$ to $y$ along $v$. If,
instead of a Riemannian manifold, we were working in ordinary Euclidean space, the endpoints $x_0$ and $y_0$ of $w_x$ and $w_y$ would constitute a rectangle with $x$ and $y$. But in a manifold,
generally these four points do not constitute a rectangle any more.
Indeed, because of curvature, the two geodesics starting along $w_x$ and $w_y$ may diverge from or converge towards each other. Thus, on a sphere (positive curvature), two meridians starting at two
points on the equator have parallel initial velocities, yet they converge at the North (and South) pole. Since the initial velocities $w_x$ and $w_y$ are parallel to each other, this effect is at
second order in the distance along the geodesics (Fig.).
Thus, let us consider the points lying at distance $\varepsilon $ from $x$ and $y$ on the geodesics starting along $w_x$ and $w_y$, respectively. In a Euclidean setting, the distance between those
two points would be $|v|$, the same as the distance between $x$ and $y$. The discrepancy from this Euclidean case is used as a definition of a curvature.
Definition(Sectional curvature). Let $(X, d)$ be a Riemannian manifold. Let $v$ and $w_x$ be two unit-length tangent vectors at some point $x \in X$. Let $\varepsilon, \delta > 0$. Let $y$ be the
endpoint of $v$ and let $w_y$ be obtained by parallel transport of $w_x$ from $x$ to $y$. Then $$ d(exp_{x} \varepsilon w_x, exp_{y} \varepsilon w_y) = \delta (1 −\frac{\varepsilon^2}{2} K(v,w)+ O(\
varepsilon^3 +\delta \varepsilon^2)) $$
when $\delta , \varepsilon \to 0$. This defines a quantity $K(v,w)$, which is the sectional curvature at $x$ in the directions $(v,w)$.
Question1 How can I derive the formula $ d(exp_{x} \varepsilon w_x, exp_{y} \varepsilon w_y) = \delta (1 −\frac{\varepsilon^2}{2} K(v,w)+ O(\varepsilon^3 +\delta \varepsilon^2)) $ from figure?
Question2 How this definition of sectional curvature can be derived from its usual definition ($K(v,w)=\frac{\langle R(v,w)w, v\rangle}{\langle v,v\rangle \langle w,w \rangle - \langle v,w \rangle ^
Thanks in advance for your time.
dg.differential-geometry manifolds curvature riemannian-geometry
2 Hi, welcome to Mathoverflow. As there are large number of user overlaps between MO and Math.StackExchange, generally we discourage immediate cross posting a question between the two forums. For
future reference, please just post your question in one of the two places. If you don't get a response after a reasonable amount of time (say, about a week), then maybe it would be worthwhile
considering cross posting to the other forum. Thanks. – Willie Wong Apr 10 '13 at 12:48
2 In the interest of not duplicating efforts, a partial answer has already been given on Math.SE: math.stackexchange.com/questions/356989/… – Willie Wong Apr 10 '13 at 12:48
@Willie Wong, Ok,thanks. – Sepideh Bakhoda Apr 10 '13 at 13:19
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closed as off-topic by Benoît Kloeckner, David White, Carlo Beenakker, Kevin P. Costello, j.c. Oct 23 '13 at 9:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Benoît Kloeckner, David White, Carlo Beenakker, Kevin P. Costello, j.c.
If this question can be reworded to fit the rules in the help center, please edit the question.
1 Answer
active oldest votes
To supplement the answer given at SE, I would mention that the definition via the Taylor expansion of the distance function is really a slick shortcut rather than a definition. It relies on
the fact that the Gaussian curvature at $x$ of the surface obtained as the image of the plane spanned by $v$ and $w$ is the same as the sectional curvature in this 2-dimensional direction.
up vote 1 So to convince yourself of the formula, you just need to understand it for surfaces. Trying to relate it to the general formula in terms of the curvature tensor is a red herring.
down vote
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More Grumbine Science
The subject line is close to a recent search query that lead someone here, and echoes a comment that's not unusual in blog comment sections. The thing about it is, it's not a very
strong question
One part of the question's failure is that it isn't really a proper statistical question. Given limits of search strings, that's no surprise. But it does show up in comments (usually in the vein of
assertion "150 years is too small a statistical sample of 4.5 billion years of climate.") where such limits don't apply.
The basis of any good question is to try to understand something. If what you are trying to understand is not statistical, then pursuing statistics is not going to help you. "What is your height?" is
marginally statistical. If you measure the length of a short metal bar many times, which I did in freshman physics lab, you'll get slightly differing answers. So you may well answer statistically
with your mean and a sample standard deviation. This is only marginally statistical, in that it is purely descriptive statistics and no hypothesis testing is involved.
Your question, however, could be non-statistical: Did it rain in my back yard last night? How much rain did my rain gauge capture? If so, you will be wasting your time if you chase after statistical
tables. More at hand, someone raising statistics in a blog debate about a non-statistical question is wasting your time.
But suppose that what you're trying to understand truly does require statistical considerations beyond minor description. What would such a question look like? One could be "Given the population of
voters in the USA, how many randomly selected voters would I need to ask a yes/no question in order to have a standard error in my sample of 3% or less as compared to asking the everybody?" or, for
more climate-related flavor "How many satellite observations with a (known) standard error of measurement would I need to find an average sea surface temperature whose standard error would be less
than 0.1 K?"
Required is a description of the 'population' -- all possible observations (the population of voters in the USA, satellite observations) -- of your 'sample' (ask some number, which we want to
determine, of voters, or satellite observations -- and what statistic you are trying to estimate (% of voters who like your candidate, mean sea surface temperature).
So, back to the original question. Does it describe the population? No. 'data' certainly doesn't limit us to anything in particular. I'll guess that it is global mean temperature, of the earth (maybe
it's Mars -- I've read interesting papers about Martian climatology), that is the point of interest. But we shouldn't have to guess, a good question is clear on what it is asking about. Does it
describe a sampling method? Not really. 150 years, well, I'll guess that this means 'take the most recent 150 years'. Most importantly, however, does it describe what statistic one is trying to
estimate? No. Again, I'll guess: that it is "Global mean temperature over the entire history of the earth."
Putting it all together, the statistical question more reasonably posed is something like "Does the last 150 years of global mean air temperatures provide a good estimate of the global mean air
temperature through the history of the earth?" (better would be to define 'good', say 'standard error within 0.2 K')
That's the statistical side. But since we're not interested solely in statistical questions regarding climate, at least not most of us, we also have to ask, "Is this a
meaningful question?" The person who did the search, I don't know what they have in mind. They could indeed have in mind a question for which the mean temperature of the planet throughout its history
is exactly the right number to answer.
Usually, though, comments about the 150 years vs. 4.5 billion surface in debates about modern climate and modern climate change. I'll have to invite contributions on what relevance the planetary mean
temperature from 4.5 billion years ago to ... oh, let's say 30 million years ago ... has to questions of current climate and climate change. Certainly one wants to know how climate has changed
through all of time, and why. I'm one such person. But at hand is those who argue that the global mean over that entire period matters.
What's important regarding human responses to climate change is the climate on human time scales. The last 150 years handily covers me back through my great-grandparents. The next 150 years will
cover my children, grandchildren, and possibly great-grandchildren. Even a 'mere' million years ago, much less 4.5 billion, there weren't any humans around to think about climate change at all. So
the 4.5 billion years is, for 'what do we do now' questions, a spectacularly large red herring.
Human society infrastructure also dictates a much shorter term (than 4.5 billion years) concern. Almost every mile of paved road in the world is less than 150 years old. Almost every mile of
railroad. Almost all port structures. Absolutely all air transport terminals are less than 150 years old. Absolutely all electrical distribution structure, all phone lines, and even more so all cell
phone towers are less than 150 years old. Coal, oil, natural gas distribution networks, again, are almost entirely or entirely less than 150 years old. Many of the world's major cities are less than
150 years old (I'll count Chicago here, as the great fire in 1871 erased so much of the city).
The climate-related concern here is that all these things were constructed based on what climate was like around the time of their construction. That climate drove what the standards would be for,
say, tolerance to flooding, tolerance to drought, high winds, high rain rates, high and low temperatures, and so on. If climate changes outside the range that the infrastructure was built for --
almost all of it globally being much less than 150 years old -- then there is a serious risk that the structure will fail when it encounters brand new climate conditions.
In this vein, then, comments about the 'we did fine in the medieval warm period' are a different flavor of red herring. Chicago, Sao Paulo, Melbourne, Johannesburg, ... didn't exist back then. They
did not 'do fine' in the medieval warm period, they never encountered it. Even cities that did, such as London, Rome, Xi'an, did so with far smaller populations than today's. London, ca. 1100, had a
population around 18,000, and is about 1000 times larger today (taking the metro area). Rome was about 20,000 around 1100 AD, vs. 200 times as large now. Xi'an was among the largest of cities in the
world then, and may have been about 500,000 around 1100 AD. But almost 20 times that today. (mostly Wikipedia figures). While a Rome or London of 10-20,000 survived a medieval warm period ok, Rome
and London of many millions have never see it before.
As usual, we can get far by asking two questions: "Is this question statistically meaningful?" "Is the question physically meaningful?" After looking at the former, regarding the original search
string, and not infrequent comment, we see that it isn't a meaningful statistical question. After rephrasing it to something that is statistically good, we check to see whether it makes physical
sense. Turns out, for the sort of thing I'm concerned about and others use it, that even the rephrased question isn't physically meaningful. Knowing that global mean temperature for the last 4.5
billion years was 20 C (I make up a number) as opposed to 15 C still doesn't tell us whether major modern cities, and their millions of residents, will be able to manage likely climate changes. Nor
does it tell us what adaptations to take, much less how expensive it would be to make those adaptations. And it doesn't tell us what the costs, both dollars and lives, would be if there were no
I'll throw this out for discussion: Are scientists in general introverted? Certainly that's the stereotype in the US. And my experiences have supported that general view. Compared to the US norm, I'm
in the middle between quiet/retiring and loud/outgoing, but at a scientific meeting, I'm well towards the outgoing end of the spectrum.
Since a fair number of non-US readers are here, additional questions are
a) Is that same stereotype common in your country (and what country is it)?
b) Do the scientists in general follow it?
This was brought to mind by some recent meetings I've been at which included non-US scientists, or a meeting that was just US folks, but not just scientists. It looked like the non-US scientists
weren't nearly as quiet/retiring as the US norm.
If anyone knows of some research on the subject, that'll be even better than anecdotes!
The first sea ice outlook from ARCUS is now out. Have a look at the full report at the link. You'll see my name there, with what turned out to be a boringly typical guess. Actually, we all seem to
have been pretty much of the same mind as to amounts, and to distributions around those amounts -- somewhere about 4.9 million km^2, with error ranges of about 0.5 million km^2.
The interesting part being that we arrive at such similar estimates for both value and variation around it by such different means. Do look at the descriptions of how groups arrived at their
predictions. I'll spend more time here on my method later. The only particularly different guess (3.2 million km^2) was from a group (Arbetter et al.) that provided 2, and labelled the other one
(4.7) their outlook prediction.
Rather like John Wilkins, I'm loath to nominate my own writing, but not about asking y'all to enter some nominations (including for my articles :-)
I've entered a link to the submission form on the right hand side of the blog, and here it is again:
All articles since 1 December 2008 are eligible. You can see the current (as I write) listing of nominated articles at Block Around the Clock, and updates will appear there routinely.
I'll also take this chance to invite comments here on which articles you've thought were the best, or worst, what you'd like to see more of, or less of, and so on.
William Connolley and I (both people who have worked professionally on sea ice) are arranging our bet regarding this year's minimum sea ice cover. I've already mentioned a bit about my prediction,
which is for this year's average sea ice extent for September, in the Arctic, as computed by the NSIDC, to be about 4.92 million km^2. To the extent that my working model is good, the standard error
of that estimate is about 0.5 million km^2.
In making that prediction, I'm taking the approach that the last two years' dramatic low covers were part of a continuing process of decline in the ice pack, though a fair portion of it still being
peculiar circumstances of Arctic weather in summer/fall 2007 and 2008. This is not the usual view in the field, where the most common take is that 2007 represents a step change in the system -- one
sort of thing going on before 2007, massive change in 2007, and things should continue more or less similarly for a number of years. (Until the next step change.)
William is taking the view that 2007 and 2008 were just bizarre years, not a fundamental change as most are taking it. The reasonable prediction in his view is to take the years 1979-2006, run a
straight line through them, and use that as your basis. I've finally done the math for it (not at all difficult in the spreadsheet but I've been running around), and his prediction for 2009 is 5.84
million km^2 (also with about 0.5 million km^2 standard error -- but I'm rounding down for his figure, and rounding up for mine. That is, my scheme makes a better fit than his.).
So William, how about even quatloos over/under 5.38 million km^2? If it's under, you pay me 50, if over, I pay you 50? If your approach is right, 84% of the time you'll win this. If mine is, it's 84%
of the time that I'll win. So, as you advised, it's a wager we each think is biased in our favor.
It's really a matter of parenting scientists, but this will appear on Father's Day. That is, suppose you're a parent and would like your child (children) to appreciate science. Not that they should
become scientists -- I don't believe that this is the only career to pursue even though I did do so. (I also had my time pursuing engineering and computer science. I ended up going with the science,
but it was a real question up through completing my undergraduate degree and receiving an engineering job offer.)
In this, I'll largely be drawing from examples my mother set. A good parent learns from all good parents, not just those of their own gender.
As a parent, perhaps the greatest thing you can support in your kids is the idea "The universe is a very interesting place." You don't need to know all the science yourself. It might even be a
benefit if you don't. In any case, it's a family article of faith that children are born scientists -- investigators of the universe around them. When they start asking about why the sky is blue,
grass green, ants are going that way, clouds are puffy rather than stringy ... cheer. You don't need to know the answers yourself. It doesn't take many years before your kids are asking questions
that nobody knows the answer to. So, maybe answer where you know the answer. Where you don't, one or more of: "That's interesting. I don't know the answer. What do you think it is? How could we
figure it out? Where would we search on the net?" is wonderful. The universe is incredibly interesting. If you're no great fan of insects, and find yourself parent (or, in my case, uncle) to a child
who thinks they're incredibly interesting, cheer their interest. Insects are very interesting, at least to folks who study them, or maybe found fireflies pretty, and so on. In any case, I learned
from my mother that you don't need to share the exact interest, and certainly don't need to know a lot about that particular thing*. The universe is interesting. That includes the parts you're not
enthusiastic about yourself.
Second major thing is -- not all ideas will work out, and that's ok. "You'll think of more ideas later, and they'll be even better." Abandoning ideas that don't work out, I mentioned earlier, is one
of the central skills of a scientist. If the idea doesn't work, don't spend your time 'reassuring' your child that it's 'ok that you failed'. They didn't fail, period. They successfully carried out
one of the most important things a scientist can do -- test an idea and realize it doesn't work. Another better response is "Ok, since this doesn't quite work, what do you think might work instead?"
Almost every idea that a scientist has, doesn't work. I doubt this is unique to science, so whether your child goes anywhere near science it's a good skill to have -- to have your idea not work, and
to take it in stride and move on to the next idea.
I didn't even remember the example myself, but in my 20s or so, mom mentioned my 'blood theory of life'. I was 4 at the time, she said. The theory was, "We all have a finite amount of blood. When it
runs out, say because of a trauma like a car accident or a blood disease like leukemia destroying healthy blood cells, you die." (I did know about leukemia at that age.) I don't exactly remember the
responses (hey, I was 4, and 4 was ... several ... years ago), but I'm positive it wasn't "my what a wonderful idea". It was, I'm sure, in the vein of "That doesn't work because bone marrow makes
more blood cells, so the supply isn't limited." Far better answer. Told me the idea didn't work, and told me something new that I could then think about more. Wow. There's even more to learn about
the universe! And, it is possible for me to learn it!
This is another major skill to encourage in your children. "You may not know or understand something now, but you can learn it." Maybe they have to devote some time or effort, but it can be done.
Things are possible, and regardless of where you are today, in time you can get where you'd like. Again, nothing unique to science about this. And, for that reason, a good reason to encourage it with
respect to science. The skill, learning more stuff, is needed for almost everything. In science, finding out more about the universe -- that your child finds interesting -- the rewards follow more
immediately from the effort than might be in other areas. In sports, if your child loves any healthy sport, support that. Skills can transfer, most especially the skill of training to do their sport
well. Later, when some other sport becomes the greater interest, this overarching skill of training to do better will transfer. Same for study, practice, learn in science transferring to any other
academic area. And, though I haven't seen it as often, between the sports and academics.
*I'll embarrass mom a little. She knew, and knows, far more than she has ever laid claim to. But, the key is regardless of how much she knew, she could never have hoped to answer all questions I
asked. Same way that regardless of how much I know, I could never hope to answer all the questions my daughter has asked. As a parent, for us both, the key was, help your child learn more. Encourage
that desire to learn more. Support the skills to learn more.
Additional support doesn't need to be expensive. Go for walks, look at the sky (works both day and night), visit the library, tour the internet with your child (perhaps at the library). An
inexpensive telescope or microscope, or rock kit, or bug collecting kit, and so on can all be nice additions. But they're not necessary. If you do get things like this, it's a good idea to match them
up to your child. A child interested in the stars will be thrilled with a reasonable telescope, while a child interested in insects ... not so much. Go with the flow here, a matter of good parenting
Chances are that your child won't become a scientist. That's fine. What they're very likely to do, however, is become interested, informed people. And that's not only fine, but vital. And you'll get
to spend time with them, and hear them being enthusiastic about their latest discovery for years while they grow up.
I'm still in 'run around' mode, but did catch up with some good comments to the blog. New comments in:
Worth a look.
Reading science is a bit different from what we're accustomed to in the rest of our lives. No, not that scientists use inscrutable jargon or math; plenty of other groups have their own jargons and
uses or abuses of math. The challenge is that scientists, in writing and speaking science, are usually much more precise in what they say.
I was reminded of this in the comments to my note about the correlation between CO2 and temperature. Still thinking about how to write so as to avoid the problem. Since, even if I do complete the
rewriting, there are many other scientists who won't be revising, I figure it's not a bad thing to write this general note.
Over in that note, I documented that there is indeed a correlation between CO2 levels and temperature. Quite a surprisingly large one, in fact. I also noted that merely running this correlation is
not how the science is being done. (In fact, the science on temperature rising because of CO2 increase, if there were enough of one, predated there being observations of the CO2 increase or
temperature rising. Conservation of energy is a powerful law.) Still, where there are people saying that there's no correlation between CO2 and temperature, there's the elementary disproof.
A catchphrase that I'd thought was very widely known (and correct, which is rarer) is "correlation is not causation". But that's particular to correlation. The more general point is, if you've
supported a particular claim (CO2 correlates with temperature), that's all that has been supported. One thing we try to do in science is to find things which are true to work with. Some things that
are true aren't very useful. But you can't reach a good scientific conclusion with things that are not true. These elementary truths are the building blocks for constructing larger true structures.
This is why the difference in reading for science vs. many other things. In many other areas, you can predict an entire argument from a short part of it (not least, because the author is arguing
rather than trying to understand -- see my note discussion vs. debate for why this is such a difference). To read science, you need to keep this in mind. Examine that one tiny building block that's
being presented at the moment very carefully, by all means. But avoid rejecting the whole thing because you think it might be used to build something that you don't like. If the block holds up to
scrutiny, then it holds up and work with it.
Turns out that I didn't have as ready access to the 'net while I was travelling last week as I'd expected to. Oh well. I was able to pass through some comments and have noted them to come back to. A
couple will likely wind up in bit bucket on grounds of being content-free. (If nothing else, a comment should betray that the author has read the post commented upon.)
This week and next include more bouncing around, so it may be a while before I get fully caught up and back to normal flow.
On the other hand, once I do get caught up, you'll have a chance to see a Sequoia picture or two, and some posts that have been percolating in the back of my mind. | {"url":"http://moregrumbinescience.blogspot.com/2009_06_01_archive.html","timestamp":"2014-04-19T22:06:32Z","content_type":null,"content_length":"143493","record_id":"<urn:uuid:621d42ac-9a02-48db-866f-005a979867b5>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00189-ip-10-147-4-33.ec2.internal.warc.gz"} |
The most serious caveat remains the fact that large M/L ratios do not uniquely imply a BH. Only step 1 of the BH search - the detection of an MDO - is nearing completion. We cannot exclude the
possibility that galaxies contain central clusters of stellar remnants or brown dwarf stars. However, old stellar populations have remarkably consistent mass-to-light ratios M/L[V] M/L[V] ^0.2
(Kormendy 1987a,b; Faber et al. 1987; Djorgovski et al. 1988; Bender et al. 1992, 1993). The increase of M/L[V] with L is due to three effects, the metallicity-luminosity correlation (Faber 1973;
more luminous galaxies are more metal-rich, so stars emit less V-band light per unit mass because of stronger absorption-line blanketing), the anisotropy-luminosity correlation (Davies et al. 1983),
and an age-luminosity correlation (Faber et al. 1992). Given that old stellar populations contain stars with very large and very small mass-to-light ratios, it is surprising that the scatter in M/L
is so small. But the small scatter is the reason why we interpret large nuclear M/L ratios as implying MDOs. Stellar densities in NGC 3115 are not outside the range observed in galaxies that lack
MDOs or in globular clusters. These have normal populations. The spectra and colors of NGC 3115 also are normal. Therefore we have no reason to expect a dramatic population gradient, and we see no
evidence of one. An extreme population gradient would be required: inside 1", the dark star cluster would have to have about five times as much mass as the visible stars. Therefore the most plausible
conclusion is that NGC 3115 contains a central BH of mass M[BH] ~ 10^9 M[].
To summarize: in NGC 3115, as in M31, the detection of an MDO is probably secure. But the arguments that we are detecting a BH are not rigorous.
4.4 M32 (M[BH] ^6 M[])
The first galaxy with convincing evidence for an MDO was the dwarf elliptical M32; it is now the most thoroughly studied BH candidate. Like M31, it was long ago known to rotate rapidly near the
center (Walker 1962). Dressler (1984) and Tonry (1984, 1987) measured its rotation curve with [*] M/L increases toward the center. He concluded that M32 contains a BH of mass M[BH] ^6 M[].
These results were strengthened by Dressler & Richstone (1988), who obtained [*] M32 with the Hale telescope. Using the techniques of Section 4.1, they showed that velocity anisotropy provides no
escape from the conclusion that an MDO is present. More detailed analysis by Richstone et al. (1990) further supported the case.
Figure 10 illustrates the kinematics.
Figure 10. Kinematics of M32 compared with the dynamical models of van der Marel et al. (1994b), Qian et al. (1995), and Dehnen (1995) kindly recalculated by the above authors for [*] = 0".20.
The conclusions of Tonry, Dressler, and Richstone have now been confirmed by three independent groups. They provide a significant iteration in spatial resolution over the discovery observations, and
they strengthen the analysis by fitting LOSVDs. Therefore M32, with M31 and NGC 3115, is among the most secure BH candidates.
Carter & Jenkins (1993) observed M32 with the WHT, a 0".45 slit, and [*] = 0".34. At this resolution, V (r) flattens out to 50 km s^-1 at ~ 1". Carter & Jenkins do not use the dispersion profile to
measure masses, but they note that a high central M/L does not follow from V (r) alone.
Van der Marel et al. (1994a) present additional WHT observations, including measurements of Gauss-Hermite coefficients. As in other rotating galaxies, h[3] and V have opposite signs; h[3] r h[4] - h
[6] are nearly zero. Van der Marel and collaborators note that Tonry's (1987) isotropic models do not fit h[3] (r); in fact, they have h[3] et al. (1994). Since h[3] r ^-1 is nearly independent of
radius (see also Figure 2 in Rix 1993). Thus, the main effect of the correction is again to lower the bulge M/L. This is confirmed by detailed analysis, as follows.
Van der Marel et al. (1994b) construct M32 models with two-integral distribution functions f (E, L[z]), where E is total energy and L[z] is the axial component of angular momentum. Without MDOs,
these models fit V (r) and h[3] (r) - h[6] (r), but they fail to fit the central dispersion gradient. In fact, they predict that r < 1". [Physically: If stars cannot climb out of their own potential
well and therefore make a cuspy profile, they must be cold; see, e.g., Binney 1980; Dehnen 1993; Tremaine et al. 1994).] Similar results are obtained with a moment equation analysis that fits V, (r)
are obtained when an MDO of mass M[BH] = (1.8 ± 0.3) x 10^6 M[] is added. These results are in good agreement with the conclusions of Tonry, Dressler, and Richstone.
Qian et al. (1995) take the moment equation models one step further by calculating their distribution functions using the contour integral method of Hunter & Qian (1993). They can then derive the
complete LOSVDs and not just their skewness. For the M[BH] = 1.8 x 10^6 M[] model, the comparison to V, h[3] - h[6] shows remarkably good agreement along all slit positions measured by van der Marel
et al. (1994b).
Dehnen (1995) has made further f (E, L[z]) models of the van der Marel et al. (1994a) observations using a slightly different technique. He recovers f (E, 0) from L[z] and used to calculate V, h[3] -
h[6]. These are compared to the observations, and the distribution function is iterated until it agrees with all observables. Slit width, pixel size, and seeing are taken into account. Dehnen's
results are closely similar to those of van der Marel et al. (1994b). In particular, he confirms that such models do not fit the kinematics unless they contain an MDO of mass M[BH] ^6 M[]. Again, the
models with an MDO provide an excellent fit to h[3](r) - h[6] (r) along the major, minor, skew, and offset axes.
Van der Marel et al. (1994b), Qian et al. (1995), and Dehnen (1995) improve on earlier analyses in several important ways. The models are properly flattened, they fit the non-Gaussian LOSVDs, and
they fit kinematic observations at a variety of slit positions, not just along the major axis. The latter point is important: as van der Marel et al. (1994b) point out, Tonry's (1984, 1987) models
explain the major-axis dispersion gradient as due to rotational line broadening, but without an intrinsic dispersion gradient, they cannot be in dynamical equilibrium along the minor axis. The f (E,
L[z]) models solve this problem.
These models also have an important shortcoming compared to the models of Dressler & Richstone (1988). The distribution functions are restricted to be functions of two integrals; in many cases, they
are required to be analytic. We can show that these models fail to fit the data without a BH, but we cannot prove that it is impossible to find a more general distribution function that succeeds
without a BH. In contrast, the maximum entropy models can be forced to have the smallest possible M/L as r -> 0. When this fails - subject to shortcomings discussed in Section 4.1 - an MDO is
required. Thus the distribution function and maximum entropy models are complementary: each has strengths that the other lacks. It is reassuring that they agree. But an exploration of extreme models
that successfully fit V, h[n] without the limitations of published maximum entropy models is still needed.
Finally, Figure 10 shows new observations of M32 obtained with the CFHT and SIS (Kormendy et al. 1995a). These provide a further improvement in spatial resolution: The slit width was 0".35, and the
seeing [*] was 0".20. At this resolution, the central rotation curve is much steeper; it peaks at V ^-1 at r ^-1. Figure 10 compares the SIS data with the models of van der Marel et al. (1994b), Qian
et al. (1995), and Dehnen (1995) as seen at the present resolution. The fit is good at large radii. But V (r) and (r) both reach higher maximum values than the models predict. Therefore the BH case
gets stronger at SIS resolution. Modeling of the SIS data is in progress.
In summary, five independent groups have observed and modeled M32. Models without MDOs consistently fail to fit the kinematics. These include maximum entropy models that were instructed to minimize
the central M/L. Successful models require that M[BH] ^6 M[]. This result has survived improvements in spatial resolution of a factor of three since the MDO discovery. Given expectations based on
Figure 10 and on the predictions of Dehnen (1995) and Qian et al. (1995), HST spectroscopy is feasible. Van der Marel et al. (1995) plan to make these observations. Meanwhile, the case for an MDO is
already strong.
4.5 NGC 4594, The Sombrero Galaxy (M[BH] ^8 M[])
The remaining stellar-dynamical BH cases (Sections 4.5, 4.6, 4.7) are weaker than those of M31, M32, and NGC 3115 because the modeling analyses have explored fewer degrees of freedom on M(r).
NGC 4594 is an almost edge-on Sa galaxy illustrated in the Hubble Atlas (Sandage 1961). Its bulge is as luminous as a giant elliptical (M[B]
Kormendy (1988d) observed NGC 4594 with the CFHT (slit width = 0".5, scale = 0".435 pixel^-1, seeing [*] = 0".40). The central kinematics reveal a nuclear disk: at r ^-1) that is significantly lower
than the bulge dispersion (^-1). The nuclear disk is also detected photometrically (Burkhead 1986, 1991; Kormendy 1988d; Crane et al. 1993b; Emsellem et al. 1994a). Its presence guarantees that the
LOSVDs are asymmetrical. But the disk is well localized and significantly brighter than the bulge. At r
The machinery of Section 4.1 was used to derive unprojected rotation velocity, velocity dispersion, and brightness profiles that bracket the bulge-subtracted data after projection and seeing
convolution. The solutions imply that the mass-to-light ratio rises from normal values M/L[V] M/L[V] > 100 near the center. (All values in this section have been corrected to the distance scale of
Table 1.) Kormendy & Westpfahl (1989) showed that the outer M/L[V] remains constant to ~ 180". So a normal old stellar population dominates the bulge over a large radius range. But an MDO is present
at r < 1"; if M/L[V] (r) = constant for the stellar population, then M[BH] ^8 M[].
The main shortcoming of the analysis was that velocity anisotropy was not explored. The rapid rotation and the nuclear disk make it unlikely that [r] >> [tangential] (Kormendy & Illingworth 1982;
Jarvis & Freeman 1985). Nevertheless, anisotropic models should be constructed.
The kinematics of NGC 4594 have been confirmed by five independent groups. Jarvis & Dubath (1988) observed the galaxy with the ESO 3.6 m telescope, a 2" slit, 1".17 pixels, and seeing [*] ^8 M[] and
``conclude that there is strong evidence that NGC 4594 contains a super-massive object, possibly a black hole or massive star cluster.'' But they calculate no M/L ratios, so they cannot tell how much
of the mass is dark. In fact, comparably luminous galaxies typically have core masses > 10^9 M[] in stars (e.g., Kormendy 1982, Table 3). In NGC 4594, the luminosity inside 3".5 is L[V] ^8 L[]
(Kormendy 1988d, Figure 10), and M/L[V],bulge
Wagner et al. (1989: ESO 2.2 m telescope, 1".87 pixels, 2" slit) were the first to examine LOSVDs. These clearly show a superposition of rapidly rotating, cold and slowly rotating, hot components.
The cross-correlation peaks at the center and at ±3".6 agree well with those illustrated in Kormendy (1994). Wagner et al. (1989) derive a seeing-corrected mass of (3 ± 1.2) x 10^9 M[] inside 3".8
radius, but like Jarvis & Dubath (1988), they do not derive M/L ratios.
The kinematics are further confirmed by Carter & Jenkins (1993: WHT, 0".34 pixels, 0".55 slit, seeing [*] M/L found by Kormendy (1988d) is largely due to the dispersion gradient. Checking anisotropy
is probably more important than Kormendy suggested.
Van der Marel et al. (1994a) took additional spectra with a 1".25 slit, 0".6 pixels, and seeing [*] = 0".47. They agree with previous results. The main new contribution is the measurement of LOSVDs.
The maximum amplitude of h[3] = -0.15 is reached at r = 6"; |h[3]| - |h[6]| then fall almost to zero at r = 10" - 20". This is in excellent agreement with the Kormendy (1988d) and Burkhead (1991)
conclusion that the nuclear disk dominates near the center but is negligible at intermediate radii.
In summary, the kinematics of NGC 4594 are well confirmed. But none of the published measurements improves on the discovery resolution.
Also, there has been little progress on modeling. Emsellem et al. (1994a) model the Kormendy data using multi-Gaussian expansions of the PSF and galaxy light distributions. Again, isotropic models do
not fit the dispersion profile unless there is an MDO of mass 5 x 10^8 M[]. However, even the MDO model rotates too slowly. Part of the problem is that Emsellem and collaborators fit the composite
(not bulge-subtracted) kinematics; this adds to the uncertainties because of the messy LOSVDs.
More definitive results on NGC 4594 should be available soon. Emsellem et al. (1994b) have obtained two-dimensional spectroscopy with 0".39 spatial sampling on the CFHT. Kormendy et al. (1995a) have
taken CFHT SIS spectra with [*] = 0".27; at this resolution, the apparent central velocity dispersion is 282 ± 8 km s^-1. They are also scheduled to observe NGC 4594 with HST in 1995 February. The
issue of whether NGC 4594 contains an MDO should be settled soon.
4.6 The Galaxy (M[BH] ^6 M[])
The center of our Galaxy is enormously complicated and well studied. Excellent reviews (Genzel & Townes 1987; Morris 1993; Genzel et al. 1994) and conferences (Backer 1987; Morris 1989; Genzel &
Harris 1994) discuss the physics in detail. Papers on a possible Galactic BH include Lynden-Bell & Rees (1971); Rees (1987); Phinney (1989); and de Zeeuw (1993). Here we summarize the AGN evidence
and gas dynamics briefly and then concentrate on the stellar-dynamical BH search.
The radio source Sgr A* (Genzel et al. 1994, Figure 2.2) is assumed to be the Galactic center; this is not certain. Sgr A* is spectacularly tiny. Lo et al. (1985) measure a radius of ^-3 arcsec = 9
AU at ^2 due to interstellar electron scattering. Strong limits require short ^-3 arcsec = 0.6 AU (Rogers et al. 1994). This is only 15 times the Schwarzschild radius of the 2 x 10^6 M[] MDO
suggested by the dynamics. It is easy to be impressed by the small size. But as an AGN, Sgr A* is feeble: its radio luminosity is 10^34 erg s^-1 ~ 10^0.4 L[]. The infrared and high-energy luminosity
is much higher (Genzel et al. 1994), but there is no compelling need for a 10^6-M[] BH. Mini-AGNs caused by stellar-mass engines are known (e.g., Mirabel et al. 1992). So, to find out whether the
Galaxy contains a supermassive BH, we need dynamical evidence.
Genzel & Townes (1987) and Genzel et al. (1994) review the dynamics of neutral and ionized gas near the center. Velocities of 100-140 km s^-1 imply masses of several x 10^6 M[] inside 1 pc (Figure 11
) it if the gas is in circular motion. This assumption is not well motivated: as discussed in Genzel et al. (1994), stellar winds from luminous young stars combine to make a wind blowing out of the
central parsec; a few hundred supernovae are thought to have occurred in the central 10^2 pc in the past 10^4-10^5 y; some noncircular motions are seen, including an expanding bubble of hot gas
(Eckart et al. 1992). It is surprising that the motions are so close to gravitational. Without confirmation from stellar dynamics, we would not dare to include the Galaxy in Table 1.
Because of optical extinction, stellar velocities are usually measured using K-band CO band heads. From the first papers (Sellgren et al. 1987; Rieke & Rieke 1988), masses of 10^6-10^7 M[] at r et
al. (1989) and Sellgren et al.(1990) measured spectra of integrated starlight in apertures of 2".7 - 20" diameter. The rotation curve rises in the inner 15" = 0.6 pc and then flattens out at V ^-1.
The projected dispersion rises from ~ 70 km s^-1 at r > 35" to 125 km s^-1 at r < 20". Interestingly, the CO band strength decreases at small radii; the true strength is consistent with zero at r <
15". The authors suggest that the atmospheres of giant stars have been modified (stripped off?) in the dense environment of the nucleus. This limits the resolution to 0.6 pc, not much better than
ground-based resolutions in M31 and M32. Neglecting projection and anisotropy and assuming that r[c] = 0.4 pc, McGinn et al. (1989) conclude that the mass distribution is inconsistent with the light
distribution: it requires M[BH] ^6 M[]. Sellgren et al. (1990) derive M[BH] ^6 M[].
The best dynamical analyses are by Kent (1992) and by Evans & de Zeeuw (1994). Kent notes that the central K-band starlight is a flattened power law, r^-1.85. He constructs a flattened, isotropic
Jeans equation model that fits the stellar and gas kinematics along the major and minor axes. This requires an MDO of mass M[BH] ^6 M[]. Evans & de Zeeuw (1994) make f (E, L[z]) models for the same
power-law density distribution; these fit the kinematics if M[BH] ^6 M[].
Kinematic measurements of a number of kinds of stars confirm these results (e.g., OH/IR stars: Lindqvist et al. 1992; He I stars: Krabbe & Genzel 1993, quoted in Genzel et al. 1994). The dynamical
mass distribution is compared to that derived from the infrared light distribution in Figure 11. Gas and stellar kinematics agree remarkably well. The conclusion that there is (1 to 3) x 10^6 M[] of
central dark matter even looks robust to the poorly known core radius of the stars.
Haller et al. (1995) have remeasured the kinematics with a 1".3 slit placed at several positions near Sgr A*. From He I lines in hot stars and CO band heads in cool stars, they find masses at r < 1
pc that are somewhat lower than those reviewed by Genzel et al. (1994). Haller and collaborators raise the possibility that some of the dark matter near the center may be extended in radius. Then M
[BH] ^6 M[].
Most recently, Krabbe et al. (1995) have obtained two-dimensional K-band spectroscopy of the central 8" x 8" at FWHM = 1" resolution. From 35 individual stellar velocities (mostly based on emission
lines), they derive ^-1 at < r > = 0.245 pc and hence M[BH] ^6 M[]. If the velocity distribution is approximately isotropic, the case for a high central mass-to-light ratio is quite strong.
Nevertheless, the Galactic center BH case is fundamentally more uncertain than those of the best candidates. The observations are sensitive to discreteness and population effects. Absorption-line
measurements are not luminosity-weighted along the line of sight because the CO bands disappear in the central 0.5 pc. It is hard to be confident that a particular kinematic tracer is distributed in
radius like the 2.2 µm light (which is used to determine d ln d ln r and the stellar mass distribution). Given the peculiar (young?, rejuvenated?) stellar population, it is not clear that M/L[K] is
constant for the stars. Also, no models have explored anisotropy. On the other hand, we can hope for better spatial resolution than in any other galaxy. The case for an massive dark object is strong
enough to be taken seriously. But further work is needed.
If the Galaxy contains a BH, it is starving in a blizzard of food. This could be embarrassing (Section 7).
Figure 11. Galactic center mass distribution (Genzel et al. 1994). The points show dynamical masses derived from various kinematic tracer populations (see text). For two possible core radii, the
dashed lines show the mass distribution in stars derived from the K-band brightness distribution assuming M/L[K] = 1.2.
4.7 NGC 3377 (M[BH] ^7 M[])
NGC 3377 is a normal elliptical galaxy illustrated in the Hubble Atlas. At M[B] = -18.8, it rotates rapidly enough to be nearly isotropic. Its core is tiny, so small radii have large luminosity
weight in projection. The axial ratio is 0.5; since no elliptical is much flatter, NGC 3377 must be nearly edge-on. Finally, the distance is only 9.9 Mpc. Therefore NGC 3377 is an excellent target
for a BH search.
Kormendy et al. (1995b, see Kormendy 1992a) observed NGC 3377 with the CFHT and resolution [*] M32: V (r) rises rapidly near the center to 100 km s^-1 and then levels off; (r) ^-1 at r ^-1 at the
center. The kinematics are confirmed by Bender et al. (1994). Kormendy et al. (1995b) show that isotropic kinematic models require an MDO of mass M[BH] ^7 M[]. They also fail to find anisotropic
models that fit without an MDO, but the failure is not large. Therefore this is a weak MDO detection. After M32, NGC 3377 is only the second elliptical galaxy with stellar-dynamical evidence for an
4.8 M33 (M[BH] ^4 M[])
One BH upper limit deserves discussion. M33 is a bulgeless Sc galaxy with a nucleus like a giant globular cluster. Kormendy & McClure (1993) observed it with the CFHT and a camera that uses a
tip-tilt mirror to partly correct seeing. At resolution [*] = 0".18, the limit on the core radius is r[c] < 0.39 pc. This is as small as the smallest cores in globular clusters. The central surface
brightness is one of the highest known (Figure 2). M33 is an excellent illustration of the fact that an unusually small and dense core is no evidence for a BH unless kinematic measurements show high
velocities (Section 3). The central velocity dispersion measured by Kormendy & McClure (1993) is only 21 ± 3 km s^-1. The central mass-to-light ratio is M/L[V] M[BH] is the limit on the core mass in
stars: M[BH] ^4 M[]. M33 is the first giant galaxy in which a dead quasar engine can be ruled out.
The central relaxation time, T[r] ^7 y, is so short that core collapse has probably occurred. Also, the center is (B - R) = 0.44 mag bluer than the rest of the nucleus. The color gradient, the F-type
spectrum (van den Bergh 1991), and the small M/L ratio imply that the nucleus contains young stars concentrated to the center. Kormendy & McClure (1993) discuss the possibility that the stellar
population has been affected by dynamical processes (e.g., stellar collisions). | {"url":"http://ned.ipac.caltech.edu/level5/Kormendy/Kormendy4.html","timestamp":"2014-04-16T19:14:53Z","content_type":null,"content_length":"83492","record_id":"<urn:uuid:3d0d2909-cda5-48c6-ad44-07f39c050bf8>","cc-path":"CC-MAIN-2014-15/segments/1398223207046.13/warc/CC-MAIN-20140423032007-00036-ip-10-147-4-33.ec2.internal.warc.gz"} |
bool behavior in Python 3000?
Steven Bethard steven.bethard at gmail.com
Tue Jul 10 21:13:38 CEST 2007
Alan G Isaac wrote:
> > Do you care to explain what is broken?
> I suppose one either finds coercion of arithmetic operations to int
> to be odd/broken or does not. But that's all I meant.
> My preference would be for the arithmetic operations *,+,-
> to be given the standard interpretation for a two element
> boolean algebra:
> http://en.wikipedia.org/wiki/Two-element_Boolean_algebra
If I understand this right, the biggest difference from the current
implementation would be that::
True + True == True
instead of:
True + True == 2
What's the advantage of that? Could you give some use cases where that
would be more useful than the current behavior?
It's much easier to explain to newcomers that *, + and - work on True
and False as if they were 1 and 0 than it is to introduce them to a two
element boolean algebra. So making this kind of change needs a pretty
strong motivation from real-world code.
More information about the Python-list mailing list | {"url":"https://mail.python.org/pipermail/python-list/2007-July/427976.html","timestamp":"2014-04-17T12:57:58Z","content_type":null,"content_length":"3619","record_id":"<urn:uuid:a0dcbeae-0fdd-46d0-afa9-0658e1290563>","cc-path":"CC-MAIN-2014-15/segments/1397609530131.27/warc/CC-MAIN-20140416005210-00193-ip-10-147-4-33.ec2.internal.warc.gz"} |
S^n\{x} isomorphic to R^n
Could someone please tell me why is $S^n-{x}$ isomorphic to $\mathbb{R}^n$? Thank you so much.
Well, that depends on your notation. The one I have seen the most is $\mathbb{S} ^n = \{ x\in \mathbb{R} ^{n+1} : \Vert x \Vert =1 \}$ in which case it is indeed true that $\mathbb{S} ^n \setminus
{e_n} \cong \mathbb{R} ^n$ To prove it, try to generalize the proof of the stereographic projection used to build the Riemann sphere which can be found in any (good) book on basic complex analysis | {"url":"http://mathhelpforum.com/differential-geometry/115874-s-n-x-isomorphic-r-n.html","timestamp":"2014-04-20T21:13:55Z","content_type":null,"content_length":"39445","record_id":"<urn:uuid:e9b29cca-81b2-435c-9ed6-11b12f7bbd05>","cc-path":"CC-MAIN-2014-15/segments/1397609539230.18/warc/CC-MAIN-20140416005219-00347-ip-10-147-4-33.ec2.internal.warc.gz"} |
Example for column rank $\neq$ row rank
up vote 12 down vote favorite
The proof that column rank = row rank for matrices over a field relies on the fact that the elements of a field commute. I'm looking for an easy example of a matrix over a ring for which column rank
$\neq$ row rank. i.e. can one find a $2 \times 3$-(block)matrix with real $2\times 2$-matrices as elements, which has different column and row ranks?
linear-algebra ra.rings-and-algebras
3 How do you define these ranks? – darij grinberg Nov 25 '10 at 19:54
1 For matrices over skew field one usually defines a left column rank, a right column rank, a left row rank and a right row rank. Then one still has the equalities left column rank = right row rank
and right column rank = left row rank. – Someone Nov 26 '10 at 8:43
add comment
2 Answers
active oldest votes
Let $D$ be a skew field and consider the sets of $2\times 1$-matrices (columns) and $1\times 2$-matrices (lines) as left vector spaces over $D$. Let $a$ and $b$ be two
non-commuting elements of $D$. Then $(a,ab)\in D(1,b)$, on the other hand $(b,ab)^{\rm T}\not\in D(1,a)^{\rm T}$.
up vote 18 down In particular the matrix $$ \left(\begin{array}{cc} 1 & b\\ a & ab \end{array} \right) $$ is not invertible, but its transpose $$ \left(\begin{array}{cc} 1 & a\\ b & ab \end
vote accepted {array} \right) $$ is invertible.
add comment
It is a classical observation due to Nathan Jacobson that a division ring such the set of invertible matrices is closed under transposition has to be a field, i.e. commutative.
The reason is simple: the matrix $\begin{pmatrix} a & b \\ c & 1 \end{pmatrix}$ is invertible if and only if $\begin{pmatrix} a - bc & 0 \\ c & 1 \end{pmatrix}$ is invertible. This
happens if and only $a - bc \neq 0$.
up vote 26 down
vote For the transpose you get the condition $a - cb \neq 0$. Hence, taking $a = cb$ and a pair of non-commuting elements $b,c$ in the division ring, you get an example of an invertible
matrix, whose transpose is not invertible.
add comment
Not the answer you're looking for? Browse other questions tagged linear-algebra ra.rings-and-algebras or ask your own question. | {"url":"https://mathoverflow.net/questions/47369/example-for-column-rank-neq-row-rank","timestamp":"2014-04-25T05:12:36Z","content_type":null,"content_length":"58728","record_id":"<urn:uuid:7132a749-0cf4-4841-bd6a-05d3a37450ce>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00127-ip-10-147-4-33.ec2.internal.warc.gz"} |
Planck Curve
Planck Curve:
The quantum theory of absorption and emission of radiation announced in 1900 by Planck ushered in the era of modern physics. He proposed that all material systems can absorb or give off
electromagnetic radiation only in "chunks" of energy, quanta E, and that these are proportional to the frequency of that radiation E = h. (The constant of proportionality h is, as noted above, called
Planck's constant.)
Planck was led to this radically new insight by trying to explain the puzzling observation of the amount of electromagnetic radiation emitted by a hot body and, in particular, the dependence of the
intensity of this incandescent radiation on temperature and on frequency. The quantitative aspects of the incandescent radiation constitute the radiation laws.
The Austrian physicist Josef Stefan found in 1879 that the total radiation energy per unit time emitted by a heated surface per unit area increases as the fourth power of its absolute temperature T
(Kelvin scale). This means that the Sun's surface, which is at T = 6,000 K, radiates per unit area (6,000/300)4 = 204 = 160,000 times more electromagnetic energy than does the same area of the
Earth's surface, which is taken to be T = 300 K. In 1889 another Austrian physicist, Ludwig Boltzmann, used the second law of thermodynamics to derive this temperature dependence for an ideal
substance that emits and absorbs all frequencies. Such an object that absorbs light of all colours looks black, and so was called a blackbody.
The wavelength or frequency distribution of blackbody radiation was studied in the 1890s by Wilhelm Wien of Germany. It was his idea to use as a good approximation for the ideal blackbody an oven
with a small hole. Any radiation that enters the small hole is scattered and reflected from the inner walls of the oven so often that nearly all incoming radiation is absorbed and the chance of some
of it finding its way out of the hole again can be made exceedingly small. The radiation coming out of this hole is then very close to the equilibrium blackbody electromagnetic radiation
corresponding to the oven temperature. Wien found that the radiative energy per wavelength interval has a maximum at a certain wavelength and that the maximum shifts to shorter wavelengths as the
temperature T is increased, as illustrated in the above figure.
Wien's law of the shift of the radiative power maximum to higher frequencies as the temperature is raised expresses in a quantitative form commonplace observations. Warm objects emit infrared
radiation, which is felt by the skin; near T = 950 K a dull red glow can be observed; and the colour brightens to orange and yellow as the temperature is raised. The tungsten filament of a light bulb
is T = 2,500 K hot and emits bright light, yet the peak of its spectrum is still in the infrared according to Wien's law. The peak shifts to the visible yellow when the temperature is T = 6,000 K,
like that of the Sun's surface.
It was the shape of Wien's radiative energy distribution as a function of frequency that Planck tried to understand. The decrease of the radiation output at low frequency had already been explained
by Lord Rayleigh (John William Strutt) in terms of the decrease, with lowering frequency, in the number of modes of electromagnetic radiation per frequency interval. Rayleigh assumed that all
possible frequency modes could radiate with equal probability, following the principle of equipartition of energy. Since the number of frequency modes per frequency interval continues to increase
without limit with the square of the frequency, Rayleigh's formula predicted an ever-increasing amount of radiation of higher frequencies instead of the observed maximum and subsequent fall in
radiative power. A possible way out of this dilemma was to deny the high-frequency modes an equal chance to radiate. To achieve this, Planck postulated that the radiators or oscillators can only emit
electromagnetic radiation in finite amounts of energy of size. At a given temperature T, there is then not enough thermal energy available to create and emit many large radiation quanta. More large
energy quanta can be emitted, however, when the temperature is raised.
Excerpt from the Encyclopedia Britannica without permission. | {"url":"http://abyss.uoregon.edu/~js/glossary/planck_curve.html","timestamp":"2014-04-20T00:38:02Z","content_type":null,"content_length":"4849","record_id":"<urn:uuid:ac6ab5ea-4f62-4397-913f-0f234ea7931c>","cc-path":"CC-MAIN-2014-15/segments/1397609537804.4/warc/CC-MAIN-20140416005217-00572-ip-10-147-4-33.ec2.internal.warc.gz"} |
Summary: BLACK-SCHOLES GOES HYPERGEOMETRIC
ABSTRACT. We introduce a general pricing formula that extends Black-Scholes' and contains as particular
cases most analytically solveable models in the literature, including the quadratic and the constant-elasticity-
of-variance (CEV) models for European and barrier options. In addition, large families of new solutions can
be found, containing as many as seven free parameters.
It has been known since the seventies that Black-Scholes pricing formulas are a special case of more
general families of pricing formulas with more than just the volatility as an adjustable parameter. The
list of the classical extensions includes affine, quadratic and the constant-elasticity-of-variance models.
These models admit up to three adjustable parameters and have found a variety of applications to solving
pricing problems for equity, foreign exchange, interest rate and credit derivatives. In a series of working
papers, the authors have recently developed new mathematical techniques that allow to go much further
and to build several families of pricing formulas with up to seven adjustable parameters in the stationary,
driftless case, and additional flexibility in the general time dependent case. The formulas extend to
barrier options and have a similar structure as the Black-Scholes formulas, the most notable difference
being that error functions (or cumulative normal distributions) are replaced by (confluent) hypergeometric
functions, i.e. the special transcendental functions of applied mathematics and mathematical physics.
Let F denote a generic financial observable which we know is driftless. Examples could be the
forward price of a stock or foreign currency under the forward measure, a LIBOR forward rate or a swap | {"url":"http://www.osti.gov/eprints/topicpages/documents/record/619/1169417.html","timestamp":"2014-04-18T11:43:30Z","content_type":null,"content_length":"8963","record_id":"<urn:uuid:ed752a55-5219-4aa7-93b0-97e26ffc9efb>","cc-path":"CC-MAIN-2014-15/segments/1397609533308.11/warc/CC-MAIN-20140416005213-00090-ip-10-147-4-33.ec2.internal.warc.gz"} |
can Matlab computer the characterisitc polynomial?
April 22nd 2010, 11:21 AM #1
Senior Member
Jun 2009
can Matlab computer the characterisitc polynomial?
Given a matrix, can Matlab computer the characteristic polynomial? If no can it computer a determinant with a $\lambda$ in it?
Compute determinant of symbolic matrix - MATLAB
April 22nd 2010, 10:26 PM #2
Grand Panjandrum
Nov 2005 | {"url":"http://mathhelpforum.com/math-software/140747-can-matlab-computer-characterisitc-polynomial.html","timestamp":"2014-04-20T00:29:39Z","content_type":null,"content_length":"33529","record_id":"<urn:uuid:5394193f-2119-4e60-ac6a-f67430083dc1>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00476-ip-10-147-4-33.ec2.internal.warc.gz"} |
Reviews: Math Curse by Jon Scieszka
Math Curse is about how math really is everywhere in our lives. The book takes everyday things like milk for your cereal and shows how many math problems you can get from one item: Quarts in a gallon
etc. Everywhere the main character looks, a math problem is there. Some of the math is rather high level for the age of the child that would be reading the book. The quadratic formula is on one page.
But it really doesn't go into the math itself; just all the places it can be found.
This book is about a little girl who has a "math curse." Everything that occurs in her life she perceives as a math word problem. This is not your typical math book! It is funny and educational. It
helps children relate math to their every day lives!
Jon Scieszka did it again. This easily became one of my favorite book. Scieszka started the book with a girl waking up in the morning and start thinking about all the math problems in her life.
Children are able to do some of the math in the book and enjoy all the nonsense Scieszka wrote. The never ending narrative helps students to see how an idea can spark another as well. Also, this is a
good book to teach students about full-circle ending writing.
This book was about a boys daily life, but it was made into word problems. Each page had a new problem that students can solve. I would read this book to my students because it was a fun book to
read. It teaches students how to solve problem. I know it would entertain them because they get to answer the questions themselves. This book would be perfect to use in a math class.
I LOVED this book. The illustrations are kind of Tim Burton-like, which gives an otherwise cutesy story this dark edge. Often students don't understand the practical use of mathematics, and so they
don't like math class, but this book shows how almost ANYTHING can be a math problem.
The title of this book will catch the attention of students. Math is a subject that many students fear. It is precise in the answers and can seem intimidating until the logic and connections are
made. While educational, the story remains fun and interesting. This is a must have for math teachers.
A little boy goes through his day and realizes that all his problems are math problems. This would be a fun book to have an interactive read-aloud and solve the problems in the book with the
I have always been very intrigued by all the flattering reviews and the seemingly fabulous reputation of author/illustrator duo [a:Jon Scieszka|27318|Jon Scieszka|http://photo.goodreads.com/authors/
1201028327p2/27318.jpg]/[a:Lane Smith|23573|Lane Smith|http://photo.goodreads.com/authors/1343416984p2/23573.jpg] since reading [b:The Stinky Cheese Man: And Other Fairly Stupid Tales|407429|The
Stinky Cheese Man And Other Fairly Stupid Tales|Jon Scieszka|http://photo.goodreads.com/books/1327955928s/407429.jpg|1814587]. Somehow I just don't have the insight required to understand the
thoughts behind the book design (which I mostly find hideous), nor do I appreciate the illustrations or understand where the humor is supposed to be coming through. Now, since I have both an
extensive background in math and a great passion for the subject, I thought maybe this book would be more within my realm.
It's a sad fact that lots of children (and adults) struggle with math and math concepts, yet our society has a greater than ever need for people with math skills or at least a good understanding of
math. So, anything that has as purpose to try to enlighten us as to the mysteries of math (which is one of the goals of this book, as far as I understand, the other being to try to make us laugh,
right?) deserves recognition. So, I first took a quick look at this book, unfortunately only to put it down thinking: "Man! No one is ever going to learn anything from this book, if ever that was the
point!" I found it so chaotic and disorganized, just page after page filled with disconnected words and numbers. But then I started thinking that maybe those chaotic pages are meant to show how some
people may actually be feeling about the math they are dealing with. It's too much, too fast and it's not served to them in a neat, orderly fashion, with appropriate visuals and helpful diagrams. So,
if that's the point the author is trying to make, that's a valid point. And where does that leave us? Well, math is this awfully complicated thing and just looking at math problems should make your
head spin and induce headache. (At least, that's what the pages of this book first did for me.)
But, can this book help us get an understanding or grasp any math concepts? Now, there's a handful of so called math problems in the book. Some of them are actual elementary arithmetic problems,
others are nonsensical thrown in for fun. A good student will probably get some satisfaction from seeing that he/she can answer all the problems. I'm not totally sure if the book will teach you
anything. That two halves make a whole? (or "hole", get it?)
My first reaction to the book was that of disappointment. Consider for instance that a large part of the book focuses on conversions. Is this, honestly, what school children are made to spend most of
their time on? How boring! And if so, wouldn't our time and money be better spent if we simply adopted the metric system??
Obviously, I'm not the target audience, but I can understand that the unconventional look of the book in itself makes it cool, and thus appealing to grade schoolers even if it doesn't appeal to me.
And once I got over my intense dislike of the esthetics of the book, there were quite a few things that I liked in it. I like how the first page (with the foreword) uses the division sign in the
design, that's neat. I like the Mayan numerals picture, I like the logical paradox in the dinner story and I like the Venn diagram on the dust jacket. The answers on the back cover are a good idea
and I like that some of the answers can be phrased as "a lot", "just because", "once in a while" or even "I don't know" to emphasize that even when doing math not everything is always black and
white, right or wrong. I also like the cupcake story as an example of thinking out of the box. And I like the bills and coins integrated into the drawings of Lincoln and Washington. So, overall, I
probably like more things than I dislike about this book, even if the same things continue to grate me. A teacher could probably use this book in a class room to motivate kids, or just to do
something different. And, in case you opened the book and felt completely overwhelmed and confused by it: Don't worry about it. The whole point of the book is to make math look like a horrid
nightmare. So, don't panic! Go to the next to last page, and rest assured that if you just give yourself enough time to gather your thoughts, things will eventually fall into place.
Finally, if you have a precocious kid/teen on your hands who really likes math, I would highly recommend [a:Martin Gardner's|7105|Martin Gardner|http://photo.goodreads.com/authors/1269645617p2/
7105.jpg] puzzle books to older kids for some juicier math problems, or even [a:Shasha's|47140|Dennis E. Shasha|http://photo.goodreads.com/authors/1221975472p2/47140.jpg] Dr. Ecco, for instance
[b:Doctor Ecco's Cyberpuzzles: 36 Puzzles for Hackers and Other Mathematical Detectives|82492|Doctor Ecco's Cyberpuzzles 36 Puzzles for Hackers and Other Mathematical Detectives|Dennis E. Shasha|
Math Curse, by Jon Scieszka, was a very fun and interesting book. Math curse was very much like science verse except instead of being a book of poetry about science we have a book full of all the
different types of math problems in word problems. Jon's idea of making these two books really was great and I think will be used in all of classes when I start teaching. I really enjoy this book and
my favorite part of this book has to be the illustrations. The math problems on each page are said in a big word problem but they all can also be seen in the background which is a great aspect that
allows any child to read this book. Another part of the book that I like was the context of the book. Most picture books have a main idea or concept but, this book really just gave a huge amount of
information that allows any student to learn a little bit about a lot of ideas. The last part of this book that I liked was character, while there was really only one main characters he went from
being just like any other kid, hating math. After his day of having a math curse the character has not only grown to like math but he has decided that math is fun. This character development was
great and realistic, but also very relatable which is great for small children. The main idea of this story is that math isn't as awful as everyone says it is.
Cute book that introduces some math concepts to lower elementary school kids.
This one left me with my head spinning, but I was smiling, so I can forgive him for that. It really was a lot to think about and his strange sense of humor pervades every page. The illusrations have
a topsy-turvy, all-consuming sort of feel and the book as a whole is funny and thought-provoking.
After the first read, this book became one of my favorites. I love the storyline, riddles, and plays on words. The topics that the book covers varies along with the skill levels. If read in an
elementary class, not everyone would understand some of the text, but everyone would understand something. I can't wait to share this book with children.
This book was great! It keeps you thinking and is interesting too. This would be great for students to read because it makes math fun.
I really loved this book. As a student who has always struggled with math until very recently I would have liked it a bit more if my teacher would have read me this book. It is fun and witty, but is
still teaching the principles of math. I also really liked that there were problems that you hardly realized were there because of the flow of the story. I would love to use this book in a future
I really enjoyed reading this book; it has so many levels to it. While it is written for children there are many parts of the book that children will not understand but an adult will. You can
appreciate this book at any age because everyone has had their struggles with math. You can relate to all of the different problems and the book really gets you thinking on another level besides just
reading and looking at the pictures.
Life becomes nothing but a series of math problems after he teacher places a "math curse" on him by saying that math can be seen everywhere... so now he sees it everywhere and it is driving him
crazy. Math covers this book from front to back and top to bottom and in very clever and witty ways. There is humor to be found on every single page and for readers of many different ages. It takes a
subject that typically causes so much anxiety and makes a fun to talk about.
It's tough work not to laugh when reading this playful, quirky, definitive math book for kids. In this book, math teacher Mrs. Fibonacci tells kids that everything can be a math problem, and now the
narrator sees math everywhere! Everything has become a comical math problem, and the frequent mathematical allusions throughout the book provide additional entertainment for any adult readers.
Perhaps the best indicator of this book's brilliance is the inclusion of the books price ($17) in binary (10001). This book would be a great inclusion to a classroom as part of morning bell work or
other quick reviews/introductions to lessons.
A young boy faces the everyday struggles of math. Everywhere he goes, he sees a math problem. It really is a cute book; I enjoyed it! It shows children that all problems can be solved, and that they
should be afraid of math like most people are.
Today in math class my teacher told us everything can be a problem. That's when I actually had a problem. Everything I do is now a problem. This book describes all his problems and makes them into a
math problem for the reader to solve. This is a great and fun book to show in a math class. It will have the children doing math problem while reading a story. The style of the book is written in a
unique way. Everything is a math problem from the price to the dedication page.
Summary: A teacher tells one of her students that you can see math in everyday life. The student becomes stressed out regarding how complication and just how much math there is in everyday life.
Personal Reaction: I wish I had read this when I was in grade school. Math was never my strong subject and I think this would have helped. The illustrations flowed well with the story and the story
line itself took a complicated subject and brought humor to it.
Classroom Extension:
1) Have students create their own math problems regarding daily activities, create a worksheet and pass it out for the class to solve.
2) Have students solve the math problems in the story.
One students gets a math curse after their teacher told them that everything can be seen as a math problem. She starts to question everything and lots of math sample calculations turn up.
A very recommendable book for children to make math interesting, rich in variety, close-to reality and colorful. It even captured me to solve all these math example since they are wonderfully thought
and funny too.
We read this book in the class and it felt like we was in a math class. We had to think logical and out of the box. I like this book and I would introduce this book to children who are in middle
school. Some people has a fear in math, but once they get through this book they might just see math from a different point of view.
I like this book because I'm a math person. I really loved how the author made everything a math problem, even the price of the book. THere was also a few riddles which was fun to solve.
This book is a fun look at how math is used in everyday life and in everything we do. The illustrations are limited to a few pictures, but the math facts, which are lettered texts, does not take away
the flow of the story. I would recommend this book in math to show how fun math can be when used in a different way.
This book brilliantly gets kids smiling and laughing at the dry subject of math. We learn that math problems are in every second of our daily lives, and that we are quickly solving math problems
without even thinking! This book could definately bring up children's confidence about math, and make them look at math in a different and fun way. This is a great example of how we can incorporate
math lessons into literature. | {"url":"http://www.librarything.com/work/121568/reviews/84894769","timestamp":"2014-04-20T17:28:21Z","content_type":null,"content_length":"75199","record_id":"<urn:uuid:6741aced-061d-45a1-a4a8-185ba97df0be>","cc-path":"CC-MAIN-2014-15/segments/1397609538824.34/warc/CC-MAIN-20140416005218-00634-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Topic: mathcounts no-trig question
Replies: 7 Last Post: Jan 10, 2012 11:35 AM
Messages: [ Previous | Next ]
mathcounts no-trig question
Posted: Jan 9, 2012 8:17 PM
Found this question on an old Mathcounts test - a portion where calculators aren't allowed.
I know the answer (3/4), which I got by using illegal methods (calculators, sines, and cosines), but I can't figure out how one could figure it out just using brain, paper, and pencil.
The question poses a 75-15-90 triangle whose hypotenuse is given as the square root of three.
The goal is to compute the area.
Any hints or suggestions would be welcome.
Guy Brandenburg, Washington, DC
Date Subject Author
1/9/12 mathcounts no-trig question Guy Brandenburg
1/9/12 Re: mathcounts no-trig question RWW Taylor
1/10/12 Re: mathcounts no-trig question Guy Brandenburg
1/9/12 Re: mathcounts no-trig question Chris L. Yuen
1/10/12 Re: mathcounts no-trig question Guy Brandenburg
1/9/12 Re: mathcounts no-trig question Lars Jensen
1/9/12 Re: mathcounts no-trig question Norman Wildberger
1/9/12 Re: [MathTalk] mathcounts no-trig question talmanl@mscd.edu | {"url":"http://mathforum.org/kb/thread.jspa?threadID=2330477","timestamp":"2014-04-16T17:32:09Z","content_type":null,"content_length":"25886","record_id":"<urn:uuid:cbec341f-138c-4d48-9089-ce372f4a109b>","cc-path":"CC-MAIN-2014-15/segments/1397609530131.27/warc/CC-MAIN-20140416005210-00350-ip-10-147-4-33.ec2.internal.warc.gz"} |
Math Goes Pop!
Hi all,
This weekend’s Father’s Day celebrations have inspired my monthly article over at the CNN Light Years blog. Here’s a sample:
There are many misconceptions about mathematicians in popular culture. For example, windows and mirrors do not make for the best writing surfaces, despite what you might assume from “A Beautiful
Mind” or “Good Will Hunting.”
Mathematicians are also frequently portrayed as painfully socially awkward. And while this is sometimes the case, the true range of personality types is much more varied. Even among the more socially
awkward, it is not uncommon for mathematicians to fall in love, marry and start a family.
What must it be like to grow up in a household with a mathematician? In the spirit of Father’s Day, I spoke with two mathematicians whose fathers were also mathematicians about what it’s like being
raised in a mathematical household
Click here for the rest! | {"url":"http://www.mathgoespop.com/tag/fathers-day","timestamp":"2014-04-19T19:54:32Z","content_type":null,"content_length":"68694","record_id":"<urn:uuid:6064a4c3-f9b1-4094-9ab1-eb786853b323>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00552-ip-10-147-4-33.ec2.internal.warc.gz"} |
TR12-173 | 8th December 2012 23:46
A Cookbook for Black-Box Separations and a Recipe for UOWHFs
We present a new framework for proving fully black-box
separations and lower bounds. We prove a general theorem that facilitates
the proofs of fully black-box lower bounds from a one-way function (OWF).
Loosely speaking, our theorem says that in order to prove that a fully black-box
construction does not securely construct a cryptographic primitive $Q$
(e.g., a pseudo-random generator or a universal one-way hash function) from a
OWF, it is enough to come up with a large enough set of functions $\mathcal{F}$ and a
parametrized oracle (i.e., an oracle that is defined for every $f \in \{0,1\}^n
\rightarrow \{0,1\}^n$)
such that $O_f$ breaks the security of the construction when instantiated with $f$
and the oracle satisfies two local properties.
Our main application of the theorem is a lower bound of $\Omega(n/\log(n))$
on the number of calls made by any fully black-box construction of a universal
one-way hash function (UOWHF) from a general one-way function. The bound holds
even when the OWF is regular, in which case it matches to a recent construction
of Barhum and Maurer. | {"url":"http://eccc.hpi-web.de/report/2012/173/","timestamp":"2014-04-18T21:13:50Z","content_type":null,"content_length":"20386","record_id":"<urn:uuid:48526238-d5ee-406c-84aa-539510fbe5c6>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.9/warc/CC-MAIN-20140416005215-00656-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
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Is the function 1/|x+y| a joint probability density function?
April 10th 2009, 10:19 AM
Is the function 1/|x+y| a joint probability density function?
Is the function $\frac{1}{|x+y|}$ a joint probability density function? That is, is $\int\hspace{-6pt}\int_{\mathbb{R}^2} \frac{1}{|x+y|} \ dA$ equal to 1 and is $\frac{1}{|x+y|}$ nonnegative?
This problem is really giving me a headache, been attempting it for a week now and the deadline's today. (Headbang)
April 10th 2009, 10:59 AM
Is the function $\frac{1}{|x+y|}$ a joint probability density function? That is, is $\int\hspace{-6pt}\int_{\mathbb{R}^2} \frac{1}{|x+y|} \ dA$ equal to 1 and is $\frac{1}{|x+y|}$ nonnegative?
This problem is really giving me a headache, been attempting it for a week now and the deadline's today. (Headbang)
no! the integral is divergent: your function is positive wherever is defined. show that the integral is $\infty$ in the first quadrant and thus it's divergent over $\mathbb{R}^2.$ | {"url":"http://mathhelpforum.com/calculus/83135-function-1-x-y-joint-probability-density-function-print.html","timestamp":"2014-04-23T20:53:13Z","content_type":null,"content_length":"6122","record_id":"<urn:uuid:591121bf-dffe-4385-b402-dc23a52055a0>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00612-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Plainfield, IL Trigonometry Tutors | {"url":"http://www.purplemath.com/Plainfield_IL_statistics_tutors.php","timestamp":"2014-04-16T22:35:03Z","content_type":null,"content_length":"24359","record_id":"<urn:uuid:7edbe5c9-b890-4f40-a8de-338ecfd9303f>","cc-path":"CC-MAIN-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00566-ip-10-147-4-33.ec2.internal.warc.gz"} |
Secondary Math and Science Education
Secondary Math and Science Education: Hearing Before the Subcommittee on Science, Research, and Technology of the Committee on Science and Technology, U.S. House of Representatives, Ninety-seventh
Congress, Second Session, May 7, 1982
U.S. Government Printing Office
, 1983 -
127 pages
We haven't found any reviews in the usual places.
References from web pages
Secondary Math and Science Education. Hearing Before the ...
ED229278 - Secondary Math and Science Education. Hearing Before the Subcommittee on Science, Research and Technology of the Committee on Science and ...
www.eric.ed.gov/ ERICWebPortal/ recordDetail?accno=ED229278
Bibliographic information | {"url":"http://books.google.com/books?id=LPN7gzxkja8C&q=elementary&source=gbs_word_cloud_r&cad=5","timestamp":"2014-04-16T10:34:36Z","content_type":null,"content_length":"99884","record_id":"<urn:uuid:766d9fe0-af5f-4a9d-a494-0f2a0bb479dd>","cc-path":"CC-MAIN-2014-15/segments/1397609523265.25/warc/CC-MAIN-20140416005203-00592-ip-10-147-4-33.ec2.internal.warc.gz"} |
Forget passwords, tricky sums are more secure
Faster, more secure logins for multimedia sites might be possible thanks to a new approach to website and database security. Boolean logins would allow thousands if not millions of users to more
quickly access the content to which they are entitled, such as music, video and images. The same approach might also reduce the risk of hackers accessing the materials illicitly.
Nikolaos Bardis of the University of Military Education, in Vari, Greece and colleagues there and at the Polytechnic Institute of Kiev, in Ukraine, have developed an innovative approach to logins,
which implements the advanced concept of zero knowledge identification.
Zero knowledge user identification solves these issues by using passwords that change for every session and are not known to the system beforehand. The system can only check their validity.
The basis of modern cryptography are analytically insoluble mathematical equations; i.e. equations that have a solution, but this solution cannot be found by a formula. Consider the modulo operation,
for which A mod B represents the remainder of the division A/B. Then assume for example, the equation 5^Xmod7 = 2. There exists no mathematical formula that can give X = … . The solution may only be
found by trying all possible values. Trying X=2: one gets 5^2=25, 25mod7 = 4. Not true. Trying X=3: one gets 5^3=125, 125mod7 = 6. Not true. Trying further X=4: one gets 5^4=625, 625mod7 = 2. Hence X
= 4 is a solution. This solution was relatively easy to find because the numbers were small.
Normally, the numbers used would have hundreds of decimal or thousands of binary digits. Hence the process is in practice safe. Algorithms like RSA, El-Gamal and DSA are based on the above principle.
One more characteristic of the equations used is that they have multiple solutions. In the above example, 5^Xmod7 = 2, another solution is X=10: 5^10mod7 = 2. The legitimate user would hence
formulate the two passwords (X =4 and X = 10). Following that, the user registers to the system giving the base (5), the modulo (7) and the result (2). When the user wants to login, they will first
use the 1st password. The system does not know if it is correct, but calculates 5^4mod7 =2, verifies the submitted password and grants access. Next time round, a different password may be used.
The basic drawback of this is the extremely high computational complexity, when numbers of hundreds of digits are raised to powers of tens of digits and the remainder of the division with another
large number is sought. Hence a different mathematical problem had to be found, for which the calculations for authentications are relatively simpler.
This problem is finding the solution of a system of non-linear Boolean equations. A solution exists but there is no easy way to find them quickly. In practice, the only solution is successive trials.
Such equations are the basis for algorithms like DES and AES. In general one may say that for a given system of non-linear Boolean equations F(X), the value X that corresponds to a given Y=F(X) can
only be determined by trials. The proposed method allows the remote user to construct a transformation F(X) and the complete set of solutions X[1], X[2], X[3], …., X[n] such that F(X1) = F(X2) = F
(X3) = … = F(Xn) =Y. The user registers to the system and submits F(X) (as a truth table) as well as Y. For the first session, the user will submit X1 as the first password. For the second session,
the username remains the same but the password is X2. Each time, the system calculates F(X) and only grants access if F(X) equals Y.
Put simply, zero-order user authentication schemes, supply the user with a special function that produces an extremely large number of different results for all its possible inputs. A set of inputs
that produce a common result is selected. These inputs are the user’s passwords. A new user registers by submitting to the system their function and the common result. The user authenticates for a
normal session using each password only once. The user provides the password at the beginning of each session. The system calculates the value produced when this password is used as input to the
function. If this is equal to the common result, then authentication is successful and access is granted. Someone that is trying to gain access without the necessary knowledge (an illicit user) will
practically have to try all possible password combinations, before reaching the correct one.
The proposed scheme has potential use in any system where malicious users have incentives to gain illegal access and perform actions they are not entitled to. The number of such systems increases
rapidly as information gains value.
The advantages of the proposed scheme depend to what it is compared. If it is compared to classic authentication systems then the proposed scheme is not faster. It is however safer, as it implements
zero knowledge identification. This means that a malicious colleague or a virus cannot obtain passwords and exploit them for illegal purposes. Additionally, passwords that have been stolen by tapping
into the communication channel may not be used since the password changes for every session.
If the comparison is done compared to existing zero knowledge schemes based on modular exponentiation, then the proposed approach accelerates authentication by about 1000 times. This is a very
important result, since the high computational cost has until now deterred the wide use of the concept of zero knowledge identification. One must take into account the fact that in modern systems the
number of users ranges to the thousands or tens of thousands and hence authentication must be quick. Apart from that, the periodic repetition of the authentication process can prohibit a criminal
from taking over an already active session.
Nikolaos Bardis, Nikolaos Doukas, & Oleksandr P. Markovskyi (2011). Fast subscriber identification based on the zero knowledge principle for multimedia content distribution Int. J. Multimedia
Intelligence and Security, 1 (4), 363-377 | {"url":"http://www.sciencetext.com/forget-passwords-tricky-sums-are-more-secure.html","timestamp":"2014-04-16T13:10:29Z","content_type":null,"content_length":"71214","record_id":"<urn:uuid:ec41d022-ec3c-4ab2-905d-6f82d90eb48d>","cc-path":"CC-MAIN-2014-15/segments/1398223207985.17/warc/CC-MAIN-20140423032007-00491-ip-10-147-4-33.ec2.internal.warc.gz"} |
NAG Library
NAG Library Routine Document
Note: this routine uses optional parameters to define choices in the problem specification. If you wish to use
settings for all of the optional parameters, then this routine need not be called. If, however, you wish to reset some or all of the settings please refer to Section 10 for a detailed description of
the specification of the optional parameters
1 Purpose
F12ADF is an option setting routine that may be used to supply individual optional parameters to
. These are part of a suite of routines that also includes:
. The initialization routine
F12AAF must
have been called prior to calling F12ADF.
2 Specification
SUBROUTINE F12ADF ( STR, ICOMM, COMM, IFAIL)
INTEGER ICOMM(*), IFAIL
REAL (KIND=nag_wp) COMM(*)
CHARACTER(*) STR
3 Description
F12ADF may be used to supply values for optional parameters to
. It is only necessary to call F12ADF for those parameters whose values are to be different from their default values. One call to F12ADF sets one parameter value.
Each optional parameter is defined by a single character string consisting of one or more items. The items associated with a given option must be separated by spaces, or equals signs
. Alphabetic characters may be upper or lower case. The string
'Vectors = None'
is an example of a string used to set an optional parameter. For each option the string contains one or more of the following items:
– a mandatory keyword;
– a phrase that qualifies the keyword;
– a number that specifies an integer or real value. Such numbers may be up to $16$ contiguous characters in Fortran's I, F, E or D format.
F12ADF does not have an equivalent routine from the ARPACK package which passes options by directly setting values to scalar parameters or to specific elements of array arguments. F12ADF is intended
to make the passing of options more transparent and follows the same principle as the single option setting routines in
Chapter E04
The setup routine
must be called prior to the first call to F12ADF and all calls to F12ADF must precede the first call to
, the reverse communication iterative solver.
A complete list of optional parameters, their abbreviations, synonyms and default values is given in
Section 10
4 References
Lehoucq R B (2001) Implicitly restarted Arnoldi methods and subspace iteration SIAM Journal on Matrix Analysis and Applications 23 551–562
Lehoucq R B and Scott J A (1996) An evaluation of software for computing eigenvalues of sparse nonsymmetric matrices Preprint MCS-P547-1195 Argonne National Laboratory
Lehoucq R B and Sorensen D C (1996) Deflation techniques for an implicitly restarted Arnoldi iteration SIAM Journal on Matrix Analysis and Applications 17 789–821
Lehoucq R B, Sorensen D C and Yang C (1998) ARPACK Users' Guide: Solution of Large-scale Eigenvalue Problems with Implicitly Restarted Arnoldi Methods SIAM, Philidelphia
5 Parameters
1: STR – CHARACTER(*)Input
2: ICOMM($*$) – INTEGER arrayCommunication Array
3: COMM($*$) – REAL (KIND=nag_wp) arrayCommunication Array
4: IFAIL – INTEGERInput/Output
6 Error Indicators and Warnings
If on entry
, explanatory error messages are output on the current error message unit (as defined by
Errors or warnings detected by the routine:
The string passed in
contains an ambiguous keyword.
The string passed in
contains a keyword that could not be recognized.
The string passed in
contains a second keyword that could not be recognized.
The initialization routine
has not been called or a communication array has become corrupted.
7 Accuracy
Not applicable.
9 Example
This example solves $Ax=\lambda Bx$ in shifted-inverse mode, where $A$ and $B$ are derived from the finite element discretization of the one-dimensional convection-diffusion operator $\frac{{d}^{2}u}
{d{x}^{2}}+\rho \frac{du}{dx}$ on the interval $\left[0,1\right]$, with zero Dirichlet boundary conditions.
The shift $\sigma$ is a real number, and the operator used in the shifted-inverse iterative process is $\mathrm{OP}={\left(A-\sigma B\right)}^{-1}B$.
9.1 Program Text
9.2 Program Data
9.3 Program Results
10 Optional Parameters
Several optional parameters for the computational routines
define choices in the problem specification or the algorithm logic. In order to reduce the number of formal parameters of
these optional parameters have associated
default values
that are appropriate for most problems. Therefore, you need only specify those optional parameters whose values are to be different from their default values.
The remainder of this section can be skipped if you wish to use the default values for all optional parameters.
The following is a list of the optional parameters available. A full description of each optional parameter is provided in
Section 10.1
Optional parameters may be specified by calling F12ADF before a call to
, but after a call to
. One call is necessary for each optional parameter.
All optional parameters you do not specify are set to their default values. Optional parameters you specify are unaltered by
(unless they define invalid values) and so remain in effect for subsequent calls unless you alter them.
10.1 Description of the Optional Parameters
For each option, we give a summary line, a description of the optional parameter and details of constraints.
The summary line contains:
• the keywords, where the minimum abbreviation of each keyword is underlined;
• a parameter value, where the letters $a$, $i\text{ and }r$ denote options that take character, integer and real values respectively;
• the default value, where the symbol $\epsilon$ is a generic notation for machine precision (see X02AJF).
Keywords and character values are case and white space insensitive.
Advisory $i$ Default $\text{}=$ the value returned by X04ABF
The destination for advisory messages.
This special keyword may be used to reset all optional parameters to their default values.
During the Arnoldi iterative process, shifts are applied internally as part of the implicit restarting scheme. The shift strategy used by default and selected by the
Exact Shifts
is strongly recommended over the alternative
Supplied Shifts
Lehoucq et al. (1998)
for details of shift strategies).
Exact Shifts
are used then these are computed internally by the algorithm in the implicit restarting scheme.
Supplied Shifts
are used then, during the Arnoldi iterative process, you must supply shifts through array arguments of
returns with
; the real and imaginary parts of the shifts are supplied in
respectively (or in
when the option
is set). This option should only be used if you are an experienced user since this requires some algorithmic knowledge and because more operations are usually required than for the implicit shift
scheme. Details on the use of explicit shifts and further references on shift strategies are available in
Lehoucq et al. (1998)
Iteration Limit $i$ Default $\text{}=300$
The limit on the number of Arnoldi iterations that can be performed before
exits. If not all requested eigenvalues have converged to within
and the number of Arnoldi iterations has reached this limit then
exits with an error;
can still be called subsequently to return the number of converged eigenvalues, the converged eigenvalues and, if requested, the corresponding eigenvectors.
The Arnoldi iterative method converges on a number of eigenvalues with given properties. The default is for
to compute the eigenvalues of largest magnitude using
Largest Magnitude
. Alternatively, eigenvalues may be chosen which have
Largest Real
Largest Imaginary
Smallest Magnitude
Smallest Real
part or
Smallest Imaginary
Note that these options select the eigenvalue properties for eigenvalues of
problems), the linear operator determined by the computational mode and problem type.
Normally each optional parameter specification is not printed to the advisory channel as it is supplied. Optional parameter
may be used to enable printing and optional parameter
may be used to suppress the printing.
Monitoring $i$ Default $\text{}=-1$
, monitoring information is output to channel number
during the solution of each problem; this may be the same as the
channel number. The type of information produced is dependent on the value of
Print Level
, see the description of the optional parameter
Print Level
for details of the information produced. Please see
to associate a file with a given channel number.
Pointers Default $\text{}=\mathrm{NO}$
During the iterative process and reverse communication calls to
, required data can be communicated to and from
in one of two ways. When
is selected (the default) then the array arguments
are used to supply you with required data and used to return computed values back to
. For example, when
returns the vector
and the matrix-vector product
and expects the result or the linear operation
to be returned in
is selected then the data is passed through sections of the array argument
. The section corresponding to
begins at a location given by the first element of
; similarly the section corresponding to
begins at a location given by the second element of
. This option allows
to perform fewer copy operations on each intermediate exit and entry, but can also lead to less elegant code in the calling program.
Print Level $i$ Default $\text{}=0$
This controls the amount of printing produced by F12ADF as follows.
$=0$ No output except error messages.
$>0$ The set of selected options.
$=2$ Problem and timing statistics on final exit from F12ABF.
$\ge A single line of summary output at each Arnoldi iteration.
$\ge If ${\mathbf{Monitoring}}>0$, Monitoring is set, then at each iteration, the length and additional steps of the current Arnoldi factorization and the number of converged Ritz values; during
10$ re-orthogonalization, the norm of initial/restarted starting vector.
$\ge Problem and timing statistics on final exit from F12ABF. If ${\mathbf{Monitoring}}>0$, Monitoring is set, then at each iteration, the number of shifts being applied, the eigenvalues and
20$ estimates of the Hessenberg matrix $H$, the size of the Arnoldi basis, the wanted Ritz values and associated Ritz estimates and the shifts applied; vector norms prior to and following
$\ge If ${\mathbf{Monitoring}}>0$, Monitoring is set, then on final iteration, the norm of the residual; when computing the Schur form, the eigenvalues and Ritz estimates both before and after
30$ sorting; for each iteration, the norm of residual for compressed factorization and the compressed upper Hessenberg matrix $H$; during re-orthogonalization, the initial/restarted starting vector;
during the Arnoldi iteration loop, a restart is flagged and the number of the residual requiring iterative refinement; while applying shifts, the indices of the shifts being applied.
$\ge If ${\mathbf{Monitoring}}>0$, Monitoring is set, then during the Arnoldi iteration loop, the Arnoldi vector number and norm of the current residual; while applying shifts, key measures of
40$ progress and the order of $H$; while computing eigenvalues of $H$, the last rows of the Schur and eigenvector matrices; when computing implicit shifts, the eigenvalues and Ritz estimates of $H$.
$\ge If Monitoring is set, then during Arnoldi iteration loop: norms of key components and the active column of $H$, norms of residuals during iterative refinement, the final upper Hessenberg matrix
50$ $H$; while applying shifts: number of shifts, shift values, block indices, updated matrix $H$; while computing eigenvalues of $H$: the matrix $H$, the computed eigenvalues and Ritz estimates.
To begin the Arnoldi iterative process,
requires an initial residual vector. By default
provides its own random initial residual vector; this option can also be set using optional parameter
Random Residual
. Alternatively, you can supply an initial residual vector (perhaps from a previous computation) to
through the array argument
; this option can be set using optional parameter
Initial Residual
Shifted Inverse Imaginary
These options define the computational mode which in turn defines the form of operation
to be performed when
returns with
and the matrix-vector product
returns with
Given a
eigenvalue problem in the form
$Ax=\lambda x$
then the following modes are available with the appropriate operator
Regular $\mathrm{OP}=A$
Shifted Inverse Real $\mathrm{OP}={\left(A-\sigma I\right)}^{-1}$ where $\sigma$ is real
Given a
eigenvalue problem in the form
$Ax=\lambda Bx$
then the following modes are available with the appropriate operator
Regular Inverse $\mathrm{OP}={B}^{-1}A$
Shifted Inverse Real with real shift $\mathrm{OP}={\left(A-\sigma B\right)}^{-1}B$, where $\sigma$ is real
Shifted Inverse Real with complex shift $\mathrm{OP}=\text{Real}\left({\left(A-\sigma B\right)}^{-1}B\right)$, where $\sigma$ is complex
Shifted Inverse Imaginary $\mathrm{OP}=\text{Imag}\left({\left(A-\sigma B\right)}^{-1}B\right)$, where $\sigma$ is complex
The problem to be solved is either a standard eigenvalue problem,
$Ax=\lambda x$
, or a generalized eigenvalue problem,
$Ax=\lambda Bx$
. The optional parameter
should be used when a standard eigenvalue problem is being solved and the optional parameter
should be used when a generalized eigenvalue problem is being solved.
Tolerance $r$ Default $\text{}=\epsilon$
An approximate eigenvalue has deemed to have converged when the corresponding Ritz estimate is within
relative to the magnitude of the eigenvalue.
Vectors Default $\text{}=\text{RITZ}$
The routine
can optionally compute the Schur vectors and/or the eigenvectors corresponding to the converged eigenvalues. To turn off computation of any vectors the option
should be set. To compute only the Schur vectors (at very little extra cost), the option
should be set and these will be returned in the array argument
. To compute the eigenvectors (Ritz vectors) corresponding to the eigenvalue estimates, the option
should be set and these will be returned in the array argument
, if
is set equal to
(as in
Section 9
) then the Schur vectors in
are overwritten by the eigenvectors computed by | {"url":"http://www.nag.com/numeric/FL/nagdoc_fl24/html/F12/f12adf.html","timestamp":"2014-04-18T11:14:45Z","content_type":null,"content_length":"50246","record_id":"<urn:uuid:d08675a8-d6aa-4740-bee2-00e9195632cf>","cc-path":"CC-MAIN-2014-15/segments/1397609533308.11/warc/CC-MAIN-20140416005213-00548-ip-10-147-4-33.ec2.internal.warc.gz"} |
Ruby: Creating destructive methods.
Here's a basic example of using Ruby to create a class with destructive methods. Though this is not as practical as just typing the actual mathematic operations themselves. It does illustrate how
destructive functions work and how to write them. Note the 'return self' in each of the destructive methods. This allows you to perform the method chains, without it the method returns a fixnum by
default and the method chaining raises an error. If anyone knows of a DRY way to perform this let me know.
class Calculator
def initialize(value)
@value = value
def multiply!(value)
@value *= value
return self
def add!(value)
@value += value
return self
def clear!
@value = 0
return self
def calculate
puts @value
c = Calculator.new(10)
c.add!(2).multiply!(4).calculate #=> 48
c.clear!.add!(5).multiply!(4).calculate #=> 20 | {"url":"http://www.donttrustthisguy.com/2005/12/30/ruby-creating-destructive-methods","timestamp":"2014-04-18T02:58:13Z","content_type":null,"content_length":"4510","record_id":"<urn:uuid:3b8dea1f-95f4-407a-968e-9609eeadca19>","cc-path":"CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00084-ip-10-147-4-33.ec2.internal.warc.gz"} |
Larkspur, CA Trigonometry Tutor
Find a Larkspur, CA Trigonometry Tutor
...I took the CBEST on 08/21/10 and achieved a total passing score of 217. Specific section scores include: Reading - Scaled Score: 77; Mathematics - Scaled Score: 65; Writing - Scaled Score: 75.
I teach Genetics every year as part of my Biology or Life Science classes, and have also included Genetics units in a Bioethics course I used to teach.
43 Subjects: including trigonometry, chemistry, Spanish, geometry
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Electrical Engineering with a 4.1/4.3 GPA. I love, love, love math, and am very enthusiastic about teaching others!
27 Subjects: including trigonometry, chemistry, physics, geometry
...But then I take them back to the beginning, find out what they missed learning, and correct that. Math is like building a brick wall, each layer relies on a solid foundation. If you didn't
learn fractions or the multiplication table, you're never going to get through Algebra.
10 Subjects: including trigonometry, calculus, geometry, algebra 1
...It’s a great tool and a big confidence booster when you can master this subject. The concepts behind Calculus are actually simple and few in number. But they are often introduced in ways that
seem complex and confusing.
18 Subjects: including trigonometry, calculus, geometry, statistics
...Math: Math is one of my favorite subjects. Having done well in all of my math classes and subjects since grade school, I am qualified to tutor students in elementary math. I even completed
math books through the 6th grade level by the time I finished 4th grade.
29 Subjects: including trigonometry, chemistry, reading, calculus
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Larkspur, CA Trigonometry Tutors | {"url":"http://www.purplemath.com/Larkspur_CA_trigonometry_tutors.php","timestamp":"2014-04-17T20:04:52Z","content_type":null,"content_length":"24243","record_id":"<urn:uuid:c94e2278-e99c-44e1-b1eb-b19c2f93a8ac>","cc-path":"CC-MAIN-2014-15/segments/1398223207046.13/warc/CC-MAIN-20140423032007-00125-ip-10-147-4-33.ec2.internal.warc.gz"} |
artificial intelligence
Archive for the ‘topics in artificial intelligence’ Category
I hadn’t heard of Biomorphs until I started wandering through The Magic Machine: A Handbook of Computer Sorcery
[ z = z^3 + c]
[ z = sin(z) + z^2 + c ]
Biomorphs were discovered by Clifford Pickover at the IBM Research Center. Biomorphs are like Mandelbrot functions in that you iterate a simple function over the complex plane. The algorithm is in
the 2 example Java files below. Several more besides these two are known to reside between -20 and 20. { z^z + z^6 + c; z^z + z^6 + c; sin(z) + e^z +c; z^5 + c; z^z + z^5 + c; . . . } You should be
able to figure out how to create them by altering the two examples in the download file.
Some people breed biomorphs and let them evolve as an artificial life form.
Source Code
biomorphs.tar.gz ( 2 Java source files for the biomorphs above )
More information:
Fractal Geometry
Dr. Clifford Pickover home page
Mad Teddy’s Fractals #2 Biomorphs
Fractals are a fascinating toy, one can easily spend an afternoon lost in Mandelbrot or Julia sets. Mathematicians were aware of fractals as early as the 1700s but it wasn’t until we had computers to
do the calculations that we really discovered fractals.
Benoit B. Mandelbrot doing research at IBM was revisiting Gaston Julia’s work with fractals (1917) when he discovered the Mandelbrot set. Fractals are simple equations that are recursively computed.
These simple equations create complex shapes.
The Mandelbrot function is z = z^2 + c. z and c are complex numbers, z is set to zero, c is the position on the x ( x, yi ) plane. You recursively compute this function to obtain the Mandelbrot
fractal. Black is for the numbers that do not escape to infinity, the other colors represent how many loops it takes to escape.
Fractals have found some use in artificial intelligence. In the world of computer games, fractals create plant life, clouds, mountains and other scenery that would not be possible in such detail.
Parkinson’s patients are diagnosed by their gait. In 2004 a sensor was developed that measures the patient’s gait, and analyzes the gait using fractals. 2002 fractals were put to use to help predict
natural disasters and better model hurricanes. More recently fractal patterns have been found in solar wind. It is hoped this information will allow us to better predict solar storms.
Fractals have been found in Jackson Pollacks paintings and are being used to try to identify real paintings from fakes. They are also being used in image compression. A more fun way to play with
fractals is to use them to predict the stock and commodity markets.
Fractals ( Mandelbrot and Julia in Java – source code )
More information:
Fractal Geometry
Math on Display, Science News Online
Genius: Benoit Mandelbrot
3D Mandelbrot images
The Fractal Geometry of Nature, Mandelbrot ( pdf/ps )
Reasoning using graphical models is rapidly gaining popularity. Not just used in reasoning they are also becoming useful in computer vision where they are used to classify objects.
Graphical models include: constraint networks, Markov random fields, belief networks, Bayes networks and influence diagrams.
Traditionally algorithms for these networks have been either inference-based or search based. While inference algorithms are not practical time wise for most problems outside the classroom some of
the search algorithms perform quite well. Combinations of these algorithms are being used to solve knowledge representation problems not otherwise practical with just inference based algorithms.
Often algorithms are tailored for each specific problem.
The advantages of this method is that all possible outcomes are known and declared in the graphical model. Also the models are easily understood by humans, and the techniques used are well
Nodes on the graph represent facts or states and may be known or unknown, hidden or visible.
Connections between nodes may be hidden or not, they may be probabilities or equations and represent connections between the data.
The graphical representation allows us to decompose the big problem into smaller simpler to solve problems.
More information:
Graphical Model Algorithms at UC Irvine has a large list of software and example problems using graphical models in knowledge reasoning.
There are some Power Point and PDF tutorials you can download from Microsoft
Video lecture on Graphical models
A list of links to several software tools for doing graphical models
Graphical knowledge representation for human detection ( pdf )
Mean Field Theory for Graphical Models
Graphical Models for Discovering Knowledge (excellent beginner’s paper )
See also:
Bayesian logic
Knowledge representation and predicate calculus
Hidden Markov models
I’ve been hearing more and more about Lua and today finally had enough free time to download, install and play with it a bit.
I downloaded the source files and they compiled and installed painlessly on the Mac, just follow the directions in the INSTALL file. There is a folder full of test applications. Open up a terminal
and type:> lua hello.lua at a command prompt to be sure you’ve gotten everything up and running.
It is an odd looking language. Hello World looks like this:
– the first program in every language
io.write(“Hello world, from “,_VERSION,”!\n”)
Bisection method for solving de looks like this:
– bisection method for solving non-linear equations
delta=1e-6 – tolerance
function bisect(f,a,b,fa,fb)
local c=(a+b)/2
io.write(n,” c=”,c,” a=”,a,” b=”,b,”\n”)
if c==a or c==b or math.abs(a-b)
local fc=f(c)
if fa*fc<0 then return bisect(f,a,c,fa,fc) else return bisect(f,c,b,fc,fb) end
– find root of f in the inverval [a,b]. needs f(a)*f(b)<0
function solve(f,a,b)
local z,e=bisect(f,a,b,f(a),f(b))
io.write(string.format(“after %d steps, root is %.17g with error %.1e, f=%.1e\n”,n,z,e,f(z)))
– our function
function f(x)
return x*x*x-x-1
– find zero in [1,2]
( * These and several other example programs are included in the Lua download. )
You can keep your programs in text format and run them as scripts ( lua program.lua ) or you can compile them using the luac compiler into binary form. ( luac -o hello hello.lua )
Lua is dynamically typed, it can call and use C functions, it has it’s own threads so you can run routines concurrently * these are not OS threads.
The key word list for Lua is surprisingly small ( and, break, do, else, elseif, end, false, for, function, if, in, local, nil, not, or, repeat, return, then, true, until, while ) and it is a case
sensitive language. All else is defined in lua_functions of which there are all the things you’d expect.
I found projects including Apache modules so you can use Lua for webscripts, Palm versions so you can write Palm programs with Lua, and more than a few games. Lua is used in game scripting and
interactive AI scripting.
See also:
Cognitive Code develops software personal assistant using Lua
More information:
Lua Reference manuals
O’Reilly article, ‘Introducing Lua’
Introduction to Lua Programming
GameDev, ‘An Introduction to Lua’
Most datamining is either done using private corporate databases, online government databases, or with web bots, spiders and scrapers. RSS has made data mining the web trivial with PERL. I’m told it
is trival with PHP also, I’m still experimenting with that.
Currently, with the help of computers, most fields of science, the government and businesses are collecting data faster than they can comb through it. Some agencies have what would be hundreds of
years worth of data if it had to be parsed by humans. So we need to use artificially intelligent datamining to sort the data, develop useful informative rules about the data, and or put it into
useful formats for us humans.
This task of artificial intelligence is often put under the category of ‘machine learning’. Sometimes a set of rules is used. The rules may be created by an expert ( domain knowledge ) or discovered
through machine learning using statistics.
Problems with this type of machine learning include coming up with an insane number of rules that are far too specific ( over fitting ) and using example data that skews the learning. Other problems
include when do you decide you have the best set of rules? At what point is your algorithm good enough? Do you want all possible out comes or is it only specific outcomes you need? An example would
be do you need 40 categories of healthy plant or only descriptions and diseases for unhealthy ones?
Datamining has four main styles of sorting through data. Classification: classes are presented and future data is to be sorted into one of the given classes. Association: associations between data
are sought. Clustering: data is sorted into clusters usually using various traits as vectors. Prediction: in which some specific information, usually numeric is to be output.
Data details for data mining is often stored in ARFF Attribute-Relation File Format
More information:
Applications of Machine Learning and Rule Induction
Machine Learning ( Theory )
UT ML Group: Text Data Mining ( several papers here )
UCI Machine Learning Repository has over 160 data sets for you to use to test and develop your AI.
See also:
Electronic cop solves crimes
Finding new diseases for known cures | {"url":"http://herselfsai.com/category/topics-in-artificial-intelligence/page/2","timestamp":"2014-04-20T13:18:56Z","content_type":null,"content_length":"33619","record_id":"<urn:uuid:ffca218d-9104-40db-9d9e-f67bc6d929da>","cc-path":"CC-MAIN-2014-15/segments/1397609538787.31/warc/CC-MAIN-20140416005218-00614-ip-10-147-4-33.ec2.internal.warc.gz"} |
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First-hitting-time problem
A hitting time is the time when a system described by a random process first crosses a certain threshold, like the Earth’s climate when it makes a transition from one locally stable state to another
locally stable state, from today’s climate to a climate closer to Snowball Earth, for example. The hitting time in this case is the time when the system or process reaches a tipping point. Another
example is the rate of a chemical reaction, the tipping point is the point when the reaction starts, and the hitting time contains information about the rate at which the reaction takes place.
First-hitting-time problems or first-passage-time problems or first-escape-time problems try pose questions and to infer information about the probability distribution of the hitting time. This
information may be used to predict and control the system under scrutiny.
Stationary diffusion processes using Fokker-Planck equations
The problem
Disclaimer for the mathematically inclined reader: We silently assume that all functions are smooth, all integrals are defined and finite, and everything converges absolutely, in short: We assume we
are physicists.
Let’s assume we have a one dimensional diffusion process - a particle - described by an autonomous Fokker-Planck equation that starts at ${x}_{0}\in ℝ$. “Autonomous” means that the Fokker-Planck
operator has no explicit time dependency, and the stochastic process is therefore stationary.
Let’s say the particle is absorbed at $a,b$ with $a<{x}_{0}<b$, and let
$p\left({x}_{t},t\mid {x}_{0},{t}_{0}=0\right)$p(x_t, t| x_0, t_0 = 0)
denote the probability to find the particle at time $t$ at ${x}_{t}$, when it was at time ${t}_{0}=0$ at ${x}_{0}$. That is, $p$ is the solution of the Fokker-Planck equation with the initial
$p\left(x,{t}_{0}=0\right)=\delta \left(x-{x}_{0}\right)$p(x, t_0 = 0) = \delta(x- x_0)
In this paragraph we will show what we can find out about the distribution of the hitting time, which in this example is the time of the absorption of the particle.
Formula for the moments of the hitting time
From the formulation of our problem we already know that the probability that the particle is not absorbed at time t, that is that the particle is still in $\left(a,b\right)$ at time t, is
${\int }_{a}^{b}p\left(x,t\mid {x}_{0},0\right)dx=:G\left({x}_{0},t\right)$\int_a^b p(x, t | x_0, 0) d x =: G(x_0, t)
When we call the time of absorption $T$, then we have for the probability distribution of $T$:
$ℙ\left(T\ge t\right)=G\left({x}_{0},t\right)$\mathbb{P}(T \ge t) = G(x_0, t)
Therefore, the mean for any function $f\left(T\right)$ is
$⟨f\left(T\right)⟩=-{\int }_{0}^{\infty }f\left(t\right)\phantom{\rule{thickmathspace}{0ex}}dG\left({x}_{0},t\right)=-{\int }_{0}^{\infty }f\left(T\right)\phantom{\rule{thickmathspace}{0ex}}{\partial
}_{t}G\left({x}_{0},t\right)\phantom{\rule{thickmathspace}{0ex}}dt$\langle f(T) \rangle = - \int_0^{\infty} f(t) \; d G(x_0, t) = - \int_0^{\infty} f(T) \; \partial_t G(x_0, t) \; d t
For the mean of the hitting time itself we therefore get:
$⟨T⟩=-{\int }_{0}^{\infty }t\phantom{\rule{thickmathspace}{0ex}}{\partial }_{t}G\left({x}_{0},t\right)\phantom{\rule{thickmathspace}{0ex}}dt={\int }_{0}^{\infty }G\left({x}_{0},t\right)\phantom{\rule
{thickmathspace}{0ex}}dt$\langle T \rangle = - \int_0^{\infty} t \; \partial_t G(x_0, t) \; d t = \int_0^{\infty} G(x_0, t) \; d t
The last part follows from integration by parts.
For the higher moments we get of course
${T}_{n}\left(x\right):=⟨{T}^{n}⟩={\int }_{0}^{\infty }\phantom{\rule{thickmathspace}{0ex}}{t}^{n-1}\phantom{\rule{thickmathspace}{0ex}}G\left({x}_{0},t\right)\phantom{\rule{thickmathspace}{0ex}}dt$
T_n(x) := \langle T^n \rangle = \int_0^{\infty} \; t^{n-1} \; G(x_0, t) \; d t
G satisfies the backward equation
Now that we have a formula for the expectation and all higher moments of the hitting time involving the function $G\left({x}_{0},t\right)$, we need more information about $G\left({x}_{0},t\right)$.
To this end, we will see that the function $G\left(x,t\right)$ satisfies the backward Fokker-Planck equation. Note that we treat the function argument $x$ now as a variable, not as a constant
starting point ${x}_{0}$.
Since our system is homogeneous in time, we have
$p\left(x,t\mid {x}_{0},0\right)=p\left(x,0\mid {x}_{0},-t\right)$p(x, t| x_0, 0) = p(x, 0| x_0, -t)
which implies that we can state the backward Fokker-Planck equation via symmetry as a forward equation:
$\frac{\partial }{\partial t}p\left(x,t\mid {x}_{0},0\right)=\left(-f\left(x\right)\frac{\partial }{\partial x}+\frac{1}{2}{g}^{2}\left(x,t\right)\frac{{\partial }^{2}}{\partial {x}^{2}}\right)p\left
(x,t\mid {x}_{0},0\right)$\frac{\partial}{\partial t} p(x, t | x_0, 0) = ( - f(x) \frac{\partial}{\partial x} + \frac{1}{2} g^2(x, t) \frac{\partial^2}{\partial x^2} ) p(x, t | x_0, 0)
When we assume that all involved functions are “nice” enough so that we may interchange differentiation and integration, we see that $G\left(x,t\right)$ satisfies this backward equation, too. The
initial condition for $G\left(x,t\right)$ is obviously
$G\left(x,t\right)=\left\{\begin{array}{ll}1,& \text{for}\phantom{\rule{thickmathspace}{0ex}}a\le x\le b\\ 0,& \text{else}\end{array}$G(x, t) = \begin{cases} 1, & \text{for } \; a \le x \le b \\ 0, &
\text{else } \end{cases}
Recursion formula for the moments
We can integrate the backwards equation over time from $0$ to $\infty$, noting that
${\int }_{0}^{\infty }{\partial }_{t}\phantom{\rule{thickmathspace}{0ex}}G\left(x,t\right)=G\left(x,\infty \right)-G\left(x,0\right)=-1$\int_0^{\infty} \partial_t \; G(x, t) = G(x, \infty) - G(x, 0)
= -1
and get a differential equation for the expectation value $T\left(x\right)$ and similarly for all higher moments ${T}_{x}\left(x\right)$:
$f\left(x\right)\frac{d}{dx}T\left(x\right)+\frac{1}{2}g\left(x\right)\frac{{d}^{2}}{d{x}^{2}}T\left(x\right)=-1$f(x) \frac{d}{ d x} T(x) + \frac{1}{2} g(x) \frac{d^2}{ d x^2} T(x) = -1
with boundary conditions $T\left(a\right)=T\left(b\right)=0$ and for the higher moments
{thickmathspace}{0ex}}\frac{{d}^{2}}{d{x}^{2}}\phantom{\rule{thickmathspace}{0ex}}{T}_{n}\left(x\right)=-n\phantom{\rule{thickmathspace}{0ex}}{T}_{n-1}\left(x\right)$f(x) \; \frac{d}{ d x} \; T_n(x)
+ \frac{1}{2} \; g(x) \; \frac{d^2}{ d x^2} \; T_n(x) = -n \; T_{n-1}(x)
Therefore, in our example, we are able to calculate the moments of the hitting time recursively.
The following textbook is a general introduction to stochastic models for practitioners, including first-hitting-time problems:
• Crispin Gardiner: Stochastic methods. A handbook for the natural and social sciences. (ZMATH)
First hitting time problems have been a research topic in mathematics of their own, for some time now, here is a textbook that provides an introduction at an elementary level, for time independent
diffusion processes, using electrostatics as an analogy:
• Sidney Redner: A guide to first-passage processes. (ZMAZH) | {"url":"http://www.azimuthproject.org/azimuth/show/First-hitting-time+problem","timestamp":"2014-04-17T18:57:08Z","content_type":null,"content_length":"31380","record_id":"<urn:uuid:eedcd330-edd6-4172-a862-1130c5190594>","cc-path":"CC-MAIN-2014-15/segments/1398223206118.10/warc/CC-MAIN-20140423032006-00109-ip-10-147-4-33.ec2.internal.warc.gz"} |
Project Euler 105: Find the sum of the special sum sets | MathBlog
Project Euler 105: Find the sum of the special sum sets in the file.
Written by Kristian on 14 July 2012
Topics: Project Euler
Problem 105 of Project Euler is closely related to Problem 103 as has been mentioned in both problem descriptions. The problem formulation for this problem is
Let S(A) represent the sum of elements in set A of size n. We shall call it a special sum set if for any two non-empty disjoint subsets, B and C, the following properties are true:
1. S(B) ≠ S(C); that is, sums of subsets cannot be equal.
2. If B contains more elements than C then S(B) > S(C).
For example, {81, 88, 75, 42, 87, 84, 86, 65} is not a special sum set because 65 + 87 + 88 = 75 + 81 + 84, whereas {157, 150, 164, 119, 79, 159, 161, 139, 158} satisfies both rules for all
possible subset pair combinations and S(A) = 1286.
Using sets.txt, a text file with one-hundred sets containing seven to twelve elements (the two examples given above are the first two sets in the file), identify all the special sum sets, A[1], A
[2], …, A[k], and find the value of S(A[1]) + S(A[2]) + … + S(A[k]).
To be completly honest, I haven’t gotten all that many new insights from solving this problem compared to Problem 103. What we need here is to check if a set of numbers fulfils both properties. In
problem 103 we generated specific functions for checking each of the properties, so I have just reused these functions in order to solve this problem. I did discover a bug in Problem 103 when I
solved this, but that has been corrected now, so I did get something from the problem but not much.
The all we need is a bit of wrapping the problem in something that can read the input file and sum up all the sets that fulfils the properties. Personally I think that solving this problem first will
make a lot more sense, since it forces you to write functions for checking if a set is valid.
The code for this problem looks as in C#
int result = 0;
int j = 0;
string[] lines = File.ReadAllLines(filename);
foreach (string line in lines) {
//Parse the line
string[] segments = line.Split(',');
int dim = segments.Length;
int[] numbers = new int[dim];
for(int i = 0; i < dim; i++){
numbers[i] = Convert.ToInt32(segments[i]);
if (VerifyRule2(numbers)) {
int[] s = MakeSubsetSums(numbers);
if (VerifyRule1(s)) {
result += numbers.Sum();
this runs in
The sum of valid sss is 73702
Solution took 33,358 ms
You can see the methods either in Problem 103, or in the source code.
Wrapping up
Having already solved problem 103, this problem was close to trivial. However, I think the problem on it’s own is a rather fun one which does require a whole lot of thinking.
You are as always welcome to take a look at the source code, and as always I would love to hear from you about how you solved the problem.
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Mapping Human Whole-Brain Structural Networks with Diffusion MRI
Understanding the large-scale structural network formed by neurons is a major challenge in system neuroscience. A detailed connectivity map covering the entire brain would therefore be of great
value. Based on diffusion MRI, we propose an efficient methodology to generate large, comprehensive and individual white matter connectional datasets of the living or dead, human or animal brain.
This non-invasive tool enables us to study the basic and potentially complex network properties of the entire brain. For two human subjects we find that their individual brain networks have an
exponential node degree distribution and that their global organization is in the form of a small world.
Citation: Hagmann P, Kurant M, Gigandet X, Thiran P, Wedeen VJ, et al. (2007) Mapping Human Whole-Brain Structural Networks with Diffusion MRI. PLoS ONE 2(7): e597. doi:10.1371/journal.pone.0000597
Academic Editor: Olaf Sporns, Indiana University, United States of America
Received: May 11, 2007; Accepted: June 1, 2007; Published: July 4, 2007
Copyright: © 2007 Hagmann et al. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction
in any medium, provided the original author and source are credited.
Funding: This work was supported by the Swiss National Science Foundation (SNSF), Professor Pierre Schnyder and Mr Yves Paternot as well as the Center for Biomedical Imaging (CIBM) of the
Geneva-Lausanne Universities and the EPFL, the foundations Leenaards and Louis-Jeantet. The sponsors had no role in the design or analysis of the study.
Competing interests: The authors have declared that no competing interests exist.
Biological neuronal networks, and in particular the human brain, are remarkable natural systems capable of complicated patterns of behavior. This capability seems possible due to the combination of
an enormous computational capacity given by a huge amount of neurons, and a highly evolved communication network [1]. To understand the mechanisms behind higher-level brain functions, a detailed
study of the individual neural cells is clearly insufficient [2]; global functional and structural properties of such a complex system need to be considered as well [3]. This requires, first of all,
a good knowledge of the network architecture of the entire brain. A graph representing the connectivity of the brain (henceforth called a ‘brain network’) can be analyzed at various scales. Probably
the most obvious is at the neuronal level, where each neuron is a separate node in the graph and physical connections between neurons are reflected by the edges. This detailed view, however, is
feasible only for the most primitive animals such as C. elegans with a brain made of 302 neurons [4]. A graph of the human brain consisting of 10^11 nodes and 10^16 edges is not only impossible to
obtain with current techniques, but it also would carry a great deal of information that is irrelevant from the global organization point of view. We must therefore resort to a different level of
granularity, where a node represents thousands or millions of neurons grouped together. Unfortunately, such available graphs are today limited to small post-mortem datasets (only 50–70 nodes) of rat
[5], cat [6], [7] and monkey [8] brains, whereas larger datasets of animal and human brains are missing [9]. In the coming years, an immense effort will be needed to map at various scales and to
create a large database of reliable information on the brain connectivity of higher order animals, especially of the human [10], [11].
Crick and Jones stated that “Clearly what is needed for a modern human brain anatomy is the introduction of some radically new techniques” [9]. In this paper we propose a methodology derived from
diffusion MRI tractography [12]–[15] to map at a millimetric scale the structural white matter connectivity of the whole brain. The resulting network consists of nodes representing small areas of
white matter–gray matter (WGM) interface, and weighted edges that capture long distance connection densities between these areas. The innovation it brings is fourfold. First, our methodology has a
relatively high resolution; the resulting networks consist of thousands of nodes, which are 1–2 orders of magnitude larger than the networks currently available (thousands versus tens of nodes). This
opens several innovative investigation possibilities. Mainly it allows us to study brain connectivity not only locally but also globally by characterizing the topological features of this large-scale
network. Such global characterizations are essential for a better understanding of brain communication. Second, our approach is non-invasive. This allows us to study the topology not only of animal
or post-mortem brains, but also, for the first time, of the living human brain. Third, for each subject we infer an individual network of the entire brain. This potentially allows us to compare
individual subjects or groups of subjects, e.g., brains from healthy controls and from patients with clinical conditions. In contrast, the datasets available to date were collected part by part from
a number of animals of the same species, and hence reflect a kind of “average” brain in the population. Fourth, our approach is efficient. It only requires performing an MRI scan on the subject
(which takes about an hour or less depending on resolution and signal-to-noise ratio of the imaging system), and to process the data on a computer.
As an illustration of our approach, we analyze the basic brain graph properties of two healthy volunteers. In particular, we study a number of distributions derived for nodes (e.g., degree, strength)
and edges (weight, length). We also answer some questions related to the topology, e.g., “Is the brain network a small world?”. With technology improvements, finer resolution and a better Signal-to
Noise Ratio (SNR), or a deeper analysis of the network, more complex and accurate network characteristics will be accessible, thus potentially contributing to the answers of some key questions in
Material and Methods
The path from diffusion MRI to a graph mapping brain connectivity is a four step process: (1) diffusion MRI acquisition, (2) white matter tractography, (3) white matter-gray matter interface
partition into Regions Of Interest (ROIs) and (4) network construction. We present a general scheme of our methodology in Figure 1. Below we first describe each step illustrated with intermediary
results. In the next section, we investigate some fundamental properties of the brain network inferred with our approach.
Figure 1. Mapping the network of brain structural connectivity with diffusion MRI is a process made of four steps.
First, Diffusion Spectrum MRI (DSI) is performed on a subject or sample. This acquisition provides a 3D diffusion function at every location in the brain. This data set is called a diffusion map. It
is shaped by the local tissue characteristics, in particular by the orientation of axonal bundles existing in the brain. Second, based on this map we generate a number of 3D curves (called fibers)
that follow the path laid by the white matter axonal bundles. Third, independently from the previous step, we use a heuristic that partitions the brain white matter-gray matter interface into small
areas of equal surface (called Regions Of Interest-ROIs) covering the whole cortex and deep cerebral nuclei boundaries. Finally, in the fourth step, we combine the output of steps two and three: the
ROIs become nodes and the fibers are transformed into edges in the resulting graph. This graph estimates the density of white matter connections between any two regions of gray matter.
Step 1: MRI acquisition
We use Diffusion Spectrum Imaging (DSI) [15], [16] . It is a diffusion MRI method that images the 3-dimensional diffusion function in every brain voxel and results in a 6-dimensional image called a
diffusion map. This new method has, contrary to Diffusion Tensor MRI (DTI) , sufficient angular resolution to map accurately the diffusion with a non-Gaussian behavior. Accordingly it can see
intra-voxel diffusion heterogeneity caused by crossing neuronal tracts, which is essential for an accurate mapping of axonal trajectories.
In the present experiment, after having obtained the informed consent of two healthy volunteers, two data sets are acquired at 3T with an Achieva (Philips, Einthoven, The Netherlands) MRI scanner
using a diffusion weighted spin echo EPI technique [17], [18]. The timing parameters of the pulse sequence are TE/TR/Δ/δ = 154/3000/47.6/35 ms, maximum diffusion gradient intensity is 80 mT/m,
yielding a maximal b value of 12000 s/mm^2 [19]. The matrix size is 128×128 and the slice number is 30. The field of view is 256×256 mm^2 and the slice thickness 3 mm, which yields a voxel size of
2×2×3 mm^3. The classical DSI scheme we use goes as follows: diffusion-weighted images covering the whole brain are acquired for 515 different values of diffusion sensitizing gradient intensity and
direction (i.e., different q-vectors) [20], comprising in q-space the points of a cubic lattice within the sphere of 5 lattice units in radius. We take q = aq[x]+bq[y]+cq[z], with a,b,c integers and
, and q[x], q[y], q[z] denoting the unit diffusion sensitizing gradient vectors in the three respective coordinate directions. Next, we process these 515 images as follows. First, we reconstruct the
3D diffusion function, or Probability Density Function (PDF) at each brain location by taking the discrete 3D Fourier transformation of the signal modulus sampled in q-space. The signal is
pre-multiplied by a Hanning window before the Fourier transformation in order to ensure a smooth attenuation of the signal at high |q| values. With this procedure and the above parameters, the PDF is
sampled over an isotropic 3-dimensional field of view of 100 µm with a nominal isotropic resolution of 10 µm. The result, called a diffusion map, is a 6-dimensional image that associates a
3-dimensional diffusion function with every brain position voxel. From this map, at each voxel, we compute an Orientation Density Function (ODF) Φ(u), by projection of the PDF in the radial
direction. If u is a 3D vector with |u| = 1, we define:(1)
where p(.) is the 3D PDF, ρ is the radius, ρ ^2dρ is the 3D volume element and the integral is evaluated as a discrete sum over the available range ρЄ[0,5]. The ODF Φ(u) is a function defined on a
discrete sphere and captures the diffusion “intensity” in every direction. It is evaluated for a set of vectors u[i] that are the vertices of a tessellated sphere that has a mean nearest-neighbour
separation about 10°.
In Figure 2 A and B we show a diffusion map, i.e., the ODF at every location in the brain. The ODFs are represented as deformed spheres with the radius proportional to Φ(u). The color-coding adds
some more clarity, with blue codes for the cranio-caudal, red for left-right and green for antero-posterior direction.
Figure 2. Tractography.
A) The result of the “diffusion MRI acquisition” step. In every voxel of a coronal slide the Orientation Density Function (ODF) captures locally for every direction the diffusion “intensity”. B) Zoom
in the centrum semi-ovale C) Each ODF is replaced by a set of vectors defining its local maxima. D) Fibers are computed following the local diffusion maxima; they are uniformly initiated over the
whole brain white matter. See also Video S1 in Supporting Information.
Step 2: White matter tractography
Tractography is a post-processing method that based on the diffusion map, constructs 3-dimensional curves of maximal diffusion coherence. These curves, called fibers, are the estimates of the real
white matter axonal bundle trajectories [21]. We use a tractography algorithm specifically designed for DSI data to create a set of such fibers for the whole brain [15], [22] which is summarized
1. Detection of the directions of maximum diffusion. At each voxel, we define a set of directions of maximum diffusion as local maxima of Φ(u) (i.e., vectors U[i] such that Φ(u[j])<Φ(U[i]) for all u
[j] adjacent to U[i] in the sampled tessellated sphere (Figure 2 C).
2. Fiber computation. We initiate the same number of fibers for every direction of maximum diffusion in every white matter voxel. For example, in a voxel with 2 directions, we initiate 30 fibers
along each direction, total 60. The starting points are chosen at random within the voxel. Next, from each such point we begin growing a fiber in two opposite directions with a fixed step of 1
mm. On entering a new voxel, the fiber growth continues along the direction of the vector U[j] (in the new voxel) whose orientation is the closest to the current direction of the fiber. If this
results in a change of direction sharper than 15°/mm, the fiber is stopped. The growth process of a valid fiber finishes when both its ends leave the white matter. The resulting fibers can be
interpreted as an estimate of the white matter axonal bundle trajectories (see Figure 2 D); in this article we use about 3 million initialization points of which only about one half to two third
connect the white-gray matter interface and are retained (See also Vie).
3. Filtering the edges. In each of our data sets we have around 1.5 to 2 million fibers. For the graph of ~1'000 nodes they translate into about 50'000 edges. The number of edges in the final
network depends on the number of initialized fibers. To investigate network properties over a wider range of connection densities we devised two ways to filter edges by varying the number of
initialized fibers or by taking into account the edge weight:
1. Random fibers. Although for every data set we generate around 3 million fibers, this is not any special number. We could as well take 100 thousand or 10 million fibers. As presented in Figure
S1 in Supporting Information, this would strongly affect the number of resulting edges. Therefore our first approach to limit the number of edges is to take a random subset of a given size
out of our 3 million fibers, which boils down to reducing the number of fiber initialization. The study of the whole spectrum of fiber numbers gives us a better view than the study based on
one, arbitrarily chosen number.
2. Top-weight edges. In the second method we consider the edges built based on all fibers, and delete only the edges with the smallest weights (according to some threshold). The heavy-tailed
distribution of edge weights guarantees that we always retain most of the “edge mass” and reject only the edges with very small weights that are possibly the result of noise. Indeed noise may
create aberrant diffusion maxima that in turn produce thin aberrant diffusion coherence paths across the data resulting into artefactual edges of small weight.
Step 3: White matter-gray matter (WGM) boundary partition into ROIs
The goal of the third step is to partition the WGM interface in a number of areas that we call Regions Of Interest (ROIs). In this step we use exclusively the 3D mask of the brain WGM interface
(i.e., the cortex and the thalamus for simplicity). The ROIs should be compact and of similar surface (counted in the number of voxels), which is a non-trivial task to achieve for the complex,
strongly folded shape of the brain. For instance, a straightforward approach would be to partition this interface according to some 3D regular lattice [23]. Unfortunately, this approach results in
large differences in ROI sizes-up to two orders of magnitude. Furthermore we do not want to partition the WGM into predefined areas like for example Brodmann's as they are too coarse (only about 50
to 55 areas) to analyze large scale network properties at high resolution. We have therefore developed a two-phase partitioning heuristic, as follows. First, we choose a WGM interface voxel at random
and iteratively connect it to the neighbouring WGM interface voxels until it reaches the desired size; this structure becomes our first ROI. Similarly, we grow other ROIs, one by one, always starting
near the ones that have already been created. We repeat this procedure until all the WGM interface is covered with ROIs. This gives us already quite a good partition, however, it can be easily
further improved. Therefore, in the second phase of our heuristic we restart the ROI growth process. This time we grow all the ROIs simultaneously, starting from the centres of gravity of the ROIs
found in the first phase. This results in a much better compactness of the ROIs with surface variations of less than 10% (See Figure S2 of Supporting Information). An example of the final result is
shown in step 3 of Figure 1 (see also Video S2 in Supporting Information).
Step 4: Network construction
Finally, in the fourth step, we combine the output of steps two and three and create the graph of brain connectivity. Every ROI constructed in step three becomes a node in the graph. We denote by ROI
(v) the ROI that is associated with the node v. Two nodes v and u are connected with an edge e = (v, u) if there exists at least one fiber f with end-points in ROI(v) and ROI(u). For each edge e we
define its length l(e) and weight w(e), as follows. Denote by F[e] the set of all fibers connecting ROI(v) and ROI(u) and hence contributing to the edge e. The length l(e) of the edge e is the
average over the lengths of all fibers in F[e], i.e., l(e) = 1/|F[e]|⋅Σ[fЄFe]l(f), where l(f) is the length of fiber f along its trajectory. The weight w(e) captures the connection density (number of
connections per unit surface) between the end-nodes of the edge e, and is defined as w(e) = Σ[fЄFe]1/l(f). The correction term l(f) in the denominator is needed to eliminate the linear bias towards
longer fibers introduced by the tractography algorithm. Indeed let us assume that an axonal bundle b exists in reality and has a length l(b). The tractography algorithm starts in some voxel of the
white matter and follows the most probable direction of a bundle. If it happens to start in a voxel that is traversed by the bundle b, the algorithm follows b until it reaches the white matter
boundary. As every voxel in the white matter is chosen as a starting point the same number of times, the longer the bundle b is, the more voxels it traverses and the more often it is followed by the
tractography algorithm, introducing a linear bias that must be corrected.
ROI size is a parameter of our methodology. On the one hand, a natural lower limit for this size is one voxel of the WGM interface. However, we prefer to combine at least several voxels into one ROI
to be sure to have a representative number of fibers connecting this ROI to the rest of the brain. On the other hand, taking ROIs that are too large results in a network of insufficient resolution
and of trivially small size. In our simulations we set the ROI size to between 8 and 64 voxels of WGM interface. This results in a weighted network of between 500 to 4000 nodes representing small
areas of WGM interface between ~250 mm^2 (64 voxels/ROI) and ~30 mm^2 (8 voxels/ROI), respectively. This graph has between 25'000 to 100'000 edges that represent axonal bundles of millimetric or
centimetric diameter. For simplicity, in the remainder of this text we analyze graphs of about 1'000 nodes. In particular, |V[1]| = 1'013 nodes and |E[1]| = 47'217 edges for suject 1, and |V[2]| =
956 and |E[2]| = 50'199 for subject 2. The two graphs were built based on about |F| = 3 million fibers generated by the tractography algorithm. Results obtained for other granularities, from |V| =
500 to 4'000 nodes, are qualitatively similar (see Figure S3 of Supporting Information).
Results and Discussion
Once the network constructed, several graph statistics characterizing the architecture of the network can be computed and examined.
Node statistics
We first turn our attention to the nodes of our graph. A basic characteristic of a node v is its degree, i.e., the number of edges incident on v. Many complex networks such as the World Wide Web,
Internet, protein networks, ecological networks or cellular networks, have been shown to follow a heavy-tailed node degree distribution [24]. In other words, they have a very significant number of
high degree nodes, called hubs. As such networks, also called “scale free”[24], are characterized by relatively short distances between any two nodes and by high robustness to random failures [25],
they seem, at first sight, to be good candidates for brain topology. Surprisingly, we find in our dataset that this is not the case. In Figure 3, we plot the node degree distribution (a), and a
closely related node strength distribution (b). (The strength s of a node v is the sum of weights of all edges incident on the node v, s(v) = Σ[e:vЄe]w(e)[26].) Although these distributions cover
more than two decades, they are roughly linear in the log-lin scale, which indicates their exponential tail. This is probably the first time that a claim about the node degree distribution of
cortical structural connectivity mapped at high spatial resolution can be made. The networks available and studied to date [27], [28] are simply too small to judge if their node distribution is
exponential, heavy tailed, or yet different. It is worth mentioning that in contrast to structural analyses, some functional brain networks have been described as scale-free [29].
Figure 3. Basic characteristics of nodes in the graph of brain connectivity.
P(d) [P(s)] is the probability that a randomly chosen node has the degree [strength] equal to d [s]. The node degree distribution (a) and node strength distribution (b) are lin-binned and plotted in
log-lin scale. Color code: subject 1 (blue circles), subject 2 (green diamonds)
A closer look at node degrees suggests that, from a developmental and energy optimization point of view, hubs do not seem to be favored. This was suggested by [30] who modeled the development of
frontal macaque cortex by a spatially embedded growing graph where preferential attachment occurs as an exponentially decaying function of spatial distance and growth limited in space. Amaral et al.
[31] modeled network growth where the node degree expansion is attenuated through node aging and energy limitations. These two models, like our measurements, resulted in networks with an
exponentially decaying distribution. Furthermore, it is quite unlikely to find hubs in the gray matter, because we know that the neuronal density does not change over orders of magnitude across the
cortex [32], [33].
Edge statistics
The edge length l distribution decays exponentially (Figure 4a), indicating that the number of long connections is small. The edge weight w distribution is much broader and close to heavy-tailed (
Figure 4b). Therefore, there are a significant number of very strong connections that are predominantly short as demonstrated in Figure 4c. This observation is in agreement with the results of other
complementary studies on the organization of the brain that suggest that brain favors, with some intriguing exceptions, locally dense communication and minimizes the number of long distance
connections [34]. For instance, by examining many alternative arrangements of macaque pre-frontal cortex, [35] showed that the layout of cortical areas minimizes the total lengths of the axons needed
to join them. A similar observation was made by [36] about the intrinsic gray matter connectivity of mice where the volume fraction of axons and dendrites seems close to optimal. The work of [37]
indicate that there is an evolutionary conserved scaling of the volume of the white matter as the 4/3 power of the volume of the gray matter, which can be explained by the fact that global geometry
of the cortex minimizes the average length of the long-distance fibers while keeping the average connection density of long-distance connection fibers constant. However recent reports suggest the
organization of neural networks is not only shaped by the minimization of total wiring length. Multiple constraints seem to be involved, not only wiring length but also the average number of
processing steps (related to the average distance between node) [38].
Figure 4. Basic characteristics of edges in the graph of brain connectivity.
(a) The distribution of edge lengths l in log-lin scale, lin-binned. (b) The distribution of edge weights w in log-log scale, log-binned. (c) Scatter plot of w vs l. The symbols are lin-binned
average values for subject 1 (blue circles) and subject 2 (green diamonds).
Network topology
Having examined separately the distributions of nodes and edges, we now discuss the topology of the graph itself. An interesting question one can ask is: “Is the brain a small world?”. The more
formal definition of a small world graph involves two metrics, clustering coefficient c and average shortest path length <sp>. We follow [39], who define the clustering coefficient c as the
probability that two randomly chosen neighbors of a node are also direct neighbors of each other, i.e., c = 1/|V|⋅Σ[vЄV]c(v), where c(v) is the number of edges interconnecting the neighbors of the
node v divided by the number of all possible edges. The average shortest path length <sp> is the average distance between any two nodes in graph. If the graph is disconnected, only the largest
connected component is considered. A graph is called a small world if it has (i) a clustering coefficient much greater than that of equivalent random graphs and (ii) the average shortest path length
<sp> is comparable with that of a random graph with the same number of nodes and edges [39].
There are two issues that we have to address before we attempt to decide if our graph G of brain connectivity is a small world. First, G is weighted. As there exists no standard way of generalizing
the clustering coefficient to weighted graphs [see e.g. [40], [41]] and it is not obvious how to interpret edge weights when computing the average shortest path length, we have decided to treat every
edge equally and apply the classic unweighted approach [39]. Second, the number of fibers that are initiated during tractography determines the density of graph G. In order to explore the effect of
connection density on our results, we exclude some of the edges by applying the two filtering techniques described above.
We present the results in Figure 5. As a reference we take a random graph not only with the same number of nodes and edges (as proposed in [39]), but also with the same degree distribution as the
brain graph. This graph was generated with the rewiring technique described in [42]. Preserving the degree distribution allows us to rule out this factor from the set of possible reasons of observed
differences between the brain and the reference topology. For any number |E'| of edges remaining after the filtering, the graph of brain connectivity has a much higher clustering coefficient than the
corresponding random graph; this is especially well pronounced for the “Top-weight edges” graph. At the same time their average shortest path lengths <sp> are comparable. Hence our measurements
suggest that the small-world model fits the brain network. Indeed, the small-world topology seems well suited for the neuronal network of the brain when thinking in evolutionary and developmental
terms. This is because it is a good compromise between full connectivity, which would be very costly in terms of wiring (i.e., brain volume) and power supply [43], and a lattice topology that impairs
massively long distance communication. Furthermore, the combination of high local clustering and small average shortest path length with efficient neural coding [44] allows for a distributed
computing where only a small fraction of local intense computation needs to be transmitted to distant regions, which may be sufficient for synchronous brain activity [45]. The small-worldness of the
brain network was already advocated before, based on relatively small networks (50-70 nodes) resulting from post-mortem tracing studies in rat, macaque monkey and cat brain regions [27], [46], [47].
In contrast, the approach presented in this article provides, for the first time, evidence for the presence of small-world topology in the structural connectivity of the human cerebral cortex.
Moreover, the one to two orders of magnitude higher resolution resulting from our method (thousands vs. tens of nodes) increases the confidence we have in the derived statistics.
Figure 5. Average shortest path <sp> and clustering coefficient c as a function of the number of edges in the brain graph |E'|.
The edges are chosen from the set of all edges E either giving the priority to the edges with high weights (“Top-weight edges”, left column), or based on a random subset of fibers (“Random fibers”,
right column). As a reference we take a random graph with the same number of nodes and edges, and the same degree distribution. Color code: subject 1 (blue circles), subject 2 (green diamonds),
random graph reference (black filled circles). The results are averaged over 10 realizations of the “random fibers” filtering and random graphs; the confidence intervals (not shown) are comparable
with the symbol size.
Intra- and inter-individual network differences
In order to test the robustness of our methodology and because of uncertainty about the ideal number of nodes for the presented methodology, we measured the brain network at 4 different node
resolutions on data set 1 (see Figure S3 of Supporting Information). We notice that for scales varying between 500 and 4000 nodes and 25'000 and 100'000 edges respectively, the global network
topology is preserved. This is a range of scales that matches the sensitivity of the method, as we do not expect to be able to accurately map tracts smaller than several milimeters in diameter, which
is presently the size of our ROIs. Pushing the network “resolution” higher by increasing the number of nodes and reducing the surface area of the ROIs would increase the quantification noise
(limiting the number of fibers per ROI), which ultimately would destroy the information contained in the network model. On the other hand, increasing the ROI size, limits the precision of the
mapping, potentially grouping together pieces of gray matter that are functionally different. At the scale we use in this study, we expect that the chance that ROI overlaps several critically
different cortices is not higher than the inaccuracy related to the matching of template atlas on our data. Notwithstanding the advantage with a fine grain method to always be able to group
arbitrarily sets of nodes in order to study connectivity patterns between for example well known functional or anatomical areas like Brodmann's.
While basic connectivity parameters differed slightly for data sets 1 and 2 (see Table 1), the global properties are quite similar. The differences that we observe in Figures 3, 4, 5 may or may not
reflect the individual properties of the subjects. Clearly, more experiments and studies are needed to be able to address the issue of between-subject variability with a high level of confidence. We
plan to address these issues in our future work.
Table 1. Network construction parameters for data set 1 and 2
The question of investigating structural network deteriorations in diseased populations like schizophrenics or demented patients is challenging and should be addressed in the future [48], [49]. The
first issue is to decide on the most representative measure of tract degradation. Should we use the connection density as presented in this article? Or are differences in connectivity better captured
through the use of the mean fractional anisotropy or the diffusion trace along a connection as is done in several DTI studies [50], [51]? If we want to capture the global network topology, the only
requirements are to use the same MRI acquisition and simulation parameters, such as the number of nodes and the way fibers are initiated. The task becomes much more challenging if our goal is to
perform an edge-by-edge comparison. The problem is twofold. First, we have to match the nodes across subjects. This requires precise cortical registration tools that work with a sub-centimetric
precision. Second, identifying significant changes when testing thousands of edges at once will either require a large cohort or strong network changes, as the significance threshold needs to account
for multiple testing.
Although our methodology yields promising results, we need to keep in mind that there are some steps prone to various kinds of noise and distortions whose effect is difficult to evaluate. First of
all, we work at a given level of granularity. The spatial and angular resolution of our diffusion MRI experiment is limited, which makes it difficult to tell much about submillimetric fiber tracts
and crossing axonal bundles separated with angles smaller than 20°. The ROIs have a given size, which automatically groups tens of thousands of neurons into a single node. Noise is also introduced
during the MRI acquisition, and the tractography algorithm might not perfectly model the relationship between water diffusion and axonal orientation. Although all these points are constantly being
improved, there will always remain a huge discrepancy between our constructed graph and the real neuronal network made of 10^11 neurons and several orders more connections.
Quality control
Nevertheless, diffusion MRI tractography is a widely used and accepted method to map axonal bundle trajectories. Furthermore it was validated experimentally to large extent in the case of DSI. First,
[21] show that the ODFs produced by DSI match accurately the fiber orientations in a phantom and follow accurately the optic tracts in the rat. Second, [52] validate the method in the monkey by
comparing DSI tractography with histological autoradiographic tracing over 10 association tracts. This study shows a remarkable agreement of results between two fundamentally different techniques. In
addition to these general arguments, we have also tested our particular data set. Figure 6 presents a qualitative impression of the type of data revealed by our method, by showing the connectivity of
part of the cortical visual system. More specifically we investigate the well-studied connections between areas V1, V2, V3, V5 and the lateral geniculate body [53]–[55]. The different visual areas
were identified manually based on the gyral anatomy and consist each of a set of ROIs. A set of well known connections was identified for the purpose of illustrating the tractography method without
claiming to be a detailed study of the visual system which would require a functional retinotopic mapping of the visual areas and an extensive search and study of the individual fiber bundles. Our
data not only reveals intermediate length connections between V1 and V2 or between V2 and V3, but also the well known long range connections such as i) the optic radiation–linking the lateral
geniculate body to V1 , ii) V1 homotopic callosal projections, which are connections that take actually their origin more at the junction between V1 and V2 [56] and iii) V2–V5. Furthermore, the
weights of these edges are by far higher than the corresponding median weights over the whole brain (see Figure S4 in Supporting Information). This gives a good level of confidence that the observed
visual connections are not caused by some random effect.
Figure 6. Visual system white matter connectivity derived from tractography: Views from the left (a), from the top (b), zoomed anterior (c) and posterior (d).
White matter–gray matter interfaces: Magenta = posterior part of the thalamus, blue and gray = right and left V1, orange = V2, cyan = V3, red = V5. Fibers: yellow = homotopic V1, red = V1–V2, white =
V2–V3, green = V2–V5, blue = lateral geniculate body–V1. See also Video S3 in Supporting Information.
In this article we have proposed a methodology for mapping networks of structural connectivity in the brain. Our approach is non-invasive, efficient, individual and of relatively high-resolution. For
the first time we can globally characterize brain connectivity with individual tract properties or network statistics in an individual living subject. Based on the analysis of two healthy subjects we
found that the graph of the human brain is a small world, but not a scale-free network. Large new areas of application are foreseen; in basic neuroscience our technique may contribute to the
discovery of the general principles regulating communication, evolution and development of the brain; in clinical neuroscience it may shed new light into diseases of disorders that involve
disruptions of anatomical brain connectivity.
Supporting Information
The number of edges in the resulting graph as a function of the number of fibers connecting two points in the gray-white mater interface. The straight line represents the y = x relation.
(0.05 MB TIF)
Histograms of ROI sizes for the number of ROIs ranging from N = 506 to 4052 in subject 1. One voxel translates to about 4 mm^2.
(0.94 MB TIF)
The results generated for all four considered scales in subject 1. The symbols in the last two rows are (as in the main paper): blue circles-“Top-weight edges”, red triangles-“Random fibers”, and
black disks-“Random graph”.
(6.92 MB TIF)
Comparison of edge weights inside the visual system with the rest of the brain. Each box plot represents all edge weights in the brain of similar white matter length. The big black dot represents the
weight of the considered connection, namely V1-V2, V2-V3, V2-V5, as well as the connections between the lateral geniculate body and V1 (LGB-V1), and between left and right V1 areas (V1^left-V1^
right). Each connection is compared with the other connections in the brain of same white matter length as short connections are usually denser that long ones. The considered connections in the
visual system are largely above their respective medians (horizontal line in within each box, whiskers represent 5th and 95th quantiles).
(0.39 MB TIF)
Whole brain tractography result in subject 1.
(5.23 MB MPG)
Partition of the white-gray matter interface in approximately 1000 ROIs.
(2.01 MB MPG)
Connections between different visual areas.
(1.65 MB MPG)
Author Contributions
Conceived and designed the experiments: PH MK XG. Performed the experiments: PH MK XG. Analyzed the data: PH MK XG PT VW RM JT. Contributed reagents/materials/analysis tools: PH. Wrote the paper: PH
Proof of principle of in vivo whole-brain fiber tracking
Posted by Mietchen | {"url":"http://www.plosone.org/article/info%3Adoi%2F10.1371%2Fjournal.pone.0000597","timestamp":"2014-04-20T23:02:51Z","content_type":null,"content_length":"187872","record_id":"<urn:uuid:5ffa5f11-260c-4068-bf30-a7d061b6cc58>","cc-path":"CC-MAIN-2014-15/segments/1397609539230.18/warc/CC-MAIN-20140416005219-00332-ip-10-147-4-33.ec2.internal.warc.gz"} |
MEANDEV estimates the Mean deviation
function R = meandev(i,DIM)
MEANDEV estimates the Mean deviation
(note that according to [1,2] this is the mean deviation;
not the mean absolute deviation)
y = meandev(x,DIM)
calculates the mean deviation of x in dimension DIM
DIM dimension
1: STATS of columns
2: STATS of rows
default or []: first DIMENSION, with more than 1 element
- can deal with NaN's (missing values)
- dimension argument
- compatible to Matlab and Octave
see also: SUMSKIPNAN, VAR, STD, MAD
[1] http://mathworld.wolfram.com/MeanDeviation.html
[2] L. Sachs, "Applied Statistics: A Handbook of Techniques", Springer-Verlag, 1984, page 253.
[3] http://mathworld.wolfram.com/MeanAbsoluteDeviation.html
[4] Kenney, J. F. and Keeping, E. S. "Mean Absolute Deviation." �6.4 in Mathematics of Statistics, Pt. 1, 3rd ed. Princeton, NJ: Van Nostrand, pp. 76-77 1962.
This function calls: This function is called by: Generated on Fri 22-May-2009 15:02:45 by m2html © 2003 | {"url":"http://biosig-consulting.com/matlab/help/NaN/meandev.html","timestamp":"2014-04-18T18:21:14Z","content_type":null,"content_length":"3657","record_id":"<urn:uuid:d15e704a-1118-4828-baa7-e2da35e2f87f>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.7/warc/CC-MAIN-20140416005215-00356-ip-10-147-4-33.ec2.internal.warc.gz"} |
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This problem's been bothering me for a while now
Hi ME101! Welcome to PF!
Once given an initial velocity, can there be a condition in which the masses keep rotating (forever) about the axis?
1 thing I found out was that the center of mass of the system needs to be above the axis of rotation at all times, right?
Yes, except you mean it needs to be
the axis of rotation … the axis is a line, and the centre of mass needs to be on that line.
The two principles of physics that you need are good ol' Newton's first law, and the principal axes of a rotating body.
Since you want the masses to rotate forever about the axis, that means the centre of mass must either rotate forever about the axis or must stay forever on the axis.
From Newton's first law, if there are no external forces on the system (ie, after you start it you just let it carry on), then the centre of mass must move at a fixed speed in a fixed direction (or
be stationary) …
so rotating about the axis is not possible.
A rotating body can only rotate without wobbling (
) if it is rotating about a "principal axis" of the body.
The principal axes of your two masses are the line joining them, and any line perpendicular to them (through the centre of mass) …
so the two masses must be lined up either along the axis of rotation (so they're just spinning on the spot), or perpendicular to it. | {"url":"http://www.physicsforums.com/showthread.php?s=d7cfdcfad4e945ce378e8f29eef8754a&p=4662104","timestamp":"2014-04-25T00:41:37Z","content_type":null,"content_length":"38432","record_id":"<urn:uuid:ebab4b8d-a9fd-412d-97e6-a1818d5fa32f>","cc-path":"CC-MAIN-2014-15/segments/1398223207046.13/warc/CC-MAIN-20140423032007-00195-ip-10-147-4-33.ec2.internal.warc.gz"} |
Topology is Connectedness
Doktor DynaSoar targeting at OMCL.mil
Mon Feb 23 17:52:35 EST 2004
On Mon, 23 Feb 2004 08:26:01 GMT, "k p Collins"
<kpaulc@[----------]earthlink.net> wrote:
} I stand on what =I've= posted.
All of it?
"You're missing some crucial data that cross-correlates
your 'time' series to the cerebellar topology.
The cerebellum is a topographically-mapped subsystem.
Any analysis must preserve, and incorporate, that mapping
if the correlations are to be meaningful."
From: "k p Collins" <kpaulc@[----------]earthlink.net>
References: <235b9607.0401210500.3ebedda5 at posting.google.com>
Subject: Re: Practical problems with correlation dimension
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Date: Wed, 21 Jan 2004 13:38:40 GMT
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"Karl" <karlknoblich at yahoo.de> wrote in message
news:235b9607.0401210500.3ebedda5 at posting.google.com...
> Hallo!
> I want to calculate the correlation dimension of a time serie.
> What I have done
> I calculated the correlation integral C(r) (number of point having a
> distance smaller than r) for different embedding dimensions. Taking
> the slopes of the curve of log C(r) against log r for the different
> embedding dimensions and plotting them against the embedding dimension
> should result in a limes of the slopes: the correlation dimension.
> My problem
> Which slope shall I take?
> In examples I saw in text books there is a nice limit of the slopes
> with higher embedding dimensions. In my data I do not know which slope
> I should take because the slope of the curve varies. If I take the
> slope at a certain value of log r I can not get a limes.
> My curves (log C(r) against log r) can be seen in
> http://karlknoblich.4t.com/korrdim.jpg
> What to do? Does anybody knows such data and how to handle it?
> Hope somebody can help!
> Karl
What I will say has not yet been accepted by others,
so keep that in mind as you consider it.
You're missing some crucial data that cross-correlates
your 'time' series to the cerebellar topology.
The cerebellum is a topographically-mapped subsystem.
Any analysis must preserve, and incorporate, that mapping
if the correlations are to be meaningful.
And, then, to continue, one has to follow this mapping into
the rest of the brain.
It's a =big= problem, but the mapping is mapped :-] through
the efforts of Neuroscientists, and all one has to do is 'grind'
through it.
There a couple of other things that make your analysis Difficult.
One is that the data is virtually always, itself, a transformation.
The other is that the activation that occurs within the cerebellum
is extremely-dynamic, with a =lot= of different inputs converging
and 'sliding' with respect to each other. There is such 'sliding'
stuff with respect to every joint in the skelleton. [These enter
into the way that the nervous system maintains it's 'awareness'
of the body's orientation in 3-D space [climbing fibers from
the inferior olive].] And this is only one set of such 'sliding-field'
stuff that occurs within the cerebellum. There are hundreds
[perhaps thousands] more.
So your analysis is Hard.
More information about the Neur-sci mailing list | {"url":"http://www.bio.net/bionet/mm/neur-sci/2004-February/057389.html","timestamp":"2014-04-16T05:15:17Z","content_type":null,"content_length":"6455","record_id":"<urn:uuid:bfb3ba66-3d95-485a-937e-0bb95d803ddc>","cc-path":"CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00315-ip-10-147-4-33.ec2.internal.warc.gz"} |
Quick PDEs Question! BVPs: time-dependent heat gen in a bar
April 21st 2009, 08:59 AM #1
Oct 2008
Quick question, what would be the general method for solving this type of problem;
(d/dt)u - (d2/dx2)u = h(x, t)
with homogeneous BCs:
u (0, t) = 0
u (L, t) = 0,
and IC u (x, 0) = f(x)
Just not sure how to deal with the RHS h(x,t) being dependent on t as well as x, (corresponding to time-dependent heat generation I think??)
What method should be used?
Quick question, what would be the general method for solving this type of problem;
(d/dt)u - (d2/dx2)u = h(x, t)
with homogeneous BCs:
u (0, t) = 0
u (L, t) = 0,
and IC u (x, 0) = f(x)
Just not sure how to deal with the RHS h(x,t) being dependent on t as well as x, (corresponding to time-dependent heat generation I think??)
What method should be used?
If $h(t,x)=0$, the we would use the usual separation of variables. Because of the fixed endpoints we would obtain
$u = \sum_{n=1}^\infty a_n e^{- \frac{n^2\pi^2}{L^2}t} \sin \frac{n \pi}{L} x$ where $a_n = \frac{2}{L} \int_0^L f(x) \sin \frac{n \pi}{L} x\,dx$
If $h(t,x) e 0$, since we still have fixed endpoints we will assume a solutuion of the form
$u = \sum_{n=1}^\infty T_n(t) \sin \frac{n \pi}{L} x$
and assume that $h(x,t)$ has a fourier since series
$h(x,t) = \sum_{n=1}^\infty h_n(t) \sin \frac{n \pi}{L} x$
Substitute into thet PDE and equate like sin terms. This will give a series of ODEs for $T_n$.
April 21st 2009, 09:38 AM #2 | {"url":"http://mathhelpforum.com/calculus/84815-quick-pdes-question-bvps-time-dependent-heat-gen-bar.html","timestamp":"2014-04-21T05:49:53Z","content_type":null,"content_length":"37129","record_id":"<urn:uuid:05366f8c-e0a4-40ae-a2e7-065cea7ede62>","cc-path":"CC-MAIN-2014-15/segments/1397609539493.17/warc/CC-MAIN-20140416005219-00143-ip-10-147-4-33.ec2.internal.warc.gz"} |
st: Mata maximum likelihood error on trivial regression
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st: Mata maximum likelihood error on trivial regression
From David Chan <david.c.chan@gmail.com>
To statalist@hsphsun2.harvard.edu
Subject st: Mata maximum likelihood error on trivial regression
Date Fri, 19 Aug 2011 12:46:10 -0400
Hi all,
I'm having problems with doing maximum likelihood in Mata. Even in the
most basic setup, I'm getting the error,
"could not calculate numerical derivatives -- flat or discontinuous
region encountered
I'm almost certain that this is not because of problems with the data
or model, but it probably has something to do with a naive programming
error. I'm simulating my own data, and I'm testing out the program on
a very simple linear regression model,
where epsilon is a random, normally distributed term, and Ey and
zeta_k are both observed regressors (both with coefficient 1).
After simulating data according to this model, simple regression,
> reg y Ey zeta_k"
works fine. Also when I code up the trivial MLE in Stata as an ado
program, I have no problems:
> program mynormal_lf
> args lnfj mu sigma
> qui replace `lnfj'=ln(normalden($ML_y1,`mu',`sigma'))
> end
> ml model lf mynormal_lf (mu: lnlos = Ey zeta_k) /sigma
> ml maximize
However, when I code it up in Mata, using the code below, I always get
the error message, even when I set the initial values at the true
> mata:
> void mydynamic_lf(transmorphic scalar ML, real rowvector b, ///
> real colvector lnfj)
> {
> real colvector y, xb
> real colvector sd
> y=moptimize_util_depvar(ML,1)
> xb=moptimize_util_xb(ML,b,1)
> j=moptimize_util_eq_indices(ML,2)
> sd=exp(b[|j|])
> lnfj=ln(normalden(y,xb,sd))
> }
> end
> ml model lf mydynamic_lf() (mu: lnlos = Ey zeta_k) /sigma
> ml maximize
I've tried renaming the variables within Mata, as well as changing the
variable declaration (e.g., calling "sd" a scalar rather than a
colvector), but nothing seems to work.
Does anyone know what I'm doing wrong?
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/ | {"url":"http://www.stata.com/statalist/archive/2011-08/msg00922.html","timestamp":"2014-04-19T07:07:43Z","content_type":null,"content_length":"8945","record_id":"<urn:uuid:69b0b095-2630-4ed3-877b-ab67bfe88496>","cc-path":"CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00081-ip-10-147-4-33.ec2.internal.warc.gz"} |
An Introduction to Population Ecology - Introduction to Population Modeling
The ability to predict the population size of a group of individuals is extremely useful to the study of ecology. It allows for the estimation of the various effects imposed upon a group by internal
and external forces. We note that the word force has a different meaning in population modeling than in physics. You can think of these forces as factors that impact the population – for example,
availability of food, spread of disease, interactions with other species. These forces can then be divided into density-independent forces and density-dependent forces.
With no forces acting upon a population, we expect the population to have simple exponentially increasing growth (Vandermeer & Goldberg, 2004). Mathematically, we expect a population function whose
rate of growth increases with the population’s size, that is,
This differential equation says that the rate of change in population size over time (dP/dt) increases by a proportional rate of growth (r) multiplied by the current population size (P). The
biological force modeled here is an example of a density-independent force, because it depends only on the population P, not on external forces such as crowding or food supply.
We know from calculus that the solution of this equation is P(t) = P[0]e^rt. The graph of this exponential function (Figure 1) shows the behavior of the population over time. Biologists call this a
J-shaped curve, and it is the foundation for all of the ecological models that we will discuss in this module (Molles, 2004; Vandermeer & Goldberg, 2004).
Figure 1. The exponential function has a J-shaped curve.
The exponential curve is best used for microorganisms, especially bacteria. Figure 2 shows the growth of a population of Escherichia coli bacteria. Notice that the population begins with just three
bacteria, and the population has doubled within 20 minutes. Within three hours, the population has increased dramatically! (To see the figure in full size and resolution, click on the reduced figure
shown here.)
Figure 2. Formation of an Escherichia coli K-12 Microcolony from
three neighboring cells, by James Shapiro and Clara Hsu.
Source: American Society for Microbiology Microbe Library
Used by permission of James Shapiro.
We can use the doubling time to calculate the rate of growth for the E. coli population. If the initial population is P[0], then the doubling time T for the population is the time required for the
population to reach 2P[0]. This requires e^rT = 2. Solving for r, we find r = ln(2)/T. Since T = 20 minutes, we have r = 0.035 min^-1.That means the bacteria population increases at a rate of
approximately 3.5% per minute, which is equivalent to 816% per hour! This is an incredibly fast rate of increase.
For microbes, this may not be a problem. The entire colony of E. coli in the bottom right of Figure 2 could easily fit on the head of a pin. However, for larger organisms, exponential growth can be
maintained for an extended length of time only under very rigid (and often unrealistic) conditions (Molles, 2004). For such organisms, logistic growth (next page) is more realistic.
1. If a population doubles every 5 hours, what is its rate of increase?
2. If a population triples every 5 hours, what is its rate of increase?
3. Suppose that the rate of increase for a particular bacterium is r = 0.24, how long does the bacterium take to double its population?
Published October, 2005
© 2005 by Brandon M. Hale and Maeve L. McCarthy | {"url":"http://www.maa.org/publications/periodicals/loci/joma/an-introduction-to-population-ecology-introduction-to-population-modeling","timestamp":"2014-04-21T13:25:52Z","content_type":null,"content_length":"102216","record_id":"<urn:uuid:5234518e-6a3a-4920-a34e-176994f526e9>","cc-path":"CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00428-ip-10-147-4-33.ec2.internal.warc.gz"} |
10th Grade Pre-Calc
Number of results: 45,246
10th Grade Pre-Calc
The demand equation for a hand-held electronic organizer is... P=5000(1-(4/4+e^-.002x)) Find the demand x for a price of (a) P=600? So strange. Thx for help
Tuesday, January 17, 2012 at 12:02am by Anthony Clay
10th Grade Pre-Calc
Don't you mean P = 5000[1 - 4/(4 + e^-.002x)] ? If P = 600 , 0.120 = [1 - 4/(4 + e^-.002x)] 4/(4 + e^-.002x)] = 0.88 4 + e^-.002x = 4.5454 e^-.002x = 0.5454 -0.002x = ln(0.5454) = -0.6061 x = 303
Tuesday, January 17, 2012 at 12:02am by drwls
oops! don't know why I put AP Calc, it's actually pre-calc....
Friday, September 18, 2009 at 8:52pm by muffy
math-10th grade
The subject is math and not 10th grade. You want math experts to look at your question and 10th grade won't get it.
Tuesday, March 30, 2010 at 10:46pm by DrBob222
10th grade
Your subject is math, not 10th grade. Do you have a specific question? (Broken Link Removed)
Tuesday, February 2, 2010 at 10:55pm by bobpursley
10th grade
The School Subject is NOT 10th grade. If you name your post correctly you will get the best help sooner. Sra
Monday, April 26, 2010 at 5:04pm by SraJMcGin
10th grade
First of all the School Subject is NOT 10th grade. Secondly, what IS your question? Sra
Sunday, May 30, 2010 at 12:32am by SraJMcGin
10th grade
10th grade does NOT help us as the School Subject. Math? Geometry? Sra
Wednesday, May 19, 2010 at 8:11pm by SraJMcGin
10th grade
The School Subject is NOT 10th grade. Please label it correctly so the right teacher sees your post and reads it. Sra
Wednesday, November 10, 2010 at 1:59am by SraJMcGin
10th Grade Chemistry
I struggle in the sciences and was wondering what I could do over the summer self studying wise to prepare for 10th grade chem?
Thursday, July 7, 2011 at 7:00pm by Wuhta
10th grade English
What are some common 10th grade action words?
Tuesday, February 24, 2009 at 4:08am by Chelcy
10th grade
First of all what is the School Subject? It is NOT 10th grade. Sra
Thursday, September 23, 2010 at 9:22pm by SraJMcGin
I'm stuck on a pre-calc problem. I need to find y as a function of x, where the constant c is a positive number for In(y-1)+In(y+1)= -x+c express your question as ln[(y-1)(y+1)] = -x+c change that to
exponential form e^(-x+c) = (y-1)(y+1) e^(-x+c) = y^2 - 1 y^2 = e^(-x+c) + 1 ...
Wednesday, May 9, 2007 at 12:36pm by Sarah
10th grade
10th grade is NOT the School Subject. Math, perhaps? Please label carefully so the right teacher will read and answer your post. Sra
Friday, December 3, 2010 at 8:49pm by SraJMcGin
Can you show me how to find the rectangular equation of r=2costheta + 3 sin theta All I need is an example to work off I was absent today from Pre-Calc and I have homework to do and no explanation
with them-Please help with this one Thank you
Monday, October 3, 2011 at 6:33pm by Leah
10th grade
what classes should i take in 10th grade?
Wednesday, May 27, 2009 at 9:23pm by katlyn
10th grade Pre APworld history
WHAT TYPE OF GOVERNMENT DID SHANG HAVE DURING 8000 B.C.E-600 C.E
Monday, September 7, 2009 at 7:38pm by Anonymous
10th grade
10th grade does NOT help as a label for the School Subject. Math or algebra will get a math teacher to read and answer you rpost. Sra
Friday, December 10, 2010 at 2:05pm by SraJMcGin
10th grade
and yeah its 10th grade chemistry
Sunday, February 8, 2009 at 8:24pm by Michael
10th grade
how i get an unssen workssheets for 10th greade?
Monday, January 11, 2010 at 12:08pm by Asa'ad
10th grade
10th Grade is NOT the School Subject. Please clearly state the School Subject so the right volunteer helps you. Sra
Friday, September 17, 2010 at 10:40pm by SraJMcGin
Pre Calc
I don't see how to do this without a calculator sin ( - pi/12) csc ( (25 pi)/12)
Friday, November 13, 2009 at 5:52pm by Pre Calc
Pre Calc
I'm obviously suppose to get - sin 2 pi ok - sin ( (24 pi)/12) I don't see how to get that though
Friday, November 13, 2009 at 5:52pm by Pre Calc
Pre-Calc-Please check
1.Write 4+6+8+10 in sigma notation: I came up with sigma in the middle 3 on top j=0 on the bottom 2. Find the 10th term of the sequence -1/5,-1/20,-1/80 this was a geometric sequence so I said a
(subscript10) = -1/5*1/4^(10-1) =-1/5*(1/4)^9 = -1/5*1/2621444 = -1/1310702
Thursday, December 15, 2011 at 6:10pm by Sandra
10th grade
The School Subject must be labeled correctly. 10th grade is not the proper School Subject. Sra
Monday, December 20, 2010 at 6:18pm by SraJMcGin
ap calc
OOPS, I meant Pre-Calc not ap Calc sorry
Saturday, October 10, 2009 at 2:45pm by MUFFY
Pre-Calc-Please check answer
1.Write 4+6+8+10 in sigma notation: I came up with sigma in the middle 3 on top j=0 on the bottom 2n+4 to the right of the signma sign 2. Find the 10th term of the sequence -1/5,-1/20,-1/80 this was
a geometric sequence so I said a(subscript10) = -1/5*1/4^(10-1) =-1/5*(1/4)^9...
Thursday, December 15, 2011 at 8:43pm by Sandra
Pre-Calc-Please check answer
1.Write 4+6+8+10 in sigma notation: I came up with sigma in the middle 3 on top j=0 on the bottom 2n+4 to the right of the signma sign 2. Find the 10th term of the sequence -1/5,-1/20,-1/80 this was
a geometric sequence so I said a(subscript10) = -1/5*1/4^(10-1) =-1/5*(1/4)^9...
Thursday, December 15, 2011 at 8:44pm by Sandra
Looks like C to me
Wednesday, April 17, 2013 at 11:05am by Steve
Calculus grade 12 exam review PLEASE HELP ASAP !!
THANK U! to be honest i really hate calculus and i didnt really pay attention in the course. im going into kin (of arts) so its a pre-req but i wont be taking any math or calc courses in university,
i dont know why its a pre req then. but thanks u so much
Thursday, June 21, 2012 at 3:52pm by Sarah Berkley Adamson
PRE-CALC still stuck
Pre-cal? V=kT V=kr^2 I=V(1/R)
Sunday, October 18, 2009 at 5:55pm by bobpursley
pre caculates is the ansewer to above is 25545872
Saturday, October 3, 2009 at 11:19am by Anonymous
pre calc
A pre-calculus answer is -0.64 + kπ radians, where k∈ℤ.
Monday, January 25, 2010 at 8:11pm by MathMate
no idea...but plz help
I just don't understand the math you're doing im in 11th grade taking pre-calc and i dont think we've ever done anything quite like that. Thank you for your help
Sunday, November 29, 2009 at 8:00pm by Ryan
Pre Calc
Does this: sqrt 3x-4=(x-4)^2-3 Multiply out to: x^4-16x^3+90^2-211x+173 I then need to put it in the graphing calc and see what the roots are.
Thursday, September 10, 2009 at 8:34pm by muffy
Pre Calc
I have no idea how to do this problem I thought you would have to do something with the diagonals but not sure... The parallel sides of a trapezoid are 4 and 10 centimeters long and the oblique sides
are 8 and 12 centimeters long. Fi9nd the angles and the area of the trapezoid
Monday, October 5, 2009 at 1:43pm by Pre Calc
i am just abt to finish my 10th grade,i am very weak at maths and science,hence i have opted to take economics. 1.is this the right choice? 2.is there goin to be maths in it? 3.what are the core
subjects involved?
Wednesday, February 23, 2011 at 1:24am by sam
10th grade
plz suggest me some good thoughts which i can use in writing compositions in my 10th board exams which is on the 3rd of march. as in slogans,or some quotations or what ever material which will make
my composition interesting. thanx!!
Wednesday, February 10, 2010 at 10:13am by muskaan
oh no Vitaliy, this person is in 7th grade pre-agebra or 8th grade pre-algebra.
Thursday, February 5, 2009 at 6:58pm by haha
Calc or Pre calc
I am having trouble doing this problem. I know how to do indefinate integrals, but I don't know how to do definate integrals. Can you show me how to do this. Evaluate 5 (x^3-2x)dx 2
Wednesday, May 14, 2008 at 9:42pm by jennifer
11th grade
The students of litchfield High school are in grades 9,10,11,12. Of the students , 1/4th are in 9th grade, 1/3rd are in 10th grade, 1/6th is in 11th grade and there are 300 in the 12th grade, how
many students are there all together?
Friday, November 6, 2009 at 12:30pm by stefanie coplin
It is math, i take a pre-calc/calc class. sorry!
Sunday, September 28, 2008 at 4:49pm by sam
10th grade Pre APworld history
I'm having a lot of trouble in this class, and tonight we have to record differences and similarities of the Ummayad and the Abbasid empires... i have read the chapter and still can not find any
differences, and we have to write of them with 3 example of specific evidence for ...
Monday, September 7, 2009 at 7:38pm by Ashley
10th grade
i need an interesting topic for a tenth grade informative essay
Monday, April 19, 2010 at 2:30pm by Dallas
Looks like D to me. sin u + sin v = 2 sin(½(u+v)) cos(½(u−v))
Wednesday, April 17, 2013 at 11:12am by Steve
Pre Calc
use properties of the trigonometric function to find the exact value of each expression. Do no use a calculator. sec ( - pi/18) * cos (37 pi)/18 I do not see how to do this problem. Every time I do
it out I get a reference angle of twenty degrees which isn't on the unit circle...
Friday, November 13, 2009 at 2:40pm by Pre Calc
12th grade
The Subject is not 10th grade, etc. If you post the title correctly (math?) you will get the correct help sooner. Sra
Tuesday, April 27, 2010 at 10:03am by SraJMcGin
Pre Calc
how did you go from the first line to the next in these lines sec(2PI +Pi/18)cos(37PI/18) sec(37pi/18)cos(37PI/18) 1/cos * cos= 1
Friday, November 13, 2009 at 2:40pm by Pre Calc
physics i suck
could you explain simpler . i get condfused easily im in 10th grade taking 12 grade classes its tough. any help will do
Saturday, January 12, 2008 at 6:46pm by ashley
10th grade (science)
For faster service, it is more important to indicate your subject rather than your grade level when posting questions.
Wednesday, March 18, 2009 at 10:38pm by drwls
8th grade
10th grade
Wednesday, April 1, 2009 at 12:35am by Anonymous
pre calc
This is a long, long way from pre-calculus. 2,4 is in quadrant I when on any axis, it is not in a quadrant. In the case of 0,3 it is on the border between quadrants I,and II.
Thursday, September 10, 2009 at 5:40pm by bobpursley
pre calc
Is there any complex numer that is equal to its conjugate? a+bi=a-bi????? precalc 11th grade a+bi=a-bi ---> 2 bi = 0 ---> b = 0 So, the imaginary part has to be zero.
Thursday, November 16, 2006 at 7:10pm by bill
Can you show me how to find the rectangular equation: r=2cos(theta) + 3 sin (theta) All I need is an example to work off I was absent today from Pre-Calc and I have homework to do and no explanation
with them-Please help with this one Thank you
Monday, October 3, 2011 at 7:49pm by Leah
- Computers - English - Foreign Languages - Health - Home Economics - Math - Music - Physical Education - Science - Social Studies GRADE LEVELS - Preschool - Kindergarten - Elementary School - 1st
Grade - 2nd Grade - 3rd Grade - 4th Grade - 5th Grade - 6th Grade - 7th Grade - ...
Friday, January 8, 2010 at 1:36pm by peter
pre algebra
my grade is 83.17 in pre algebra in college, but the final exam is worth 300 points. If I got a B or B- or a C on the final, what would my grade be overall?
Thursday, March 10, 2011 at 9:20pm by Amber shall
pre algebra
5 1/8 = 4 9/8 1 7/8 = 1 7/8 subtract to get 3 2/8 = 3 1/4 Not sure I'd call this pre-algebra; just 4th grade math to me. Still, I guess all arithmetic is pre-algebra.
Tuesday, January 22, 2013 at 3:49pm by Steve
I just failed the 10th grade by 1 credit. And I feel like I could do much better in the 11th and 12 grade. Is it possible for me to still go to college?
Wednesday, July 16, 2008 at 6:56pm by Ben
Pre calculus? This is stock Algebra II. put the equations in the form of ax^2+bx+c=0 then factor. for instance, c is already in that form. 2x^2+x-6=0 (2x-3)(x+2)=0 x= 3/2 x=-2 do the others the same
Monday, January 25, 2010 at 5:18pm by bobpursley
Pre Calc
h t t p : / / w w w . j i s k h a . c o m / d i s p l a y . c g i ? i d = 1 2 5 8 1 5 2 7 6 5
Friday, November 13, 2009 at 6:52pm by Pre Calc
pre calc
its 2.4
Monday, January 21, 2008 at 11:59am by Julie
Sunday, October 5, 2008 at 7:12pm by Karren
Sunday, October 5, 2008 at 7:12pm by bobpursley
pre calc
okay thanks
Monday, February 9, 2009 at 11:44am by kate
pre calc
Tuesday, March 31, 2009 at 8:19pm by Reiny
and "c"?
Monday, May 25, 2009 at 10:40pm by Anonymous
Pre Calc
How did you get n?
Wednesday, August 26, 2009 at 11:37am by Megan
Pre Calc
How did you get n?
Wednesday, August 26, 2009 at 11:37am by Megan
pre calc
thank you!
Sunday, August 30, 2009 at 3:30pm by gab
pre calc
Tuesday, September 8, 2009 at 6:09pm by bobpursley
pre calc
is 4,5, and 6 as well
Tuesday, September 8, 2009 at 6:09pm by jaz
:) Thanks!
Sunday, September 20, 2009 at 9:01pm by MUFFY
You're welcome!
Thursday, September 24, 2009 at 8:59pm by MathMate
Thanks :D
Friday, October 16, 2009 at 10:49pm by Muffy
pre calc
What is x^2+y^2=4
Thursday, October 29, 2009 at 6:14pm by bobpursley
pre calc
Wednesday, December 2, 2009 at 9:11pm by kim
pre calc
Thursday, January 21, 2010 at 7:21pm by drwls
Wednesday, February 17, 2010 at 8:24pm by Anonymous
Pre Calc
Tuesday, May 18, 2010 at 3:17pm by Keith
Pre Calc!
Thank you!
Tuesday, September 14, 2010 at 7:17pm by Hayli
Pre Calc!
You're welcome!
Tuesday, September 14, 2010 at 7:17pm by MathMate
Thank you for your help
Wednesday, December 8, 2010 at 11:15pm by Amy
pre calc
4* 2^x 2^y = 4*2^(x+y) but 4=2^2 so 2^(2+x+y)
Tuesday, December 21, 2010 at 6:32pm by Damon
Pre Calc
Monday, February 14, 2011 at 10:32pm by Giselle
Wednesday, February 16, 2011 at 7:10pm by megan
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2-2 = 0
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x=1, y=-4, z=-5
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"He drew the picture with the parabola at points (-6,0) (6,0) (0,36)" The point (0,6) is not on the parabola, (0,36) is. The parabola is given as: y(x)=36-x² Given the rectangle is inside the
parabola and above (I think) the x-axis, we define the four corners of the ...
Saturday, October 3, 2009 at 11:19am by MathMate
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globally hyperbolic Lorentzian manifold
globally hyperbolic Lorentzian manifold
Riemannian geometry
Basic definitions
Further concepts
A Lorentzian manifold is called globally hyperbolic if it admits a well-defined time evolution from initial data of physical fields on it.
There are several equivalent definitions of global hyperbolicity. A simple one is:
In this form the characterization of global hyperbolicity appears for instance in the paragraph at the bottom of page 211 in (HE). The equivalence of this to more traditional definitions is (HE,
prop. 6.6.3) together with (HE, prop. 6.6.8), due to (Geroch1970). The latter in fact implies the following stronger statement:
A standard textbook exposition is section 6.6 of
• Hawking, Ellis, The large-scale structure of Space-Time Cambridge (1973)
The fact that a single Cauchy surface implies a foliation by Cauchy surfaces is due to
The refinement of this statement to a smooth splitting is in
• Antonio N. Bernal, Miguel Sánchez, On smooth Cauchy hypersurfaces and Geroch’s splitting theorem (arXiv:gr-qc/0306108v2)
Revised on September 9, 2013 21:18:59 by
Urs Schreiber | {"url":"http://www.ncatlab.org/nlab/show/globally+hyperbolic+Lorentzian+manifold","timestamp":"2014-04-20T08:51:00Z","content_type":null,"content_length":"19118","record_id":"<urn:uuid:5b6d2f8a-2640-4219-b3fa-21d20450698c>","cc-path":"CC-MAIN-2014-15/segments/1398223205137.4/warc/CC-MAIN-20140423032005-00394-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Topic: ekpirical histogram of a density function
Replies: 1 Last Post: Jun 8, 2012 11:54 PM
Messages: [ Previous | Next ]
Re: ekpirical histogram of a density function
Posted: Jun 8, 2012 11:54 PM
kumar vishwajeet
How is that a 50 dimensional array? It seems to me to be a 2 dimensional array where each row of X could contain one of your X1, X2, etc. vectors. So your X matrix would be 10000 rows by 50 columns.
That is a 2D matrix. To get the mean of each row, you'd do
meanOfRows = mean(X, 2); % Take mean of each row going along the 50 columns.
If you want the probability density function of X I'm not sure why you're doing the covariance. Why not just take the hist of the whole thing and normalize:
pdfX = hist(X) / numel(X);
Date Subject Author
6/8/12 ekpirical histogram of a density function kumar vishwajeet
6/8/12 Re: ekpirical histogram of a density function ImageAnalyst | {"url":"http://mathforum.org/kb/message.jspa?messageID=7834907","timestamp":"2014-04-19T04:41:12Z","content_type":null,"content_length":"17511","record_id":"<urn:uuid:b0637aca-447b-41db-9127-e2ab93af726a>","cc-path":"CC-MAIN-2014-15/segments/1397609535775.35/warc/CC-MAIN-20140416005215-00644-ip-10-147-4-33.ec2.internal.warc.gz"} |
Constrained Simultaneous and Near-Simultaneous Embeddings
Frati, Fabrizio and Kaufmann, Michael and Kobourov, Stephen G. (2008) Constrained Simultaneous and Near-Simultaneous Embeddings. In: Graph Drawing 15th International Symposium, GD 2007, September
24-26, 2007, Sydney, Australia , pp. 268-279 (Official URL: http://dx.doi.org/10.1007/978-3-540-77537-9_27).
Full text not available from this repository.
A geometric simultaneous embedding of two graphs G_1=(V_1,E_1) and G_2=(V_2,E_2) with a bijective mapping of their vertex sets gamma : V_1 rightarrow V_2 is a pair of planar straight-line drawings
Gamma _1 of G_1 and Gamma _2 of G_2, such that each vertex v_2=gamma(v_1) is mapped in Gamma _2 to the same point where v_1 is mapped in Gamma _1, where v_1 in V_1 and v_2 in V_2. In this paper we
examine several constrained versions and a relaxed version of the geometric simultaneous embedding problem. We show that if the input graphs are assumed to share no common edges this does not seem to
yield large classes of graphs that can be simultaneously embedded. Further, if a prescribed combinatorial embedding for each input graph must be preserved, then we can answer some of the problems
that are still open for geometric simultaneous embedding. Finally, we present some positive and negative results on the near-simultaneous embedding problem, in which vertices are not mapped exactly
to the same but to "near" points in the different drawings.
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The popular American TV show "Lost" made many esoteric references to the number "108" (a series that I never watched personally)
Apparently, the series includes quite a few references to Buddhism. For example, the sum of "the numbers" (4, 8, 15, 16, 23, and 42) is 108, and "the numbers" must be entered into a computer every
108 minutes. One hundred eight is also the number of days "the Oceanic 6" have spent on the island. And Jacob tells Hurley to rotate the mirror in the lighthouse to a heading of 108°.
The following is a collection of information about the number "108" (reference links included):
One hundred [and] eight (or nine dozen) is an abundant number and a semiperfect number. It is a tetranacci number.
It is the hyperfactorial of 3 since it is of the form
108 is a number that is divisible by the value of its φ function, which is 36. 108 is also divisible by the total number of its divisors (12), hence it is a refactorable number.
In Euclidean space, the interior angles of a regular pentagon measure 108 degrees each.
There are 108 free polyominoes of order 7.
In base 10, it is a Harshad number and a self number.
The equation Golden ratio
Importance in Astronomy and Astrology
The earth cycle is supposed to be of 2160 years = 20 x 108. The distance between the Earth and Sun is 108 times the diameter of the Sun. The diameter of the Sun is 108 times the diameter of the
Earth. The distance between the Earth and Moon is 108 times the diameter of the Moon. The universe is made up of 108 elements according to ancient texts. The current periodic table claims a few more
than 108.
There are 12 constellation and 9 arc segments. 9 times 12 equal 108. The 9 planets travelling through the 12 signs constitute the whole of existence. 9 x 12 = 108. The 27 nakshatras or lunar
constellations spread over the 4 elements – fire, earth, air, water or the 4 directions – north, south, east, and west. This also constitutes the whole of existence. 27 x 4 = 108.
Source: http://humanityhealing.net (http://s.tt/13wja)
Religion and the arts
The number 108 is considered sacred in many Eastern religions and traditions, such as Hinduism, Buddhism, Jainism, Sikhism and connected yoga and dharma based practices. The individual numbers 1, 0,
and 8 represent one thing, nothing, and everything (infinity). 108 represents the ultimate reality of the universe as being (seemingly paradoxically) simultaneously One, emptiness, and infinite.
Gaudiya Vaishnavism, there are 108 gopis of Vrindavan. Recital of these names, often accompanied by counting of 108-beaded Mala, is considered sacred and often done during religious ceremonies. The
recital is called namajapa. Accordingly, a mala usually has beads for 108 repetitions of a mantra.
The distance of Sun from Earth divided by diameter of Sun and distance of Moon from Earth divided by diameter of Moon is approximately equal to 108. It is claimed that the great sires of Vedanta knew
this relationship and thus 108 is a very important number in Vedantic chantings.
Likewise, Tibetan Buddhist malas or rosaries (Tib. ཕྲེང་བ Wyl. phreng ba, "Trengwa") are usually 108 beads;^[1] sometimes 111 including the Guru Bead(s), reflecting the words of the Buddha called in
Tibetan the Kangyur (Wylie: Bka'-'gyur) in 108 volumes. Zen priests wear juzu (a ring of prayer beads) around their wrists, which consists of 108 beads.^[2]
Japa Mala
, or
Japa beads
, made from
wood. Consisting of 108 beads in total + the head bead.
The Lankavatara Sutra has a section where the Bodhisattva Mahamati asks Buddha 108 questions^[3] and another section where Buddha lists 108 statements of negation in the form of "A statement
concerning X is not statement concerning X".^[4] In a footnote, D.T. Suzuki explains that the Sanskrit word translated as "statement" is "pada" which can also mean "foot-step" or "a position." This
confusion over the word "pada" explains why some have mistakenly held that the reference to 108 statements in the Lankavatara refer to the 108 steps that many temples have.^[5]
In some schools of Buddhism it is believed that there are 108 defilements.^[6] In Japan, at the end of the year, a bell is chimed 108 times in Buddhist temples to finish the old year and welcome the
new one. Each ring represents one of 108 earthly temptations a person must overcome to achieve nirvana.
Ancient artifacts
The pre-historic monument Stonehenge is about 108 feet in diameter. [3]
Other References
In the neo-Gnostic teachings of Samael Aun Weor, an individual has 108 chances (lifetimes) to eliminate his egos and transcend the material world before "devolving" and having the egos forcefully
removed in the infradimensions.^[7]
Martial arts
• According to Marma Adi and Ayurveda, there are 108 pressure points in the body, where consciousness and flesh intersect to give life to the living being.^[8]
• The Chinese school of martial arts agrees with the South Indian school of martial arts on the principle of 108 pressure points.^[9]^[10]
• 108 number also figures prominently in the symbolism associated with karate, particularly the Gōjū-ryū discipline. The ultimate Gōjū-ryū kata, Suparinpei, literally translates to 108. Suparinpei
is the Chinese pronunciation of the number 108, while gojūshi of Gojūshiho is the Japanese pronunciation of the number 54. The other Gōjū-ryū kata, Sanseru (meaning "36") and Seipai ("18") are
factors of the number 108.^[2]
• The 108 of the Yang long form and Wing Chun, taught by Yip Man having 108 movements are noted in this regard.^[5]
• Several different Taijiquan long forms consist of 108 moves.
• Paek Pal Ki Hyung, the 7th form taught in the art of Kuk Sool Won, translates literally to "108 technique" form. It is also frequently referred to as the "eliminate 108 torments" form. Each
motion corresponds with one of the 108 Buddhist torments or defilements.
In literature
• In Homer's Odyssey, the number of suitors coveting Penelope, wife of Odysseus.
• There are 108 outlaws in the Chinese classic Water Margin/Outlaws of the Marsh.
• There are 108 love sonnets in Astrophil and Stella, the first English sonnet sequence by Sir Philip Sidney.
In other fields
• An official Major League Baseball baseball has 108 stitches.^[11]
• In the Manga and Anime series Shikabane Hime Aka and Shikabane Hime Kuro (Corpse Princess), the shikabane himes must hunt 108 evil undead corpses in order to ascend to heaven.
• The atomic number of hassium.
• The number of Mbit/s of a non-standard extension of IEEE 802.11g wireless network using channel bonding.
• There are 108 Stars of Destiny to collect in the video game Suikoden distributed by Konami for PlayStation
• 108 is the name of a community of and for open source developers, created by Red Hat.^[12]
• There are 108 cards in a deck of UNO cards.
• There are 108 Code Crowns in Digimon Xros Wars.
• Volume expansion of freezing water is roughly 108%.
EMRI - Emergency Services
108 degrees Fahrenheit is also the internal temperature at which the human body's vital organs begin to fail from overheating.In India, 108 (1-0-8) is the toll-free emergency telephone number. Calls
are handled by GVK EMRI (GVK Emergency Management and Research Institute), the only professional Emergency Service Provider in India today, handling medical, police and fire emergencies.
Highways and Roads
OTHER SIGNIFICANCES OF THE NUMBER 108
The Indian Subcontinent rosary or set of mantra counting has 108 beads. 108 has been a sacred number in the Indian Subcontinent for a very long time. This number is explained in many different ways.
The ancient Indians were excellent mathematicians and 108 may be the product of a precise mathematical operation (e.g. 1 power 1 x 2 power 2 x 3 power 3 = 108) which was thought to have special
numerological significance.
Powers of 1, 2, and 3 in math: 1 to 1st power=1; 2 to 2nd power=4 (2x2); 3 to 3rd power=27 (3x3x3). 1x4x27=108
Sanskrit alphabet: There are 54 letters in the Sanskrit alphabet. Each has masculine and feminine, shiva and shakti. 54 times 2 is 108.
Sri Yantra: On the Sri Yantra there are marmas where three lines intersect, and there are 54 such intersections. Each intersections has masculine and feminine, shiva and shakti qualities. 54 x 2
equals 108. Thus, there are 108 points that define the Sri Yantra as well as the human body.
9 times 12: Both of these numbers have been said to have spiritual significance in many traditions. 9 times 12 is 108. Also, 1 plus 8 equals 9. That 9 times 12 equals 108.
Heart Chakra: The chakras are the intersections of energy lines, and there are said to be a total of 108 energy lines converging to form the heart chakra. One of them, sushumna leads to the crown
chakra, and is said to be the path to Self-realization.
Marmas: Marmas or marmastanas are like energy intersections called chakras, except have fewer energy lines converging to form them. There are said to be 108 marmas in the subtle body.
Time: Some say there are 108 feelings, with 36 related to the past, 36 related to the present, and 36 related to the future.
Astrology: There are 12 constellations, and 9 arc segments called namshas or chandrakalas. 9 times 12 equals 108. Chandra is moon, and kalas are the divisions within a whole.
Planets and Houses: In astrology, there are 12 houses and 9 planets. 12 times 9 equals 108.
Gopis of Krishna: In the Krishna tradition, there were said to be 108 gopis or maid servants of Krishna.
1, 0, and 8: 1 stands for God or higher Truth, 0 stands for emptiness or completeness in spiritual practice, and 8 stands for infinity or eternity.
Sun and Earth: The diameter of the sun is 108 times the diameter of the Earth.
Numerical scale: The 1 of 108, and the 8 of 108, when added together equals 9, which is the number of the numerical scale, i.e. 1, 2, 3 ... 10, etc., where 0 is not a number.
Smaller divisions: The number 108 is divided, such as in half, third, quarter, or twelfth, so that some malas have 54, 36, 27, or 9 beads.
Islam: The number 108 is used in Islam to refer to God.
Jain: In the Jain religion, 108 are the combined virtues of five categories of holy ones, including 12, 8, 36, 25, and 27 virtues respectively.
Sikh: The Sikh tradition has a mala of 108 knots tied in a string of wool, rather than beads.
Chinese: The Chinese Buddhists and Taoists use a 108 bead mala, which is called su-chu, and has three dividing beads, so the mala is divided into three parts of 36 each.
Stages of the soul: Said that Atman, the human soul or center goes through 108 stages on the journey.
Meru: This is a larger bead, not part of the 108. It is not tied in the sequence of the other beads. It is the quiding bead, the one that marks the beginning and end of the mala.
Dance: There are 108 forms of dance in the Indian traditions.
Pythagorean: The nine is the limit of all numbers, all others existing and coming from the same. ie: 0 to 9 is all one needs to make up an infinite amount of numbers. | {"url":"http://lawrencerspencer.com/tag/108/","timestamp":"2014-04-18T15:39:27Z","content_type":null,"content_length":"59662","record_id":"<urn:uuid:a16177c8-1587-45a9-bad8-d1c0a6d3c4c4>","cc-path":"CC-MAIN-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00011-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Examine if f is primitive of g?
g=e^x(x+1)^2 - Homework Help - eNotes.com
Examine if f is primitive of g?
You need to test if `f(x)` is the primitive of `g(x)` , hence, you need to check if `f'(x) = g(x)` , such that:
`f'(x) = (e^x(x^2+1))'`
You need to differentiate the function with respect to `x` , using the product rule, such that:
`f'(x) = (e^x)'(x^2 + 1) + e^x*(x^2 + 1)'`
`f'(x) = e^x*(x^2+1) + e^x*(2x)`
Factoring out `e^x` yields:
`f'(x) = e^x*(x^2 + 2x + 1)`
You need to notice that the factor `x^2 + 2x + 1` represents the expansion of squared binomial `(x + 1)^2` , such that:
`f'(x) = e^x*(x + 1)^2`
Comparing the equation of `f'(x)` with the equation of `g(x)` yields that they coincide, hence, the function `f(x)` represents the primitive of `g(x)` .
Join to answer this question
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[Numpy-discussion] bug in numpy.mean() ?
Zachary Pincus zachary.pincus@yale....
Tue Jan 24 13:05:50 CST 2012
> You have a million 32-bit floating point numbers that are in the
> thousands. Thus you are exceeding the 32-bitfloat precision and, if you
> can, you need to increase precision of the accumulator in np.mean() or
> change the input dtype:
>>>> a.mean(dtype=np.float32) # default and lacks precision
> 3067.0243839999998
>>>> a.mean(dtype=np.float64)
> 3045.747251076416
>>>> a.mean(dtype=np.float128)
> 3045.7472510764160156
>>>> b=a.astype(np.float128)
>>>> b.mean()
> 3045.7472510764160156
> Otherwise you are left to using some alternative approach to calculate
> the mean.
> Bruce
Interesting -- I knew that float64 accumulators were used with integer arrays, and I had just assumed that 64-bit or higher accumulators would be used with floating-point arrays too, instead of the array's dtype. This is actually quite a bit of a gotcha for floating-point imaging-type tasks -- good to know!
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Here's the question you clicked on:
Select the best possible first step to solving the following system by first eliminating the y variable. 3x + 8y = -7 2x + 12y = -3 A.Multiply the first equation by 12 and multiply the second
equation by 8. B.Multiply the first equation by -12 and multiply the second equation by -8. C.Multiply the first equation by 3 and multiply the second equation by -2. D.Multiply the first equation by
3 and the second equation by 7. Need some help plzzz!!!
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C.Multiply the first equation by 3 and multiply the second equation by -2.
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Math Forum Discussions - Re: Matheology § 224
Date: Mar 25, 2013 1:22 PM
Author: fom
Subject: Re: Matheology § 224
On 3/25/2013 6:47 AM, WM wrote:
> For every and for all in my example are equivalent. Every and all
> elements of the inductive set of FISONs can be removed without
> changing the union, iff actual infinity exists.
This might be a plausible statement if one
confuses a closure property with the definition
of an operation.
But, only one of us is confused.
That, of course, is me since the theory of monotone
inclusive crayon marks is generally undefined except
for the "creation myth" of Hanching.
An alleged mathematical theory with a creation myth. This
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How to Conquer the Times Table, Part 3
Sep 12 2011
How to Conquer the Times Table, Part 3
Photo of Javier times 4, by Javier Ignacio Acuña Ditzel, via flickr.
If you remember, we are in the middle of an experiment in mental math. We are using the world’s oldest interactive game — conversation — to explore multiplication patterns while memorizing as little
as possible. Talk through these patterns with your student. Work many, many, many oral math problems together. Discuss the different ways you can find each answer, and notice how the number patterns
connect to each other.
So far, we have mastered the times-1 and times-10 families and the Commutative Property (that you can multiply numbers in any order).
The Doubles
What else is relatively simple? Does your student know the doubles? Doubles are often considered easy, because children do so much counting and addition with numbers less than 20. Even if your child
finds the doubles tricky, a little focused practice should fix these facts in mind.
• Here is your first memory task: Learn the doubles!
Practice doubling big numbers, too. Use silly numbers to help:
38 $\times$ 2
= double 3 tens + double 8
= “sixty-sixteen”
= 76
2 $\times$ 56
= double 5 tens + double 6
= “tenty-twelve”
= 112
Go back and forth, inventing double-puzzles for each other:
3½ $\times$ 2
= double 3 + double ½
= 6 + 1
= 7
2 $\times$ 47,000
= (double 4 tens + double 7) thousand
= “eighty-fourteen” thousand
= 94,000
Review Game: Once Through the Deck
The best way to practice the math facts is through the give-and-take of conversation, orally quizzing each other and talking about how you might figure the answers out. But occasionally you may want
a simple, solitaire method for review. Here’s how:
• Shuffle a deck of math cards and place it face down on the table in front of you.
• Flip the cards face up, one at a time.
• For each card, say (out loud) the product of that number times the number you want to practice.
• Don’t say the whole equation, just the answer.
• Go through the deck as fast as you can.
• But don’t try to go so fast that you have to guess! If you are not sure of the answer, stop and figure it out.
Brian at The Math Mojo Chronicles demonstrates the game in this video, which my daughter so thoroughly enjoyed that she immediately ran to find a deck of cards and practiced her times-4 facts. (It’s
funny, sometimes, what will catch a child’s interest.)
You can use Once Through the Deck as a final check that your student knows the fact family well enough to mark it on your chart. Remember to mark both the row and the column!
The Times-4 Family
Notice that the answers in the times-4 row are exactly double the answers in the times-2 row. Can your child see why that makes sense? If you have two of something, and you replicate two more of it,
then you would have four of that thing, whether it is minions or cookies or numbers.
This means you do not need to memorize the times-4 facts. Just double the number to get the times-2 answer, and then double it again. For example, $7 \times 4$ would mean seven doubled, which is 14,
and then that answer doubled again:
7 $\times$ 4
= 7 $\times$ 2 $\times$ 2
= 14 $\times$ 2
= 28
Practice double-doubling a bunch of numbers. Can you use the double-double trick to figure out something like $53 \times 4$ ?
53 $\times$ 4
= 53 $\times$ 2 $\times$ 2
= 106 $\times$ 2
= 212
This may take a little more time to practice — but that is okay! Quiz each other with unusual numbers, using double-doubling to get the answer. Test yourself with Once Through the Deck, and when you
are ready, mark the chart.
The Times-8 Family
In the same way that times-4 was the double of times-2, it makes sense that times-8 is the double of times-4. If you have four of something, and you replicate four more of it, then you will have
eight of the thing in all. It doesn’t matter whether the thing is books or aliens or numbers. Even fractions: If you have four 1/16 size slices of pizza, and you take four more pieces, then you will
have 8/16 of the pizza.
If you need to calculate something times eight, you can double the something, then double again — that makes four times — then double it once more for your final answer:
6 $\times$ 8
= 6 $\times$ 2 $\times$ 2 $\times$ 2
= 12 $\times$ 2 $\times$ 2
= 24 $\times$ 2
= 48
I find it helpful to count on three fingers, to make sure I don’t forget any of the doublings. Remember to experiment with big numbers, too. Can you double-double-double to figure out $132 \times 8$
132 $\times$ 8
= 132 $\times$ 2 $\times$ 2 $\times$ 2
= 264 $\times$ 2 $\times$ 2
= 528 $\times$ 2
= 1056
It may take a few days or even weeks of practice before you and your student feel comfortable with these. Take all the time you need, and when you both are able to mentally double-double-double
almost any number the other can pose, mark off the times-8 column and row on the chart.
The Times-5 Family
Your daughter can probably count by fives, but many children get confused when trying to skip-count large multiplication problems. A more reliable number pattern for times-5 calculations uses the
doubles in reverse. Two fives make ten, so any even number of fives will make exactly half that number of tens:
6 fives = 3 tens
18 fives = 9 tens
24 fives = 12 tens
450 fives = 225 tens
All the odd numbers times five will come out “somethingty-five,” and you can predict what the “somethingty” will be by looking at the next lower even number.
7 fives = (6 + 1) fives = 3 tens + 5
25 fives = (24 + 1) fives = 12 tens + 5
109 fives = (108 + 1) fives = 54 tens + 5
Practice intensively on these until you can both get the answer right every time. Test yourself with a round of Once Through the Deck, and then mark them off.
Wow! Look back and see how much you have learned. You started with 144 facts, and you have narrowed it down to only 21 — and all you had to memorize so far was the doubles! Your child could probably
memorize the last 21 facts without too much trouble, but let’s see if we can find a few more patterns to make them easier.
This is the fourth post in my Times Table Series, which will eventually be included in one of my math books. To be continued…[Go to part 4.]
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Have more fun on Let’s Play Math! blog:
5 comments on “How to Conquer the Times Table, Part 3”
1. I absolutely love the information in your blog. I’m a math tutor that developed my own method of teaching multiplication tables drawing from various sources. I’m constantly adding to it, and have
added your tips on extending the doubling principal for times 4 and times 8 (when I think about it, this is the technique Greg Tang shows in his book “The best of times” but I never employed it
before. Just like students, teachers need something new to spark their interest).
I’m working with a student right now that seems to have become stuck on times 8. We’ve been working on it for over a month now (off and on) and he just doesn’t seem to be able to recall that when
he can’t remember a times 8 fact, that he can double 3 times. I’m just wondering if you (or anyone else with similar experience) have advice on whether to continue working on it until he gets it,
or if it’s better to skip it, move on, then come back to try it again later.
In addition to “learning facts” I do practical applications, like multiplication war and DAMULT dice. He always gets stuck on times 8 no matter the application where it comes up. He resorts to
guessing, or thinks that he has to double 2 times or 4 times when prompted “remember how we multiply by 8?”. Any further advice or suggestions would be appreciated!
2. I have had success with the “backing off” method when my kids hit a wall. Sometimes just forgetting about the topic for awhile lets it simmer in the student’s unconscious mind, and when we come
back to it in a month or two, everything flows easily.
But I’m not sure that would work so well in this case. I’m assuming the student needs to use times-8 facts in his schoolwork and can’t ignore them. Is he mature enough to catch himself when he
starts guessing, so that if you give him a fool-proof technique, he could slow down and force himself to use it?
If so, here’s the technique:
* Write down 8 copies of whatever number you are multiplying times 8, arranged like the dots on a domino. You are going to add these numbers up in pairs to find out how much all 8 of them are
To calculate 13 times 8, I would write:
* Then circle the numbers in pairs, writing their sum right next to the circle. This is the first doubling. Now you only have 4 numbers:
* Now circle those numbers in pairs, and write their sum next to the circle. This is the second doubling, and it leaves you with only 2 numbers to add:
* Finally, add the last two numbers (the third doubling) to get your answer.
It takes some scratch paper and the discipline to resist wild guessing, but it will always work!
3. Russian Peasant Multiplication would also work. It’s quicker, but it’s more abstract, which means it may be harder for a child to remember. In each step, you double one number and cut the other
in half:
13 times 8 =
26 times 4 =
52 times 2 =
104 times 1 = 104
4. Thanks for the tips! The student is in grade 6, but I’m not covering school work with him, I went back to just going over basic facts. Like most students I tutor, adding/subtracting isn’t a
problem, but (basic) multiplication is, and doesn’t become noticeable until they start on long division (which is when the parent gives me a call to get help).
He’s a high energy little guy, and definitely a visual/auditory learner. Trying to use beads and cups to show grouping or having him write number bonds went disastrously. I like your first tip,
and I’ll give it a try. If he rebels, I’m going to skip times 8 for now. We still have to work on times 3, 6 and 7. When an 8 times fact comes up that he’s stuck on, I’ll do what I always do,
which is patiently explain it to him again. It’ll sink in eventually.
Thanks for all the resources on your blog, I can get lost in it reading everything and following the links. I look forward to your book coming out! | {"url":"http://letsplaymath.net/2011/09/12/how-to-conquer-the-times-table-part-3/","timestamp":"2014-04-16T16:42:11Z","content_type":null,"content_length":"86191","record_id":"<urn:uuid:5bb40532-d214-4a5b-a407-fcf07668e5d8>","cc-path":"CC-MAIN-2014-15/segments/1397609524259.30/warc/CC-MAIN-20140416005204-00267-ip-10-147-4-33.ec2.internal.warc.gz"} |
Area in Polar Coordinates
April 2nd 2008, 12:32 PM #1
Nov 2007
Area in Polar Coordinates
Find the area of the region which is inside the polar curve :
r = 3 cos( θ)
and outside the curve :
r = 2 − 1 cos( θ)
Please help!
draw the curves, to see how it looks (looks like a moon)
then calculate where the curves meet (set r=r, calculate phi1 and phi2, insert phi into the equation and calculate r1 and r2)
then integrate from r1 to r2 and from phi1 to phi2, using
area=integral r*dr*dphi
edit: sorry, my wrong. It's two circles one inside the other, one has radius 3/2 the other 1/2, so the area should be
pi*(9/4 - 1/4) = pi*2
I still don't understand, what should I integrate and what method should I use to find the area enclosed?
sorry I was wrong with the edit.
You should start with drawing the curves, to see what you are calculating.
You draw them first as simple cosine (linear), and then kinda curl them around the middle of the coordinate system.
Then you calculate the intersection points of the curves.
Then you can integrate with the intersection points as borders of integration.
Like this:
April 2nd 2008, 12:51 PM #2
Apr 2008
April 2nd 2008, 06:02 PM #3
Nov 2007
April 3rd 2008, 12:23 AM #4
Apr 2008 | {"url":"http://mathhelpforum.com/calculus/32981-area-polar-coordinates.html","timestamp":"2014-04-20T00:29:35Z","content_type":null,"content_length":"35785","record_id":"<urn:uuid:36650554-8bc6-423b-a596-95d81bfdbe63>","cc-path":"CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00499-ip-10-147-4-33.ec2.internal.warc.gz"} |
[Numpy-discussion] np.void from 0d array + subclassing
Robert Kern robert.kern@gmail....
Thu Dec 17 17:35:26 CST 2009
On Thu, Dec 17, 2009 at 16:11, Pierre GM <pgmdevlist@gmail.com> wrote:
> On Dec 17, 2009, at 10:16 AM, Francesc Alted wrote:
>> A Thursday 17 December 2009 15:16:29 Pierre GM escrigué:
>>> All,
>>> * What is the most efficient way to get a np.void object from a 0d
>>> structured ndarray ?
>> I normally use `PyArray_GETITEM` C macro for general n-d structured arrays. I
>> suppose that this will work with 0-d arrays too.
> Francesc, you're overestimating my knowledge of C... Can we stick to the Python implementation ?
> Here's the catch: IIUC, each individual element of a nD structured array is a void, provided the element can be accessed, ie that n>0. A 0D array cannot be indexed, so I don't know how capture the object below. The sad trick I found was to do a .reshape(1)[0], but that looks really overkill...
Robert Kern
"I have come to believe that the whole world is an enigma, a harmless
enigma that is made terrible by our own mad attempt to interpret it as
though it had an underlying truth."
-- Umberto Eco
More information about the NumPy-Discussion mailing list | {"url":"http://mail.scipy.org/pipermail/numpy-discussion/2009-December/047519.html","timestamp":"2014-04-16T18:57:15Z","content_type":null,"content_length":"3936","record_id":"<urn:uuid:e16b4a0c-3d59-42a0-ad1e-82544d3244ef>","cc-path":"CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00283-ip-10-147-4-33.ec2.internal.warc.gz"} |
Inside the Windows Vista Disk Encryption Algorithm
Suppose an attacker is attacking two identical hard drives, one encrypted with AES-CBC + Elephant diffuser and the other encrypted only by AES-CBC. We are going to give the attacker the tweak key in
Figure 1; this means the attacker can now perform the diffusion layer for any plaintext. In other words, the diffuser layer becomes transparent to the attacker. All that is left now for the attacker
is to attack the AES-CBC layer, which is the same problem he has when attacking the other hard drive (encrypted using only AES-CBC mode). This means that, although we helped the attacker
significantly by providing him with the tweak key, he still has to attack the AES-CBC layer. This shows that attacking the AES-CBC + Elephant diffuser is not easier than attacking just AES-CBC. Thus,
the AES-CBC + Elephant diffuser is at least as secure as AES-CBC in fact, it has better statistical characteristics because it passes four statistical tests, while AES-CBC passes only one.
Note that this security proof sketch is valid for all the values of AC and BC, but we do not recommend using AC less than 2 or BC less than 1 to have good statistical properties. Passing Test1
implies that the AES-CBC + Elephant diffuser does not suffer from a bit-flipping attack, as does AES-CBC mode.
Using the same methodology, it can be proven that the AES-ECB + Elephant diffuser is at least as secure as AES-ECB, and by providing the attacker with the tweak key in Figure 2, he will have to
attack the AES-ECB layer. AES-ECB + Elephant diffuser passes all four statistical tests, while AES-ECB passes only one test. This shows that the AES-ECB + Elephant diffuser has better statistical
properties than AES-ECB. By XORing the plaintext with the drive sector key, the 512-byte block becomes dependent on the sector. And by adding the counter in Figure 2, each 16-byte block (within the
512) is dependent on its position. This is to reduce the possibility of replay attacks.
To sum it up, in this article, we studied the new AES-CBC + Elephant diffuser the Windows Vista disk encryption algorithm. Our study shows that this cipher possesses good diffusion properties that
can reduce manipulation attacks. And from our study, we set a lower bound of the number of cycles used by each of the two diffusers; these values are AC=2 and BC=1. With these values, the cipher will
not lose its statistical properties. Moreover, it still will be at least as secure as AES-CBC mode. On the other hand, the use of these values increases the total performance of the cipher by about
18 percent.
We also propose a new cipher AES-ECB + Elephant diffuser that has similar properties to the AES-CBC + Elephant diffuser. From our study, we set a lower bound of the number of cycles used by each of
the two used diffusers; these values are AC=2 and BC=2. These values can be used and the cipher will not lose its statistical properties and still will be at least as secure as AES-ECB mode. On the
other hand, the use of these values increases the total performance of the cipher by about 14 percent.
While the AES-ECB + Elephant diffuser is about the same speed as the AES-CBC + Elephant diffuser when a single processor is used, it is about 60-100 percent faster than the AES-CBC + Elephant
diffuser (depending on the values of AC and BC used) when a dual-core processor is used. This speed up is inversely proportional to the number of processor cores used. The complete source code
accompanying this article is available electronically (for demonstration purposes only, contact Microsoft for development use); see www.ddj.com/code/. | {"url":"http://www.drdobbs.com/security/inside-the-windows-vista-disk-encryption/201804187?pgno=4","timestamp":"2014-04-18T23:30:52Z","content_type":null,"content_length":"95162","record_id":"<urn:uuid:7d60f0b2-cfa1-4967-9bfc-dc1f7533ada5>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00021-ip-10-147-4-33.ec2.internal.warc.gz"} |
Quantum Computing Takes Center Stage In Wake of NSA Encryption Cracking
In the late 1990′s while I was with Ascend Communications, I participated in the creation of the “point-to-point tunneling protocol” (PPTP) with engineers at Microsoft and Cisco Systems, now an
Internet Engineering Task Force (IETF) industry standard. PPTP is the technical means for creating the “virtual private networks” we use at UBC, by encrypting “open” Internet packets with IPSEC 128
bit code, buried in public packets. It was an ingenious solution, enabling private Internet traffic that we assumed would last for a very long time. It was not to be, as we now know. Most
disturbing, in the 1990′s the US Congress debated giving the government the key to all encryption, which was resoundingly defeated. Now, the NSA appears to have illegally circumvented this
prohibition and cracked encryption anyway. But this discussion is not about the political, legal and moral issues, which are significant. In this post I am more interested in exploring the question:
“So now what do we do?” There may be an answer on the horizon, and Canada is already a significant participant in the potential solution.
As it happens, Canada is already at the forefront of quantum computing, a critically important new area of research and development, that has significant future potential in both computing and
cryptography. I have previously written about Vancouver-based D-Wave, which has produced commercial systems that have been purchased by Google and Lockheed Martin Aerospace. The Institute for
Quantum Computing in Waterloo, Ontario is the other major center of quantum computing research in Canada. Without taking a major diversion to explain quantum mechanics and its applications in
computing and cryptography, there is a great PBS Nova broadcast, available online, which provides a basic tutorial. The Economist article below, also does an admirable job of making this area
understandable, and the role that the Waterloo research centre is playing in advancing cryptography to an entirely new level.
We need to insure that Canada remains at the forefront of this critically important new technology.
The solace of quantum
Eavesdropping on secret communications is about to get harder
At the moment cryptography concentrates on making the decrypting part as hard as possible. The industry standard, known as RSA (after its inventors, Ron Rivest, Adi Shamir and Leonard Adleman, of the
Massachusetts Institute of Technology), relies on two keys, one public and one private. These keys are very big numbers, each of which is derived from the product of the same two prime numbers.
Anyone can encrypt a message using the public key, but only someone with the private key can decrypt it. To find the private key, you have to work out what the primes are from the public key. Make
the primes big enough—and hunting big primes is something of a sport among mathematicians—and the task of factorising the public key to reveal the primes, though possible in theory, would take too
long in practice. (About 40 quadrillion years with the primes then available, when the system was introduced in 1977.)
Since the 1970s, though, the computers that do the factorisation have got bigger and faster. Some cryptographers therefore fear for the future of RSA. Hence the interest in quantum cryptography.
Alice, Bob and Werner, too?
The most developed form of quantum cryptography, known as quantum key distribution (QKD), relies on stopping interception, rather than preventing decryption. Once again, the key is a huge number—one
with hundreds of digits, if expressed in the decimal system. Alice sends this to Bob as a series of photons (the particles of light) before she sends the encrypted message. For Eve to read this
transmission, and thus obtain the key, she must destroy some photons. Since Bob will certainly notice the missing photons, Eve will need to create and send identical ones to Bob to avoid detection.
But Alice and Bob (or, rather, the engineers who make their equipment) can stop that by using two different quantum properties, such as the polarities of the photons, to encode the ones and zeros of
which the key is composed. According to Werner Heisenberg’s Uncertainty Principle, only one of these two properties can be measured, so Eve cannot reconstruct each photon without making errors. If
Bob detects such errors he can tell Alice not to send the actual message until the line has been secured.
One exponent of this approach is ID Quantique, a Swiss firm. In collaboration with Battelle, an American one, it is building a 700km (440-mile) fibre-optic QKD link between Battelle’s headquarters in
Columbus, Ohio, and the firm’s facilities in and around Washington, DC. Battelle will use this to protect its own information and the link will also be hired to other firms that want to move
sensitive data around.
QuintessenceLabs, an Australian firm, has a different approach to encoding. Instead of tinkering with photons’ polarities, it changes their phases and amplitudes. The effect is the same, though: Eve
will necessarily give herself away if she eavesdrops. Using this technology, QuintessenceLabs is building a 560km QKD link between the Jet Propulsion Laboratory in Pasadena, California, which
organises many of NASA’s unmanned scientific missions, and the Ames Research Centre in Silicon Valley, where a lot of the agency’s scientific investigations are carried out.
A third project, organised by Jane Nordholt of Los Alamos National Laboratory, has just demonstrated how a pocket-sized QKD transmitter called the QKarD can secure signals sent over public data
networks to control smart electricity grids. Smart grids balance demand and supply so that electricity can be distributed more efficiently. This requires constant monitoring of the voltage, current
and frequency of the grid in lots of different places—and the rapid transmission of the results to control centres. That transmission, however, also needs to be secure in case someone malicious wants
to bring the system down.
In their different ways, all these projects are ambitious. All, though, rely on local fixed lines to carry the photons. Other groups of researchers are thinking more globally. To do that means
sending quantum-secured data to and from satellites.
At least three groups are working on this: Thomas Jennewein and his team at the Institute for Quantum Computing in Waterloo, Canada; a collaboration led by Anton Zeilinger at the University of Vienna
and Jian-Wei Pan at the University of Science and Technology of China; and Alex Ling and Artur Ekert at the Centre for Quantum Technologies in Singapore.
Dr Jennewein’s proposal is for Alice to beam polarisation-encoded photons to a satellite. Once she has established a key, Bob, on another continent, will wait until the satellite passes over him so
he can send some more photons to it to create a second key. The satellite will then mix the keys together and transmit the result to Bob, who can work out the first key because he has the second.
Alice and Bob now possess a shared key, so they can communicate securely by normal (less intellectually exhausting) terrestrial networks. Dr Jennewein plans to test the idea, using an aircraft rather
than a satellite, at some point during the next 12 months.
An alternative, but more involved, satellite method is to use entangled photon pairs. Both Dr Zeilinger’s and Dr Ling’s teams have been trying this.
Entanglement is a quantum effect that connects photons intimately, even when they are separated by a large distance. Measure one particle and you know the state of its partner. In this way Alice and
Bob can share a key made of entangled photon pairs generated on a satellite. Dr Zeilinger hopes to try this with a QKD transmitter based on the International Space Station. He and his team have been
experimenting with entanglement at ground level for several years. In 2007 they sent entangled photon pairs 144km through the air across the Canary Islands. Dr Ling’s device will test entanglement in
orbit, but not send photons down to Earth.
If this sort of thing works at scale, it should keep Alice and Bob ahead for years. As for poor Eve, she will find herself entangled in an unbreakable quantum web.
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Math Help
September 10th 2009, 09:39 PM #1
Junior Member
Oct 2008
Hi, can somebody help me with his problem, I have no idea how to do it. Thanks!
A cube has an edge length of 7 cm. If it is divided up into 1-cm cubes, how many 1-cm would there be?
What I think: 7 * 8 = 56 since a cube has 8 edges...
i think its 7*7*7, cause 1cm cubes have volume of 1, and volume of big cube is like, 7*7*7
How many 1-cm *what* would there be? Edges? Cubes? Faces?
Count the edges on a cube again. I rather believe there may possibly be more than 8 ...
September 10th 2009, 10:24 PM #2
Sep 2009
September 10th 2009, 10:51 PM #3 | {"url":"http://mathhelpforum.com/pre-calculus/101614-pre-cal.html","timestamp":"2014-04-19T01:00:11Z","content_type":null,"content_length":"30369","record_id":"<urn:uuid:01232079-04f5-41df-9374-d8fb381e493c>","cc-path":"CC-MAIN-2014-15/segments/1397609535535.6/warc/CC-MAIN-20140416005215-00525-ip-10-147-4-33.ec2.internal.warc.gz"} |
Derivations of semisimple Lie algebras and the abstract Jordan decomposition
I learned the material in this post from the book by Humphreys on Lie algebras and representation theory.
Recall that if ${A}$ is any algebra (not necessarily associative), then the derivations of ${A}$ form a Lie algebra ${Der(A)}$, and that if ${A}$ is actually a Lie algebra, then there is a
homomorphism ${\mathrm{ad}: A \rightarrow Der(A)}$. In this case, the image of ${\mathrm{ad}}$ is said to consist of inner derivations.
Theorem 1 Any derivation of a semisimple Lie algebra ${\mathfrak{g}}$ is inner.
To see this, consider ${\mathrm{ad}: \mathfrak{g} \rightarrow D :=Der(\mathfrak{g})}$; by semisimplicity this is an injection. Let the image be ${D_i}$, the inner derivations. Next, I claim that $
{[D, D_i] \subset D_i}$. Indeed, if ${\delta \in D}$ and ${\mathrm{ad} x \in D_i}$, we have
$\displaystyle [\delta, \mathrm{ad} x] y = \delta( [x,y]) - [x, \delta(y)] = [\delta(x),y] = (\mathrm{ad}(\delta(x)))y.$
In other words, ${[\delta, \mathrm{ad} x] = \mathrm{ad}(\delta(x))}$. This proves the claim.
Consider the Killing form ${B_D}$ on ${D}$ and the Killing form ${B_{D_i}}$ on ${D_i}$. The above claim and the definition as a trace shows that ${B_D|_{D_i \times D_i} = B_{D_i}}$.
Now, if ${D_i^{\perp}}$ is the orthogonal complement to ${D_i}$ under the form ${B_D}$, we must have
$\displaystyle D_i + D_i^{\perp} = D$
but also, because of nondegeneracy of ${B_{D_i}}$,
$\displaystyle D_i \cap D_i^{\perp} = \{ 0 \}.$
I claim now ${D_i^{\perp} = \{0\}}$. If ${\delta \in D_i^{\perp}}$, then for any ${x \in \mathfrak{g}}$,
$\displaystyle [\delta, \mathrm{ad}(x)] = \mathrm{ad}(\delta x) \in D_i \cap D_i^{\perp} = \{ 0 \}$
because ${D_i^{\perp}}$ is an ideal (by invariance of the Killing form again). By semisimplicity, ${\delta x = 0}$, and since ${x}$ was arbitrary, we find ${\delta }$ is the zero derivation. So ${D =
Abstract Jordan decomposition
Now assume our fields are algebraically closed.
Proposition 2 Let ${A}$ be a finite-dimensional algebra. Let ${\delta: A \rightarrow A}$ be a derivation, and regard ${\delta}$ as an element of ${End(A)}$ to take its Jordan decomposition ${\
delta = S + N}$. Then ${S,N}$ are derivations of ${A}$ as well.
It is clearly enough to prove ${S}$ is a derivation. There may be confusion caused by my using ${\delta}$ to refer to a specific
The idea is to write ${A}$ as a direct sum
$\displaystyle A = \bigoplus_{\lambda} A_{\lambda}$
where the ${A_\lambda}$ are ${\delta}$-invariant subspaces with ${(\delta-\lambda)}$ acting nilpotently on them. Then ${S}$ acts on ${A_{\lambda}}$ by ${\lambda}$.
It is enough to check that for any ${a \in A_{\lambda}, b \in A_{\mu}}$,
$\displaystyle a b \in A_{\lambda + \mu}.$
For then
$\displaystyle S(ab) = (\lambda + \mu)(ab) = (Sa) b + a(Sb).$
Now indeed, we have a binomial-like formula
$\displaystyle (\delta - \lambda - \mu)^n(ab) = \sum_{i=0}^n \binom{n}{i} (\delta - \lambda)^i a \cdot (\delta - \mu)^{n-i} b$
that can be checked by induction on ${n}$. It shows that ${ab}$ is annihilated by some high power of ${\delta - \lambda - \mu}$.
Theorem 3 (Abstract Jordan decomposition) Let ${\mathfrak{g}}$ be a semisimple Lie algebra. Then we can write any ${x \in \mathfrak{g}}$ uniquely as ${x = s + n}$ where ${\mathrm{ad} s}$ is
semisimple, ${\mathrm{ad} n}$ is nilpotent, and ${s,n}$ commute with each other and with every element of ${\mathfrak{g}}$ that commutes with ${x}$.
We imbed ${\mathfrak{g}}$ as a subalgebra ${D_i}$ of ${gl(\mathfrak{g})}$ via the ${\mathrm{ad}}$ mapping. Then ${\mathrm{ad} x \in gl(\mathfrak{g})}$ splits into semisimple and nilpotent parts, ${\
mathrm{ad} x = S + N}$. Then ${S,N}$ are derivations of ${\mathfrak{g}}$ by the proposition, and inner ones by Theorem 1, so we get
$\displaystyle \mathrm{ad} x = \mathrm{ad} s + \mathrm{ad} n$
and semisimplicity implies then ${x = s + n}$. Since
$\displaystyle \mathrm{ad} [s,n] = [\mathrm{ad} s, \mathrm{ad} n] = 0,$
we find ${[s,n]=0}$. The commuting properties are thus seen to follow from the corresponding ones of the normal Jordan decomposition. Uniqueness follows by the uniqueness of the Jordan decomposition
in ${End(\mathfrak{g})}$.
Note that if we have decompositions ${x = s+n, y = s' + n'}$, and ${x,y}$ commute, then
$\displaystyle x+ y = (s+s') + (n + n')$
is the Jordan decomposition for ${x+y}$.
This by itself is not all that interesting. But it turns out to be the case that semisimple elements (i.e., those whose nilpotent part is zero) in a Lie algebra act by semisimple endomorphisms in any
representation. We need to talk about complete reducibility first though. | {"url":"http://amathew.wordpress.com/2010/01/30/derivations-of-semisimple-lie-algebras-and-the-abstract-jordan-decomposition/","timestamp":"2014-04-17T18:37:11Z","content_type":null,"content_length":"65379","record_id":"<urn:uuid:4df658a4-5cf6-485f-98cb-fbbca7c4d671>","cc-path":"CC-MAIN-2014-15/segments/1397609530895.48/warc/CC-MAIN-20140416005210-00660-ip-10-147-4-33.ec2.internal.warc.gz"} |
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PERT advantages and disadvantages
I would like someone to discuss, in details, the advantages and disadvantages of PERT, and whether PERT is still relevant and should still be used in today's projects.
By the way, how old is PERT? When was it invented, and by whom?
PERT was developed in the late 50s by Booz Allen Hamilton to manage the Polaris Missle Project which was considered very ambitious for its time. This is what the PM history texts credit and I'm sure
it evolved over time since its very logical. I adopted PERT-like estimating techniques long before I knew about the existence of PERT.
The main strength of PERT is that it uses a multiple estimate approach that allows one to compute a level of confidence that the estimate will be met. You can apply it to both cost and budget
PERT is still relevant today and often used informally by project managers not knowing the details. These project managers ask estimating questions like "what's the worst case?" or "if everything
goes well, will your estimate improve?". They put together estimate ranges when planning the budget and timeline to let the project sponsor know there will be some variability in any big project. | {"url":"http://www.projectmanagementquestions.com/4087/pert-advantages-and-disadvantages","timestamp":"2014-04-16T13:05:28Z","content_type":null,"content_length":"22960","record_id":"<urn:uuid:d4421689-c748-4d03-a02f-f00bda604851>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00456-ip-10-147-4-33.ec2.internal.warc.gz"} |
The Rational Class Program [HOMEWORK HELP]
October 18th, 2011, 09:50 AM
The Rational Class Program [HOMEWORK HELP]
This is for my AP Computer Science class that I am struggling with. Please help me.
Assignment Purpose:
The primary purpose of this lab is to demonstrate knowledge of creating a class with object methods, instantiate multiple objects of the created class, and then call the object methods from the
main program method.
Write a program with a Rational class. The purpose of the Rational class is to manipulate rational number operations. A rational number is a number that can be expressed in the form A / B where A
and B are both whole numbers (no fractions or decimals) and B 0.
This program will utilize GUI input and output windows. For this assignment you are not expected to create this GUI code yourself. Your main concern is to create and use the Rational class. The
Rational class is quite involved and will be developed over two separate assignments. This first assignment will just get the ball rolling.
The main method is provided for you and needs to be used as shown. You are also provided with a getGCF method of the Rational class which will return the Greatest Common Factor of 2 integers. You
will find this useful in writing other methods of the Rational class. Your mission is to complete the Rational class that is used by the Lab08MATH02java program.
Here is the Code that they give you;
Code :
// MathLab02st.java
// The Rational Class Program I
// This is the student, starting version of the MathLab02 assignment.
import javax.swing.JOptionPane;
public class Lab08MATH02st
public static void main (String args[])
String strNbr1 = JOptionPane.showInputDialog("Enter Numerator ");
String strNbr2 = JOptionPane.showInputDialog("Enter Denominator ");
int num = Integer.parseInt(strNbr1);
int den = Integer.parseInt(strNbr2);
Rational r = new Rational(num,den);
JOptionPane.showMessageDialog(null,r.getNum()+"/"+r.getDen()+" equals "+r.getDecimal());
class Rational
// Rational
// getNum
// getDen
// getDecimal
// getRational
// getOriginal
// reduce
private int getGCF(int n1,int n2)
int rem = 0;
int gcf = 0;
rem = n1 % n2;
if (rem == 0)
gcf = n2;
n1 = n2;
n2 = rem;
while (rem != 0);
return gcf;
80 Point Version Specifics
Your Rational class needs to declare two data attributes: num for numerator and den for denominator. Only one constructor is required, which uses two parameters entered at the keyboard. The first
parameter is the numerator and the second parameter is the denominator. The Rational class requires three additional methods, which are getNum, getDen and getDecimal. Method getNum returns the
integer numerator, getDen returns the integer denominator and the getDecimal method returns a real number decimal value of the fraction. For example, if the numerator is 3 and the denominator is
4, getDecimal will return 0.75
80 (and 90) Point Version Output 1
The GUI windows appear one after the other. They do not all show up simultaneously as shown below. The windows will display on top of the current desktop and they are smaller than the windows
shown for this sample execution.
80 (and 90) Point Version Output 2
90 Point Version Specifics
The 90-point version adds the getRational method. This method returns a String representation of the fraction. For example, if the numerator is 3 and the denominator is 4, getRational will return
Concatenation Hint:
You probably know that String variables/values can be concatenated together.
Example: "John" + "Smith" = "JohnSmith"
What you may not know is that other data types can be concatenated with Strings as well.
Example: "John" + 19 = "John19"
This shows an int being concatenated to the end of a String.
Even though the output of the 90 point version is identical to the 80 point version (see previous page), the showMessageDialog statement will need to be changed in the main method for the 90
point version to work properly. (See below.) Now a single call to getRational replaces the 2 calls to methods getNum and getDen.
90 Point Version
Code :
public static void main (String args[])
String strNbr1 = JOptionPane.showInputDialog("Enter Numerator ");
String strNbr2 = JOptionPane.showInputDialog("Enter Denominator ");
int num = Integer.parseInt(strNbr1);
int den = Integer.parseInt(strNbr2);
Rational r = new Rational(num,den);
JOptionPane.showMessageDialog(null,r.getRational() + " equals " + r.getDecimal());
100 Point Version Specifics
The 100-point version adds the getOriginal and reduce methods as well as firstNum and firstDen variable attributes. The constructor also needs to be changed. This version of the lab assignment
reduces the fraction, if possible. The output displays something like 15/20 reduces to 3/4. Without additional variables, the original values of the numerator and denominator will be lost. You
need to achieve the following sequence. The Rational constructor initializes all variables and then reduces the fraction. The reduce method needs getGCF to insure maximum reduction.
As with the 90 point version, the showMessageDialog statement will need to be changed in the main method for this program to work properly. (See below.)
Code :
public static void main (String args[])
String strNbr1 = JOptionPane.showInputDialog("Enter Numerator 1");
String strNbr2 = JOptionPane.showInputDialog("Enter Denominator 2");
int num = Integer.parseInt(strNbr1);
int den = Integer.parseInt(strNbr2);
Rational r = new Rational(num,den);
JOptionPane.showMessageDialog(null,r.getOriginal() + " equals " +
r.getDecimal() + "\n and reduces to " + r.getRational());
Attachment 797 Here si the original Word Doc that describes the lab.
Here I the Java file I am provided with.Attachment 798
If you can help me figure this out I would be most appreciative.
October 18th, 2011, 12:31 PM
Re: The Rational Class Program [HOMEWORK HELP]
Please do not multi-post the same question - your other post has been removed.
And I recommend reading the getting help link in my signature, and the following:
Your post is very long, and with the number of posts here many (including myself) don't have the time to immerse themselves with it. | {"url":"http://www.javaprogrammingforums.com/%20object-oriented-programming/11694-rational-class-program-%5Bhomework-help%5D-printingthethread.html","timestamp":"2014-04-19T10:04:43Z","content_type":null,"content_length":"12494","record_id":"<urn:uuid:d8718b5c-b0a6-4948-bd79-08df33a02e26>","cc-path":"CC-MAIN-2014-15/segments/1397609537097.26/warc/CC-MAIN-20140416005217-00288-ip-10-147-4-33.ec2.internal.warc.gz"} |
1053 Pendulum
TimeLimit: 1 Second MemoryLimit: 32 Megabyte
Totalsubmit: 4 Accepted: 1
Consider a pendulum hanging on a string from a hook on a wall. When pushed, this pendulum will swing back and forth. Now imagine other hooks on the wall, placed in the path of our pendulum's string.
The pendulum will bend around them, possibly even loop around them. In general, it will follow a much more complex path than before. After some time, the pendulum's motion will repeat, the pendulum
will follow a periodic orbit. What we would like you to do is to compute the distance travelled by the pendulum as it completes one cycle of the orbit.
More formally, we place a cartesian coordinate system on the wall. The pendulum's string is affixed at the origin (0, 0). As usual, the x-axis points to the right and the y-axis points upwards. The
string of the pendulum has a length of r. The pendulum is released at position (-r, 0) and therefore starts swinging to the right. Furthermore, there are n additional hooks distributed over the plane
which may influence the path of the pendulum.
In our ideal world, the following assumptions are true:
• The diameters of the hooks and of the string are zero.
• The pendulum loses no energy (e.g. by friction).
• The pendulum will never hit a hook, only its string will.
• The pendulum's string is made of some futuristic material that only bends where it touches a hook but is otherwise rigid.
Your program should simulate the movement of the pendulum and output the spatial length of the periodic orbit that it finally enters. As you may remember from physics: due to gravity, the pendulum
will never reach a height greater than the one it started from! That is, it will never get above the x-axis. It will either reach its initial height again or circle endlessly around a hook in the
The input file contains several test cases. Each case begins with a line containing an integer n (the number of hooks, 1 <= n < 500) and a real r (the length of the pendulum's string). The following
n lines each contain two integers specifying the x- and y-coordinate of the corresponding hook.
The file ends with a case having r = 0, which should not be processed.
For each case output a line containing the number of the case (`Pendulum #1', `Pendulum #2', etc.).
Then print a line that contains the distance which the pendulum travels for completing one cycle of its periodic orbit. Do not count the distance travelled to reach the starting point of the orbit.
(Adhere to the format shown in the output sample.) The distance should be exact to two digits to the right of the decimal point.
Output a blank line after each test case.
Sample Input
2 16.0
3 -4
-3 -4
1 18.0
5 -12
Sample Output
Pendulum #1
Length of periodic orbit = 87.66
Pendulum #2
Length of periodic orbit = 31.42
Southwestern European Regional Contest ETH Zürich,November 16, 1996 | {"url":"http://acm.hrbeu.edu.cn/index.php?act=problem&id=1053","timestamp":"2014-04-20T13:27:54Z","content_type":null,"content_length":"7917","record_id":"<urn:uuid:e6f6980b-af2c-471b-a235-63685a77216c>","cc-path":"CC-MAIN-2014-15/segments/1397609538787.31/warc/CC-MAIN-20140416005218-00384-ip-10-147-4-33.ec2.internal.warc.gz"} |
Claude Shannon
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Claude Elwood Shannon (April 30, 1916 – February 24, 2001) was an American mathematician, communication theorist, and cryptographer known as "the father of information theory".^[1]^[2]
Shannon is famous for having founded information theory with one landmark paper published in 1948. But he is also credited with founding both digital computer and digital circuit design theory in
1937, when, as a 21-year-old master's student at MIT, he wrote a thesis demonstrating that electrical application of Boolean algebra could construct and resolve any logical, numerical relationship.
It has been claimed that this was the most important master's thesis of all time.^[3] Shannon contributed to the field of cryptanalysis during World War II and afterwards, including basic work on
code breaking.
Shannon was born in Petoskey, Michigan. His father, Claude Sr (1862–1934), a descendant of early New Jersey settlers, was a self-made businessman and for a while, Judge of Probate. His mother, Mabel
Wolf Shannon (1890–1945), daughter of German immigrants, was a language teacher and for a number of years principal of Gaylord High School, Michigan. The first 16 years of Shannon's life were spent
in Gaylord, Michigan, where he attended public school, graduating from Gaylord High School in 1932. Shannon showed an inclination towards mechanical things. His best subjects were science and
mathematics, and at home he constructed such devices as models of planes, a radio-controlled model boat and a wireless telegraph system to a friend's house half a mile away. While growing up, he
worked as a messenger for Western Union. His childhood hero was Thomas Edison, who he later learned was a distant cousin. Both were descendants of John Ogden, a colonial leader and an ancestor of
many distinguished people.^[4]^[5]
Boolean theory
In 1932 he entered the University of Michigan, where he took a course that introduced him to the works of George Boole. He graduated in 1936 with two bachelor's degrees, one in electrical engineering
and one in mathematics. Later he began his graduate studies at the Massachusetts Institute of Technology (MIT), where he worked on Vannevar Bush's differential analyzer, an analog computer.^[6]
While studying the complicated ad hoc circuits of the differential analyzer, Shannon saw that Boole's concepts could be used to great utility. A paper drawn from his 1937 master's thesis, A Symbolic
Analysis of Relay and Switching Circuits,^[7] was published in the 1938 issue of the Transactions of the American Institute of Electrical Engineers. It also earned Shannon the Alfred Noble American
Institute of American Engineers Award in 1940. Howard Gardner, of Harvard University, called Shannon's thesis "possibly the most important, and also the most famous, master's thesis of the century."
In this work, Shannon proved that Boolean algebra and binary arithmetic could be used to simplify the arrangement of the electromechanical relays then used in telephone routing switches, then
expanded the concept and also proved that it should be possible to use arrangements of relays to solve Boolean algebra problems. Exploiting this property of electrical switches to do logic is the
basic concept that underlies all electronic digital computers. Shannon's work became the foundation of practical digital circuit design when it became widely known among the electrical engineering
community during and after World War II. The theoretical rigor of Shannon's work completely replaced the ad hoc methods that had previously prevailed.
Vannevar Bush suggested that Shannon, flush with this success, work on his dissertation at Cold Spring Harbor Laboratory, funded by the Carnegie Institution headed by Bush, to develop similar
mathematical relationships for Mendelian genetics, which resulted in Shannon's 1940 PhD thesis at MIT, An Algebra for Theoretical Genetics.^[8]
In 1940, Shannon became a National Research Fellow at the Institute for Advanced Study in Princeton, New Jersey. At Princeton, Shannon had the opportunity to discuss his ideas with influential
scientists and mathematicians such as Hermann Weyl and John von Neumann, and even had the occasional encounter with Albert Einstein. Shannon worked freely across disciplines, and began to shape the
ideas that would become information theory.^[9]
Wartime research
Shannon then joined Bell Labs to work on fire-control systems and cryptography during World War II, under a contract with section D-2 (Control Systems section) of the National Defense Research
He met his wife Betty when she was a numerical analyst at Bell Labs. They married in 1949.^[10]
For two months early in 1943, Shannon came into contact with the leading British cryptanalyst and mathematician Alan Turing. Turing had been posted to Washington to share with the US Navy's
cryptanalytic service the methods used by the British Government Code and Cypher School at Bletchley Park to break the ciphers used by the German U-boats in the North Atlantic.^[11] He was also
interested in the encipherment of speech and to this end spent time at Bell Labs. Shannon and Turing met at teatime in the cafeteria.^[11] Turing showed Shannon his seminal 1936 paper that defined
what is now known as the "Universal Turing machine"^[12]^[13] which impressed him, as many of its ideas were complementary to his own.
In 1945, as the war was coming to an end, the NDRC was issuing a summary of technical reports as a last step prior to its eventual closing down. Inside the volume on fire control a special essay
titled Data Smoothing and Prediction in Fire-Control Systems, coauthored by Shannon, Ralph Beebe Blackman, and Hendrik Wade Bode, formally treated the problem of smoothing the data in fire-control by
analogy with "the problem of separating a signal from interfering noise in communications systems."^[14] In other words it modeled the problem in terms of data and signal processing and thus heralded
the coming of the information age.
His work on cryptography was even more closely related to his later publications on communication theory.^[15] At the close of the war, he prepared a classified memorandum for Bell Telephone Labs
entitled "A Mathematical Theory of Cryptography," dated September, 1945. A declassified version of this paper was subsequently published in 1949 as "Communication Theory of Secrecy Systems" in the
Bell System Technical Journal. This paper incorporated many of the concepts and mathematical formulations that also appeared in his A Mathematical Theory of Communication. Shannon said that his
wartime insights into communication theory and cryptography developed simultaneously and "they were so close together you couldn’t separate them".^[16] In a footnote near the beginning of the
classified report, Shannon announced his intention to "develop these results ... in a forthcoming memorandum on the transmission of information."^[17]
Postwar contributions
In 1948 the promised memorandum appeared as "A Mathematical Theory of Communication", an article in two parts in the July and October issues of the Bell System Technical Journal. This work focuses on
the problem of how best to encode the information a sender wants to transmit. In this fundamental work he used tools in probability theory, developed by Norbert Wiener, which were in their nascent
stages of being applied to communication theory at that time. Shannon developed information entropy as a measure for the uncertainty in a message while essentially inventing the field of information
The book, co-authored with Warren Weaver, The Mathematical Theory of Communication, reprints Shannon's 1948 article and Weaver's popularization of it, which is accessible to the non-specialist. This
was the first model of communication and is known as the Shannon–Weaver model
Shannon's concepts were also popularized, subject to his own proofreading, in John Robinson Pierce's Symbols, Signals, and Noise.
Information theory's fundamental contribution to natural language processing and computational linguistics was further established in 1951, in his article "Prediction and Entropy of Printed English",
proving that treating whitespace as the 27th letter of the alphabet actually lowers uncertainty in written language, providing a clear quantifiable link between cultural practice and probabilistic
Another notable paper published in 1949 is "Communication Theory of Secrecy Systems", a declassified version of his wartime work on the mathematical theory of cryptography, in which he proved that
all theoretically unbreakable ciphers must have the same requirements as the one-time pad. He is also credited with the introduction of sampling theory, which is concerned with representing a
continuous-time signal from a (uniform) discrete set of samples. This theory was essential in enabling telecommunications to move from analog to digital transmissions systems in the 1960s and later.
He returned to MIT to hold an endowed chair in 1956.
Hobbies and inventions
Outside of his academic pursuits, Shannon was interested in juggling, unicycling, and chess. He also invented many devices, including rocket-powered flying discs, a motorized pogo stick, and a
flame-throwing trumpet for a science exhibition^[citation needed]. One of his more humorous devices was a box kept on his desk called the "Ultimate Machine", based on an idea by Marvin Minsky.
Otherwise featureless, the box possessed a single switch on its side. When the switch was flipped, the lid of the box opened and a mechanical hand reached out, flipped off the switch, then retracted
back inside the box. In addition he built a device that could solve the Rubik's cube puzzle.^[4]
He is also considered the co-inventor of the first wearable computer along with Edward O. Thorp.^[18] The device was used to improve the odds when playing roulette.
Legacy and tributes
Shannon came to MIT in 1956 to join its faculty and to conduct work in the Research Laboratory of Electronics (RLE). He continued to serve on the MIT faculty until 1978. To commemorate his
achievements, there were celebrations of his work in 2001, and there are currently six statues of Shannon sculpted by Eugene L. Daub: one at the University of Michigan; one at MIT in the Laboratory
for Information and Decision Systems; one in Gaylord, Michigan; one at the University of California, San Diego; one at Bell Labs; and another at AT&T Shannon Labs.^[19] After the breakup of the Bell
system, the part of Bell Labs that remained with AT&T was named Shannon Labs in his honor.
Robert Gallager has called Shannon the greatest scientist of the 20th century. According to Neil Sloane, an AT&T Fellow who co-edited Shannon's large collection of papers in 1993, the perspective
introduced by Shannon's communication theory (now called information theory) is the foundation of the digital revolution, and every device containing a microprocessor or microcontroller is a
conceptual descendant of Shannon's 1948 publication:^[20] "He's one of the great men of the century. Without him, none of the things we know today would exist. The whole digital revolution started
with him."^[21]
Shannon developed Alzheimer's disease, and spent his last few years in a Massachusetts nursing home. He was survived by his wife, Mary Elizabeth Moore Shannon; a son, Andrew Moore Shannon; a
daughter, Margarita Shannon; a sister, Catherine S. Kay; and two granddaughters.^[10]^[22]
Shannon was oblivious to the marvels of the digital revolution because his mind was ravaged by Alzheimer's disease. His wife mentioned in his obituary that had it not been for Alzheimer's "he would
have been bemused" by it all.^[21]
Other work
Shannon and his famous electromechanical mouse Theseus (named after Theseus from Greek mythology) which he tried to have solve the maze in one of the first experiments in artificial intelligence
Shannon's mouse
Theseus, created in 1950, was a magnetic mouse controlled by a relay circuit that enabled it to move around a maze of 25 squares. Its dimensions were the same as an average mouse.^[2] The maze
configuration was flexible and it could be modified at will.^[2] The mouse was designed to search through the corridors until it found the target. Having travelled through the maze, the mouse would
then be placed anywhere it had been before and because of its prior experience it could go directly to the target. If placed in unfamiliar territory, it was programmed to search until it reached a
known location and then it would proceed to the target, adding the new knowledge to its memory thus learning.^[2] Shannon's mouse appears to have been the first artificial learning device of its
Shannon's computer chess program
In 1950 Shannon published a paper on computer chess entitled Programming a Computer for Playing Chess. It describes how a machine or computer could be made to play a reasonable game of chess. His
process for having the computer decide on which move to make is a minimax procedure, based on an evaluation function of a given chess position. Shannon gave a rough example of an evaluation function
in which the value of the black position was subtracted from that of the white position. Material was counted according to the usual relative chess piece relative value (1 point for a pawn, 3 points
for a knight or bishop, 5 points for a rook, and 9 points for a queen).^[23] He considered some positional factors, subtracting ½ point for each doubled pawns, backward pawn, and isolated pawn.
Another positional factor in the evaluation function was mobility, adding 0.1 point for each legal move available. Finally, he considered checkmate to be the capture of the king, and gave the king
the artificial value of 200 points. Quoting from the paper:
The coefficients .5 and .1 are merely the writer's rough estimate. Furthermore, there are many other terms that should be included. The formula is given only for illustrative purposes. Checkmate
has been artificially included here by giving the king the large value 200 (anything greater than the maximum of all other terms would do).
The evaluation function is clearly for illustrative purposes, as Shannon stated. For example, according to the function, pawns that are doubled as well as isolated would have no value at all, which
is clearly unrealistic.
The Las Vegas connection: information theory and its applications to game theory
Shannon and his wife Betty also used to go on weekends to Las Vegas with M.I.T. mathematician Ed Thorp,^[24] and made very successful forays in blackjack using game theory type methods co-developed
with fellow Bell Labs associate, physicist John L. Kelly Jr. based on principles of information theory.^[25] They made a fortune, as detailed in the book Fortune's Formula by William Poundstone and
corroborated by the writings of Elwyn Berlekamp,^[26] Kelly's research assistant in 1960 and 1962.^[3] Shannon and Thorp also applied the same theory, later known as the Kelly criterion, to the stock
market with even better results.^[27] Over the decades, Kelly's scientific formula has become a part of mainstream investment theory^[28] and the most prominent users, well-known and successful
billionaire investors Warren Buffett,^[29]^[30] Bill Gross^[31] and Jim Simons use Kelly methods. Warren Buffett met Thorp the first time in 1968. It's said that Buffett uses a form of the Kelly
criterion in deciding how much money to put into various holdings. Also Elwyn Berlekamp had applied the same logical algorithm for Axcom Trading Advisors, an alternative investment management
company, that he had founded. Berlekamp's company was acquired by Jim Simons and his Renaissance Technologies Corp hedge fund in 1992, whereafter its investment instruments were either subsumed into
(or essentially renamed as) Renaissance's flagship Medallion Fund. But as Kelly's original paper demonstrates, the criterion is only valid when the investment or "game" is played many times over,
with the same probability of winning or losing each time, and the same payout ratio.^[32]
The theory was also exploited by the famous MIT Blackjack Team, which was a group of students and ex-students from the Massachusetts Institute of Technology, Harvard Business School, Harvard
University, and other leading colleges who used card-counting techniques and other sophisticated strategies to beat casinos at blackjack worldwide. The team and its successors operated successfully
from 1979 through the beginning of the 21st century. Many other blackjack teams have been formed around the world with the goal of beating the casinos.
Claude Shannon's card count techniques were explained in Bringing Down the House, the best-selling book published in 2003 about the MIT Blackjack Team by Ben Mezrich. In 2008, the book was adapted
into a drama film titled 21.
Shannon's maxim
Shannon formulated a version of Kerckhoffs' principle as "the enemy knows the system". In this form it is known as "Shannon's maxim".
Awards and honors list
See also
Further reading
• Claude E. Shannon: A Mathematical Theory of Communication, Bell System Technical Journal, Vol. 27, pp. 379–423, 623–656, 1948. [1] [2]
• Claude E. Shannon and Warren Weaver: The Mathematical Theory of Communication. The University of Illinois Press, Urbana, Illinois, 1949. ISBN 0-252-72548-4
• Rethnakaran Pulikkoonattu — Eric W. Weisstein: Mathworld biography of Shannon, Claude Elwood (1916–2001) [3]
• Claude E. Shannon: Programming a Computer for Playing Chess, Philosophical Magazine, Ser.7, Vol. 41, No. 314, March 1950. (Available online under External links below)
• David Levy: Computer Gamesmanship: Elements of Intelligent Game Design, Simon & Schuster, 1983. ISBN 0-671-49532-1
• Mindell, David A., "Automation's Finest Hour: Bell Labs and Automatic Control in World War II", IEEE Control Systems, December 1995, pp. 72–80.
• David Mindell, Jérôme Segal, Slava Gerovitch, "From Communications Engineering to Communications Science: Cybernetics and Information Theory in the United States, France, and the Soviet Union" in
Walker, Mark (Ed.), Science and Ideology: A Comparative History, Routledge, London, 2003, pp. 66–95.
• Poundstone, William, Fortune's Formula, Hill & Wang, 2005, ISBN 978-0-8090-4599-0
• Gleick, James, The Information: A History, A Theory, A Flood, Pantheon, 2011, ISBN 9780375423727
Shannon videos
External links
Template:IEEE Medal of Honor Laureates 1951-1975
Edit General subfields and scientists in Cybernetics
K1 Polycontexturality, Second-order cybernetics
K2 Catastrophe theory, Connectionism, Control theory, Decision theory, Information theory, Semiotics, Synergetics, Sociosynergetics, Systems theory
K3 Biological cybernetics, Biomedical cybernetics, Biorobotics, Computational neuroscience, Homeostasis, Medical cybernetics, Neuro cybernetics, Sociocybernetics
William Ross Ashby, Claude Bernard, Valentin Braitenberg, Ludwig von Bertalanffy, George S. Chandy, Joseph J. DiStefano III, Heinz von Foerster, Charles François, Jay Forrester,
Cyberneticians Buckminster Fuller, Ernst von Glasersfeld, Francis Heylighen, Erich von Holst, Stuart Kauffman, Sergei P. Kurdyumov, Niklas Luhmann, Warren McCulloch, Humberto Maturana, Horst
Mittelstaedt, Talcott Parsons, Walter Pitts, Alfred Radcliffe-Brown, Robert Trappl, Valentin Turchin, Francisco Varela, Frederic Vester, John N. Warfield, Kevin Warwick, Norbert Wiener
Template:Winners of the National Medal of Science | {"url":"http://psychology.wikia.com/wiki/Claude_Shannon?oldid=142231","timestamp":"2014-04-18T21:44:57Z","content_type":null,"content_length":"154016","record_id":"<urn:uuid:33e661a9-7f63-4e66-9808-2fa7e543b136>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.9/warc/CC-MAIN-20140416005215-00649-ip-10-147-4-33.ec2.internal.warc.gz"} |
Corality Financial Group
The Loan Life Cover Ratio (LLCR) is one of the most commonly used debt metrics in Project Finance. Unlike period-on-period measures such as the Debt Service Cover Ratio (DSCR) it provides an analyst
with a measure of the number of times the cashflow over the scheduled life of the loan can repay the outstanding debt balance.
How many times more than the senior debt liability is the projects asset base (on a discounted basis), are the operational cashflows?
Interpretation of the Loan Life Cover Ratio (LLCR) – Example 1
An LLCR of 2.00x means that the Cashflow Available for Debt Service (“CFADS”), on a discounted basis, is double the amount of the outstanding debt balance.
Interpretation of the Loan Life Cover Ratio (LLCR) – Example 2
An LLCR of 1.00x means that the CFADS, on a discounted basis, is exactly equal to the amount of the outstanding debt balance. The movement of a key variable to achieve an LLCR of 1.00x is an
important measure of the strength of the project economics, often referred to as the ‘LLCR break-even’. A typical example is analysis of a toll road where the analysis could be ‘the project achieves
a break-even LLCR at 38% reduction of patronage from the Base Case’. In comparison the DSCR breakeven might only be 20%.
LLCR Definition
Generally the LLCR is calculated as:
LLCR = NPV [ CFADS over Loan Life ] / Debt Balance b/f
The Discount Rate used in the NPV calculation is usually the Cost of Debt, also known as the Weighted Average Cost of Debt.
Variations in LLCR Definition
From time to time Borrowers request and Lenders allow other ‘assets’ to be either included in the numerator or excluded from the denominator to reflect instances where there will be other cash
deposits available to the lender in the event of default rather than just the NPV of the Cashflow.
For example it is not uncommon to find the balance of the project’s cash account, or the Debt Service Reserve Account (‘DSRA’) added to the numerator or netted from the numerator. Extreme caution
needs to be applied when assessing the economics of a project where the LLCR is supported with cash account balances
When DSRA is included, the Loan Life Coverage Ratio shall then be calculated as:
LLCR = (NPV [ CFADS over Loan Life ] + DSRA/c Balance b/f ) / Debt Balance b/f
LLCR Calculation
The LLCR does not pick up weak periods as it essentially represents a discounted average. For this reason, if the Project has steady cashflows with Credit Foncier repayment, a good rule of thumb is
that the LLCR should be roughly equal to the average DSCR.
The example below illustrates a Project with a steady CFADS and Credit Foncier repayment. The example illustrates that the average DSCR is very close to the LLCR’s in this situation.
Screenshot #1: Example of LLCR and DSCR calculation
Common Errors and Oversights made when Modelling the LLCR
Algebraically the LLCR is a simple calculation, however it is also a calculation which is prone to error. Below are some of the frequently encountered mistakes:
• The NPV in the numerator has not calculated with the wrong base time so the LLCR values are all on different ‘time basis’.
• Incorrect use of the (X)NPV function in Excel.
• The discount rate has not been calculated as cost of debt, which includes interest and margin (and if appropriate PRI).
• The Discount Rate, which is usually the ‘Average Cost of Debt’ is overcomplicated rather than calculated simply as:
= Total Interest (for all tranches) / Total Debt Balance B/f (for all tranches)
• The formulas are too complicated making them hard to edit if the definition requires variation or to validate.
• The definition of the LLCR in the model is not clear and has not been validated against the debt term sheet.
• LLCR covenants are used to trigger cash sweeps, while at the same time including interest on reserve accounts / cash balances in the model resulting in a circular model.
• The inclusion of CFADS has not been correctly capped at the end of the loan life.
• CFADS has not been clearly presented in the cash waterfall and the LLCR is incorrectly referring to some other line in the waterfall.
• The discounting of CFADS is calculated incorrectly by confusing compound/simple interest rates.
• Adjustment has not been made and an annual discount rate is being applied to quarterly cashflows, or vice versa.
Caution needs to be applied when adding the DSRA/c Balance B/f into the numerator. Remember that the DSRA/c Balance shall be added to the NPV (CFADS) line, and not to be discounted as in CFADS. This
clause must be carefully checked in the definition of LLCR in the Term Sheet.
Checking the Calculation of LLCR
Some good rules of thumb to check if LLCT has been calculated correctly:
• For steady cashflows with a Credit Foncier repayment the Average DSCR should be very close to the LLCR.This is a good rule of thumb to cross-check the LLCR and/or the average DSCR. This is not
always the case especially in highly sculpted or exotic repayment scenarios.
• If CFADS is constant throughout and a Credit Foncier repayment is being adopted, then the calculated LLCR should also be constant.Please refer to the Screenshot #2, where the CFADS is USD 100
Million throughout and the Credit Foncier repayment is adopted. The LLCR is 1.66x throughout.
Screenshot #2: Example of constant LLCR | {"url":"http://www.corality.com/tutorials/llcr-loan-life-coverage-ratio","timestamp":"2014-04-18T08:40:08Z","content_type":null,"content_length":"58599","record_id":"<urn:uuid:80ce0577-1ddd-4f1c-93ef-d68c094191fb>","cc-path":"CC-MAIN-2014-15/segments/1397609533121.28/warc/CC-MAIN-20140416005213-00183-ip-10-147-4-33.ec2.internal.warc.gz"} |
Free splittings of one-relator groups
up vote 15 down vote favorite
Roughly speaking, I want to know whether one-relator groups only have 'obvious' free splittings.
Consider a one-relator group $G=F/\langle\langle r\rangle\rangle$, where $F$ is a free group. Is it true that $F$ splits non-trivially as a free product $A * B$ if and only if $r$ is contained in
a proper free factor of $F$?
1. One direction is obvious. It is clear that if $r$ is contained in a proper free factor then $G$ splits freely. (We think of $\mathbb{Z}\cong\langle a,b\rangle/\langle\langle b\rangle\rangle$ as
an HNN extension of the trivial group, so it's not really a counterexample, even though it might look like one.)
2. A quick search of the literature suggests that the isomorphism problem for one-relator groups is wide open. (I'd be interested in any details that anyone may have.)
3. There is no decision-theoretic obstruction. Magnus famously solved the word problem for one-relator groups. Much more recently, Nicholas Touikan has shown that, for any finitely generated group,
if you can solve the word problem then you can compute the Grushko decomposition. So one can algorithmically determine whether a given one-relator group splits. If the answer to my question is
'yes' then one can use Whitehead's Algorithm to find this out comparatively quickly.
4. When I first considered this question, it seemed to me that the answer was obviously 'yes' - I don't see how there could possibly be room in a presentation 2-complex for a 'non-obvious' free
splitting. But a proof has eluded me, and of course many seemingly obvious facts about one-relator groups are extremely hard to prove.
gr.group-theory gt.geometric-topology geometric-group-theory combinatorial-group-theor
add comment
4 Answers
active oldest votes
I think Grushko plus the Freiheitssatz does the trick. Suppose that $G=A\ast B$ is a 1-relator group which splits as a free product non-trivially. By Grushko, $rank(G)=rank(A)+rank(B)=
m+n$, where $rank(A)=m, rank(B)=n$. If $G$ is not free, then by Grushko there is a 1-relator presentation $\langle x_1,\ldots, x_m ,y_1,\ldots, y_n | R\rangle$, such that $\langle x_1,\
ldots, x_m \rangle =A \leq G, \langle y_1, \ldots, y_n \rangle=B \leq G$ (for this, one has to use the strong version of Grushko that any 1-relator presentation is Nielsen equivalent to
up vote 9 one of this type). Suppose that $R$ is cyclically reduced, and involves a generator $x_1 \in F_m$. Then $\langle x_2,\ldots , x_m, y_1, \ldots, y_n\rangle$ generates a free subgroup of $G$
down vote by the Freiheitssatz. But this implies that $B = \langle y_, \ldots, y_n\rangle$ is free. Moreover, if $G$ involves one of the generators $y_i$, then one sees that $A=\langle x_1,\
accepted ldots,x_m\rangle$ is also free, and therefore $G=A\ast B$ is free, a contradiction. So $R \in F_m$, as required.
Great - thanks, Ian! The strengthened version of Grushko's Theorem was the ingredient I was missing. – HJRW Jun 2 '10 at 0:10
add comment
I just came across the following, which is Prop. II.5.13 of Lyndon-Schupp.
Proposition. Let $G = \langle x_1, \ldots , x_n: r\rangle$ where $r$ is of minimal length under $Aut(\langle x_1, \ldots , x_n\rangle)$ and contains exactly the generators $x_1,\ldots ,
x_k$ for some $0 \leq k \leq n$. Then $G \cong G_1*G_2$ where $G_1 = \langle x_1, \ldots , x_k:r\rangle$ is freely indecomposable and $G_2$ is free with basis $x_{k+1}, \ldots ,x_n$.
up vote 6
down vote Unless I'm missing something, up to isomorphism you can assume that your relator has minimal length. If it is not contained in a free factor of $G$, then $k=n$ in the Proposition, hence
$G = G_1$ is freely indecomposable.
Excellent! I thought it must be in Lyndon and Schupp somewhere, but I didn't manage to find it. Thanks! – HJRW Jun 29 '10 at 21:02
add comment
Your question is reminiscent of Jaco's lemma. A special case of Jaco's lemma applies to a 2-handle attached to the boundary of a (3-dimensional) handlebody $H$ (which has free fundamental
group). If the boundary curve $J\subset \partial H$ along which the handle is attached is "disk-busting", that is, $\partial H-J$ is incompressible (and therefore $\pi_1$-injective) in $H$,
then the manifold obtained has $\pi_1$-injective boundary (and therefore does not split as a free product). This condition is easily seen to be equivalent to the conjugacy class of $J$ not
up vote belonging to any free factor of $\pi_1(H)$. So this answers your question in this very special case (also note that this works for "orbifold" handles attached along $J$). It's not clear
3 down whether Jaco's method might apply in your case, but it might be worth having a look (there are other proofs and generalizations of it too which you can find through Mathscinet). In
vote particular, his argument might also apply if the word is only virtually geometric.
add comment
I'm not sure whether it helps here, but your question reminds me of the Freiheitssatz. As you probably know, this is the theorem of Magnus that says that if $G=\langle x_1,\ldots,x_n|r\
rangle$ and $r$ involves the generator $x_n$, then the elements $x_1,\ldots,x_{n-1}$ generate a free group of rank $n-1$ inside $G$. Certainly your assumption implies this hypothesis with
respect to any basis $x_1,\ldots,x_n$.
up vote 2
down vote Also, I feel like we don't want to count HNN extensions as free products -- can't we just exclude the example of $\mathbb{Z}=\langle a,b|b\rangle$ by fiat? There aren't any other such
counterexamples, right?
Thanks for the answer Tom. Yes, I know about the Freiheitssatz; I don't see how it helps with this question directly. It's certainly conceivable that the Magnus-rewriting-and-induction
proof scheme of the Freiheittsatz could be applied to prove what I want. Unfortunately, I don't have a strong enough feeling for that method to know what's possible and what's not. – HJRW
Jun 1 '10 at 15:54
Regarding your second paragraph: from the Bass--Serre Theory point of view, we're really talking about actions on trees with trivial edge stabilizers; this boils down to free products or
HNN extensions over the trivial group. A group G is an HNN extension over the trivial group if and only if it's a non-trivial free product with a $\mathbb{Z}$ factor or $G\cong\mathbb{Z}
$. In my remark, I wanted to give a 'natural' reason why $\mathbb{Z}$ wasn't a counterexample, without mentioning actions on trees. I don't think there's anything natural about excluding
something by fiat. – HJRW Jun 1 '10 at 15:59
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Continuous cohomology of semi-simple Lie group.
up vote 5 down vote favorite
Let $G$ be a real connected semi-simple Lie group. Let $M$ be a finite dimensional representation of it. Are there general criteria when the continuous cohomology groups $H_{cont}^q(G,M)$ vanish?
A situation of particular interest for me is $G=SO^+(n-1,1)$, namely the connected Lorentz group, and $M$ is the standard representation of it. Is it true that the first continuous cohomology $H^1_
{cont}(G,M)=0$ ?
lie-groups group-cohomology cohomology
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3 Answers
active oldest votes
As pointed out by Konrad, this follows from the generalisation of van Est's theorem to the continuous case (due to Hochschild and Mostow); namely, that $$ H_c^m(G,M) \cong H^m(\mathfrak
{g},\mathfrak{k};M) $$ where $\mathfrak{g}$ is the Lie algebra of $G$ and $\mathfrak{k}$ is the Lie algebra of the maximal compact subgroup of $G$.
For the case in question, $m=1$, $\mathfrak{g} = \mathfrak{so}(n-1,1)$ and $\mathfrak{k}=\mathfrak{so}(n-1)$. I will take $n>2$.
According to Chevalley and Eilenberg, the cohomology $H^m(\mathfrak{g},\mathfrak{k};M)$ is computed from 'horizontal' 'equivariant' cochains in $C^m(\mathfrak{g},M)$, where 'horizontal'
means that the cochain vanishes whenever any of its entries belongs to $\mathfrak{k}$ and 'equivariant' means with respect to the action of $\mathfrak{k}$.
up vote 6
down vote Now for the algebras in question, $\mathfrak{g}$ breaks up as $\mathfrak{k} \oplus V$ under the action of $\mathfrak{k}$, where $V$ is the fundamental vector representation of $\
accepted mathfrak{k}$, whereas $M = V \oplus \mathbb{R}$, with $\mathbb{R}$ the trivial one-dimensional representation.
Since $$ C^0(\mathfrak{g},\mathfrak{k};M) = M^{\mathfrak{k}} $$ it follows that $$ \dim C^0(\mathfrak{g},\mathfrak{k};M) = 1~. $$ The differential $\delta: C^0 \to C^1$ is injective,
since if $T \in M^{\mathfrak{k}}$ ($T$ is 'timelike' hence the notation) $$ \delta T(X) = X \cdot T $$ which does not vanish identically.
On the other hand, $$ C^1(\mathfrak{g},\mathfrak{k};M) = \text{Hom}(V,M)^{\mathfrak{k}} $$ is again one-dimensional, hence $C^1 = \delta C^0$ and hence $H^1 =0$.
Great! Thank you very much!! – semyon alesker Oct 25 '11 at 10:38
You are welcome! If I may ask, in what context do you need this calculation? – José Figueroa-O'Farrill Oct 25 '11 at 10:42
To be short, I want to understand when invariant vectors in a quotient representation of the Lorentz group can be lifted to invariant vectors in the given representation. – semyon
alesker Oct 25 '11 at 18:58
This problem comes from my interest in the theory of valuations on convex sets. These are fintiely additive measures on convex sets, see e.g. these survey arxiv.com/PS_cache/arxiv/pdf
/1008/1008.0287v3.pdf It is of interest, both intrinsically to the theory and sometimes for applications in integral geometry, to classify valuations invariant under various groups.
Right now my student is trying to classify valuations invariant under the Lorentz group. Vanishing of the first cohomology is sufficient for this purpose. – semyon alesker Oct 25 '11
at 19:00
Many thanks for this. I'm always happy to see groups I like appear in seemingly unrelated work. – José Figueroa-O'Farrill Oct 25 '11 at 20:38
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I think that for $G$ a connected compact Lie group, we have $$ H^m_{cont}(G,M) = 0 $$ for $m>0$.
This follows from the van Est-isomorphism $$ H^m_{cont}(G,M) \cong H^m(g,k; M), $$ which holds for $G$ a connected Lie group, $K$ a maximal compact subgroup and $g$, $k$ the Lie algebras
up vote 4 of $G$, $K$, and the thing on the right hand side is the relative Lie algebra cohomlogy. For $K=G$ in the compact case, this relative Lie algebra cohomology is identically zero.
down vote
A good place to look is Stasheff's "Continuous cohomology of groups and classifying spaces".
The group in the question is not compact, though. Still, the calculation of $H^1(\mathfrak{g},\mathfrak{k};M)$ is not hard in this case and one finds that indeed it vanishes. – José
Figueroa-O'Farrill Oct 24 '11 at 17:50
Oh, oops, of course you're right, José. – Konrad Waldorf Oct 24 '11 at 17:54
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I think that it is true if you use "smooth" instead of continuous.
up vote 3 down
Van Est's original theorem was indeed for smooth cohomology, but I believe that this was extended to the continuous case by Hochschild and Mostow in ams.org/mathscinet-getitem?mr=
147577 – José Figueroa-O'Farrill Oct 24 '11 at 17:53
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Not the answer you're looking for? Browse other questions tagged lie-groups group-cohomology cohomology or ask your own question. | {"url":"http://mathoverflow.net/questions/78999/continuous-cohomology-of-semi-simple-lie-group","timestamp":"2014-04-16T13:37:49Z","content_type":null,"content_length":"66991","record_id":"<urn:uuid:35df5e77-5042-4f60-8f36-060d167a63ec>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00315-ip-10-147-4-33.ec2.internal.warc.gz"} |
Hausdorff maximal principle
Hausdorff maximal principle
The Hausdorff maximal principle is a version of Zorn's lemma, equivalent to the usual version and thus (given excluded middle) equivalent to the axiom of choice.
Statement and proofs
Given a poset (or proset) $S$, let a chain in $S$ be a subset $A$ of $S$ which, as a sub-proset, is totally ordered. A chain $A$ is maximal (as a chain) if the only chain that $A$ is contained in is
$A$ itself.
Theorem (Hausdorff maximal principle)
Every chain in a proset is contained in a maximal chain.
We will use Zorn's lemma. Let $P$ be a proset and let $C \subseteq P$ be a chain. Consider the collection $\mathcal{C}$ of chains in $P$ that contain $C$, ordered by inclusion. If $\{C_\alpha\}_{\
alpha \in A} \subseteq \mathcal{C}$ is a family totally ordered by inclusion, then the union $\bigcup_\alpha C_\alpha$, with the order coming from $P$, is also totally ordered: any two elements $x \
in C_\alpha, y \in C_\beta$ are comparable in $max(C_\alpha, C_\beta)$. The hypotheses for Zorn’s lemma therefore obtain on $\mathcal{C}$, and we conclude that $\mathcal{C}$ has a maximal element,
which is clearly maximal in the collection of all chains.
Proof of converse
Conversely, suppose that the Hausdorff maximal principle holds; we will prove Zorn’s lemma. Suppose given a poset (or preorder) $P$ such that every chain in $P$ has an upper bound. Since $\empty$ is
a chain, the Hausdorff maximal principle implies that $P$ contains a maximal chain $C$; let $x$ be an upper bound of $C$. Then $x$ is maximal: if $x \leq y$, then $C = C \cup \{y\}$ by maximality of
$C$; therefore $y \in C$ and hence $y \leq x$ since $x$ is an upper bound of $C$.
Revised on June 5, 2009 02:03:02 by
Toby Bartels | {"url":"http://www.ncatlab.org/nlab/show/Hausdorff+maximal+principle","timestamp":"2014-04-21T02:02:48Z","content_type":null,"content_length":"20912","record_id":"<urn:uuid:84341be8-5b7a-4b18-bf28-66e4538b8e4a>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00475-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Confused with Complex Vector Space notation
November 26th 2012, 11:38 PM #1
Oct 2009
Confused with Complex Vector Space notation
I've attached a picture with the problem statement. I've done some least squares problems with matrices, but I don't think that will work here.
I thought I'd just find the orthogonal projection of f(x) onto W directly, but in order to do that I need basis vectors for W.
I don't really understand what the vector space W is here. Is it a subspace of V, where V is all continuous functions with a function value between -pi and pi? Or is it the input in e^ix which is
between -pi and pi?
If I could get some help clarifying that vector space notation, I believe I will be able to solve the problem.
Last edited by gralla55; November 27th 2012 at 01:07 AM. Reason: Forgot to upload the problem, haha
Re: Confused with Complex Vector Space notation
W is a vector space of functions, the smallest containing the functions x->e^ix and x->e^-ix (all functions that can be written ae^ix+be^-ix), so it's a subspace of V.
Re: Confused with Complex Vector Space notation
All right, thank you for replying! But is V the vector space of all complex functions where the input is between -pi and pi?
Re: Confused with Complex Vector Space notation
Yes, V the vector space of all continuous complex functions where the input is between -pi and pi
November 27th 2012, 01:26 AM #2
Junior Member
Nov 2012
November 27th 2012, 01:53 AM #3
Oct 2009
November 27th 2012, 07:27 AM #4
Junior Member
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Multiwavelets on the Interval
Bin Han and Qing-Tang Jiang
Smooth orthogonal and biorthogonal multiwavelets on the real line with their scaling function vectors being supported on $[-1,1]$ are of interest in constructing wavelet bases on the interval $[0,1]$
due to their simple structure. In this paper, we shall present a symmetric $C^2$ orthogonal multiwavelet with multiplicity $4$ such that its orthogonal scaling function vector is supported on $[-1,1]
$, has accuracy order $4$ and belongs to the Sobolev space $W^{2.56288}$. Biorthogonal multiwavelets with multiplicity $4$ and vanishing moments of order $4$ are also constructed such that the primal
scaling function vector is supported on $[-1,1]$, has the Hermite interpolation properties and belongs to $W^{3.63298}$ while the dual scaling function vector is supported on $[-1,1]$ and belongs to
$W^{1.75833}$. A continuous dual scaling function vector of the cardinal Hermite interpolant with multiplicity $4$ and support $[-1,1]$ is also given. Based on the above constructed orthogonal and
biorthogonal multiwavelets on the real line, both orthogonal and biorthogonal multiwavelet bases on the interval $[0,1]$ are presented. Such multiwavelet bases on the interval $[0,1]$ have symmetry,
small support, high vanishing moments, good smoothness and simple structures. Furthermore, the sequence norms for the coefficients based on such orthogonal and biorthogonal multiwavelet expansions
characterize Sobolev norm $\|.\|_{W^s([0, 1])}$ for $s\in (-2.56288, 2.56288)$ and for $s\in (-1.75833, 3.63298)$, respectively.
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[SOLVED] How to change the base of y=A(1.3)^-t
April 27th 2009, 12:13 PM #1
Apr 2009
[SOLVED] How to change the base of y=A(1.3)^-t
The question says:
" $y=A(1.3)^{-t}$
By changing the base of the exponent to e, express y in terms of A, e and t."
I tried:
$y=A(1.3)^{-t}\Rightarrow\frac{y}{A}=1.3^{-t} \Rightarrow log_{1.3}\frac{y}{A}=-t\Rightarrow\frac{\ln\frac{y}{A}}{\ln1.3}=-t$
$\Rightarrow\ln\frac{y}{A}=-t\ln 1.3\Rightarrow\ln y-\ln A=\ln1.3^{-t}\Rightarrow\ln y=\ln A+\ln1.3^{-t}$
$\Rightarrow\ln y=\ln A(1.3)^{-t}\Rightarrow y=A(1.3)^{-t}$
It just goes round in a circle. What am I doing wrong?
Thanks in advance.
Changing exponential bases
Taking the last equality on your second line:
$\Rightarrow\ln\frac{y}{A}=-t\ln 1.3\Rightarrow\ln y-\ln A=\ln1.3^{-t}\Rightarrow\ln y=\ln A+\ln1.3^{-t}$
You can simplify like this:
$y=e^{ln A + ln (1.3^{-t})}=e^{lnA} * e^{-t*ln1.3}$
Another way of thinking of it is to solve the following for x:
$e^{x}=1.3$ to get $x=ln{1.3}$.
Then, instead of using 1.3 as your base, use $e^{ln{1.3}}$
It may seem strange to leave your answer in this form, without simplifying the e to the "ln" power, but there could be a variety of reasons for it. For example, now you know that $ln{1.3}$ is the
rate of decay in a continuous decay model.
Last edited by pflo; April 27th 2009 at 02:23 PM.
Thanks a lot, that's a really great answer. : )
April 27th 2009, 01:20 PM #2
April 27th 2009, 01:41 PM #3
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intersecting line segments definition
Best Results From Wikipedia Yahoo Answers Youtube
From Wikipedia
Line segment
In geometry, a line segment is a part of a line that is bounded by two end points, and contains every point on the line between its end points. Examples of line segments include the sides of a
triangle or square. More generally, when the end points are both vertices of a polygon, the line segment is either an edge (of that polygon) if they are adjacent vertices, or otherwise a diagonal.
When the end points both lie on a curve such as a circle, a line segment is called a chord (of that curve).
If V\,\! is a vector space over \mathbb{R} or \mathbb{C}, and L\,\! is a subset of V,\,\! then L\,\! is a line segment if L\,\! can be parameterized as
L = \{ \mathbf{u}+t\mathbf{v} \mid t\in[0,1]\}
for some vectors \mathbf{u}, \mathbf{v} \in V\,\!, in which case the vectors \mathbf{u} and \mathbf{u+v} are called the end points of L.\,\!
Sometimes one needs to distinguish between "open" and "closed" line segments. Then one defines a closed line segment as above, and an open line segment as a subset L\,\! that can be parametrized as
L = \{ \mathbf{u}+t\mathbf{v} \mid t\in(0,1)\}
for some vectors \mathbf{u}, \mathbf{v} \in V\,\!.
An alternative, equivalent, definition is as follows: A (closed) line segment is a convex hull of two points.
• A line segment is a connected, non-emptyset.
• If V is a topological vector space, then a closed line segment is a closed set in V. However, an open line segment is an open set in V if and only if V is one-dimensional.
• More generally than above, the concept of a line segment can be defined in an ordered geometry.
In proofs
In geometry, it is sometimes defined that a point B is between two other points A and C, if the distance AB added to the distance BC is equal to the distance AC.
In an axiomatic treatment of Geometry, the notion of betweenness is either assumed to satisfy a certain number of axioms, or else defined in terms of an isometry of a line (used as a coordinate
Segments play an important role in other theories. For example, a set is convex if the segment that joins any two points of the set is contained in the set. This is important because it transforms
some of the analysis of convex sets to the analysis of a line segment.
Line segment intersection - Wikipedia, the free encyclopedia
ISBN 3-540-65620-0. Chapter 2: Line Segment Intersection, pp.19 44. ... For the simple case of testing the intersection of two line segments: Solution by ...
From Yahoo Answers
Question:it is said that a quadrilateral is formed by joining 4 line segments but a quadrilateral can also be formed by joining 4 lines(by intersecting them) then why only line segments is given or
used in the definition of a quadrilateral?pls friends give me a answer fast it is very necessary
Answers:. Quadrilateral is a closed figure. Now if you wish to have a closed figure having four sides, these sides must be line segments. <<<<<<<<<<<<<<<<<<<<<< .
Question:Just wondering.
Answers:Yes, it's always a segment bisector.
Answers:A line segment is a part of a line that is bound by two distinct end points, and contains every point on the line between its endpoints.
Question:write the word that describes the lines or line segments. 1.the strings on a guitar 2.the marks left by a skidding car 3.sidewalks on opposite sides of a street 4.the segments that make up a
+ sign 5.the wires suspended between telephone poles 6.the hands of a clock at 9:00 P.M. 7the trunks of grown trees in a forest. please help. i dont understand this. :(
Answers:1) Parallel 2) Parallel 3) Parallel 4) Perpendicular 5) Parallel 6) Perpendicular 7) Parallel If the instructions are written as you wrote them, then it means pick the word that describes the
relationship of the lines in each example. Like with the guitar strings or the sidewalk, these lines run in the same direction so they are parallel. Since a + sign has opposite lines, it is
perpendicular, same with 9:00 because the hour hand is horizontal and the minute hand is vertical. Make sense?
From Youtube
Segment Intersection :demonstrations.wolfram.com The Wolfram Demonstrations Project contains thousands of free interactive visualizations, with new entries added daily. Line segments are said to
intersect when they cross each other. This can be determined mathematically by analyzing the endpoints of the line segments. Drag the endpoints to explore cases of intersecting line segments. In this
Demonstration, the segments... Contributed by: Ed Pegg Jr
Lines and Segments that Intersect Circles :Learn about Lines and Segments that Intersect Circles | {"url":"http://www.edurite.com/kbase/intersecting-line-segments-definition","timestamp":"2014-04-21T15:21:29Z","content_type":null,"content_length":"72775","record_id":"<urn:uuid:93e1fd92-214d-4c13-9a3f-c93dd33cd515>","cc-path":"CC-MAIN-2014-15/segments/1397609540626.47/warc/CC-MAIN-20140416005220-00448-ip-10-147-4-33.ec2.internal.warc.gz"} |
Normal Distribution
Hi robinsonov;
Please start your own thread from now on in "Help Me" that way you will get faster help.
Here is how I do it. Start with the obvious fraction.
Now you have to get a whole number over another whole number.
If you moved the decimal point in 1.0021 four places to the right
you would have 10021 Whatever you do to the top of the fraction you have to do to the bottom. So
You can now make sure that it is in lowest terms. | {"url":"http://www.mathisfunforum.com/viewtopic.php?pid=179636","timestamp":"2014-04-20T00:51:54Z","content_type":null,"content_length":"29139","record_id":"<urn:uuid:4599c09f-3fd8-4318-b42c-dc94bd88e56e>","cc-path":"CC-MAIN-2014-15/segments/1397609537804.4/warc/CC-MAIN-20140416005217-00459-ip-10-147-4-33.ec2.internal.warc.gz"} |
Exact first order differential equations problems
February 18th 2009, 08:41 AM
Exact first order differential equations problems
I'm totally unsure of where to put this topic - sorry itll probably need moved.(Worried)
Ok I'm doing a further maths A-level right now but I'm also doing an ordinary maths A-level at the same time. This causes a LOT of problems with my further maths, because I have not covered some
of the topics in ordinary maths yet.
Anyway i'm having particular problems with Exact 1st order differential equations. It's probably something simple enough that im not getting but I dont understand how:
y + x(dy/dx) = x(cubed)
(d/dx)(xy) = x(cubed)
In my notes my teacher just wrote (y + x(dy/dx) is exactly the derivative of ?)
What is the question mark.
At the moment we are really rushing through the course, because i started the course late in the year, so doubtless I'll need more help from this forum. (Worried)
Oh and also is there any way to get proper maths notation here. its gonna get hard when it gets to things like integral signs.
February 18th 2009, 10:05 AM
Chris L T521
I'm totally unsure of where to put this topic - sorry itll probably need moved.(Worried)
Ok I'm doing a further maths A-level right now but I'm also doing an ordinary maths A-level at the same time. This causes a LOT of problems with my further maths, because I have not covered some
of the topics in ordinary maths yet.
Anyway i'm having particular problems with Exact 1st order differential equations. It's probably something simple enough that im not getting but I dont understand how:
y + x(dy/dx) = x(cubed)
(d/dx)(xy) = x(cubed)
In my notes my teacher just wrote (y + x(dy/dx) is exactly the derivative of ?)
What is the question mark.
At the moment we are really rushing through the course, because i started the course late in the year, so doubtless I'll need more help from this forum. (Worried)
It may be hard to see at first, but if you treat y as a function of x, $y+x\frac{\,dy}{\,dx}=y\left(1\right)+x\frac{\,dy}{ \,dx}=y\frac{\,d}{\,dx}\!\left(x\right)+x\frac{\,d y}{\,dx}$, which has
the form of a product rule application. We can conclude that this is the same as $\frac{\,d}{\,dx}\left(xy\right)$.
Does this make sense?
Oh and also is there any way to get proper maths notation here. its gonna get hard when it gets to things like integral signs.
The forum uses LaTeX. You can find out how to use it here. In order to generate the images, the code must be put within [tex][/tex] tags. You can also put your mouse cursor over the LaTeX
generated images to see the code that was used to produce those images.
February 18th 2009, 10:37 AM
That does help a lot, and I now get that question - but more complex questions are very tricky to understand . I only get one hour per week of this - so its pretty hectic, and my teacher does not
get time to go over much
Now confused about how
2xy (dy/dx) + y(squared) = x (cubed)
becomes 4xy (squared) = x^4 +c
My thoughts behind this are: if
: d/dx (x)(y) = y(dx/dx) +x(dy/dx)
Then : d/dx (2xy)(y(squared) ) = y (squared) (1) + (2xy)(dy/dx)
So : (2x)(y(cubed)) = (x^4)/4 + c
And my answer is 8xy(cubed) = x^4 +c
Which... is wrong :( I would assume that my product is wrong yea?
Thanks. :)
February 18th 2009, 10:50 AM
$\frac{\,d(y^2)}{\,dx}=2y \frac{\,dy}{\,dx}$.
$2xy \frac{\,dy}{\,dx} + y^2=x \: 2y \frac{\,dy}{\,dx} + y^2 \left(1\right)=x \: \frac{\,d(y^2)}{\,dx}+y^2 \:\frac{\,dx}{\,dx}$, which has the form of a product rule application. We can conclude
that this is the same as $\frac{\,d}{\,dx}\left(xy^2\right)$.
Special tribute to Chris (Rofl)
February 18th 2009, 11:23 AM
Ah. I didnt know
$\frac{\,d(y^2)}{\,dx}=2y \frac{\,dy}{\,dx}$
wow :)
Gah these are only the most basic of questions of this type and im struggling - my teacher im sure hasnt taught me enough for this (Headbang) I have been working for hours on these first 3
questions of the 18 i have been set. last question before I go - could you explain how
(dy/dx) sin x + y cos x = tan x
becomes y sin x = ln (secx) + c
thanks once again for the help.
February 18th 2009, 11:40 AM
This is a basic rule but if you have never seen this relation I don't know how you can perform your exercises ....
The general rule is $\frac{\,d(y^n)}{\,dx}=ny^{n-1} \frac{\,dy}{\,dx}$
Gah these are only the most basic of questions of this type and im struggling - my teacher im sure hasnt taught me enough for this (Headbang) I have been working for hours on these first 3
questions of the 18 i have been set. last question before I go - could you explain how
(dy/dx) sin x + y cos x = tan x
becomes y sin x = ln (secx) + c
thanks once again for the help.
$\frac{d(y\:\sin x)}{dx} = \frac{dy}{dx}\:\sin x + y\:\frac{d(\sin x)}{dx} = \frac{dy}{dx}\:\sin x + y \:\cos x$
And $\tan x = \frac{\sin x}{\cos x} = -\frac{\frac{d(\cos x)}{dx}}{\cos x}$ therefore integration gives $- ln(|\cos x|) + c$
February 18th 2009, 12:07 PM
Ah yes written in a different way than im used to, but essentially the same now I see(Giggle). But that has helped me tonnes - seriously thanks a lot :) I think i can get the rest from here on...
maybe :) | {"url":"http://mathhelpforum.com/differential-equations/74274-exact-first-order-differential-equations-problems-print.html","timestamp":"2014-04-19T05:07:52Z","content_type":null,"content_length":"14239","record_id":"<urn:uuid:9f068324-9c9e-44a2-95dc-5efa015a359b>","cc-path":"CC-MAIN-2014-15/segments/1397609535775.35/warc/CC-MAIN-20140416005215-00063-ip-10-147-4-33.ec2.internal.warc.gz"} |
High Voltage Power Supply - UltraVolt®
AP-6: Thermal Management of UltraVolt HVPSs
download as PDF
Any electronic device's long-term reliability is directly related to how the overall system design allows it to dissipate any generated heat properly. Likewise, the long life and reliability of
UltraVolt high-voltage power supplies (HVPSs) will be enhanced through proper thermal management.
Due to the laws of physics, the process of converting one voltage level to another cannot be done without energy loss; that loss, the power used to operate the power supply itself, is usually
dissipated as heat. UltraVolt HVPSs have excellent efficiency ratings, ranging from 75% to 92% at full output power (depending upon the model), delivering full rated voltage at nominal operating
conditions. The remaining fraction of the HVPS's input power is converted to heat. Methods of properly dissipating this heat can be found in this application note.
The operating efficiency of a power supply — one of the key variables in thermal calculations — can be determined in different ways. The Acceptance Test Procedure (ATP) that came with your UltraVolt
power supply contains a few common operating points and related data. The ATP data can be used to determine the supply's operating efficiency at various points. Alternately, the efficiency of a
particular UltraVolt HVPS under specific operating conditions can be obtained from our Customer Service Department. Typical operating efficiency is also listed in the DC Efficiency vs. Input Voltage
Range graph of the power supply's data sheet. This graph shows only typical values for the various models in that series of power supplies.
The analysis and several sample calculations in this application note are divided between Units without user-supplied heat sinks (and -H option units), and Units with user-supplied heat sinks (and
other cooling methods).
Units without user-supplied heat sinks (and -H option units):
An UltraVolt 2A12-P4 is continuously delivering its nominal 2mA at 2kV (PDCout = 4W) with a nominal 12VDC input. The input current is reported as 390mA (PDCin = 4.68W), based on the ATP at this
operating point (nominal output power, nominal 12V input voltage).
We can determine the heat energy dissipated as
Pheat = PDCin – PDCout
PDCin = VinIin
and the unit's efficiency is
PDCout / PDCin.
Solving for PDCin yields
PDCin = VinIin = ((12V)(0.39A))
= 4.68W.
So, Pheat = PDCin – PDCout = (4.68W) – (4W)
= 0.68W.
Thus, efficiency = PDCout / PDCin = (4W) / (4.68W)
= 86%.
Therefore, this power supply will dissipate 0.68W of heat when operated as per the discussed operating conditions. At this operating point, its DC efficiency is 86%.
For an “A” Series UV HVPS (for example, the 2A24-P4) with the standard plastic case, no extra heat sinking, and all surfaces exposed to free air, calculate the temperature rise of the power supply's
case under these conditions. Assume the HVPS is mounted in sockets on the printed circuit board (PCB).
First, the total surface area (SA) of the standard 4W “A” Series must be calculated. The SA can be calculated from simple geometry, since the HVPS has 6 sides consisting of 3 sets of 2 equal sides.
(Dimensions can be found in the “A” Series datasheet, where l represents Length, w is Width, and h is Height.) Solving for surface area we find,
SA = 2 (lw + lh + wh) = 2 ((3.7in)(1.5in) + (3.7in)(0.77in) + (1.5in)(0.77in))
= 19.11in^2.
It is universally accepted that 1W will raise 1in^2 approximately 100°C for all surface area exposed to free, still air (assumptions made for simplicity). Since we are dissipating 0.68W,
∆T = (Pheat / SA)(100°C • in^2 / W) = (0.68W / 19.11in^2)(100°C • in^2 / W)
= 3.56°C.
Finally, assuming the enclosure within which the UV HVPS is placed has a continuous operating temperature of 40°C (Tambient), we have a final UV HVPS case temperature of 43.56 °C ( ∆T + Tambient).
Thankfully, this value is comfortably below the recommended maximum “A” Series operating temperature of 65°C. In practice, the actual case temperature may be slightly different than the expected
value, depending upon the mounting of the power supply and the heat dissipation properties of that mounting.
The power supply will also have certain areas that are warmer and other areas that are cooler because certain components within the unit dissipate more heat than others. These warm areas are
generally referred to as “hot spots.” Note, flush-mounting this power supply onto a PCB will hinder any convection cooling along the surface of the power supply that faces the circuit board. To
account for this, the SA calculation of this particular example is carried out by summing the 5 exposed sides, ignoring the cooling effect of the mounting surface. Alternately, if the power supply is
PCB-mounted using sockets, then the mounting side will allow slight cooling by convection; this should be accounted for in the calculations using a scaling factor. A discussion of mounting methods
for UltraVolt high-voltage power supplies appears in Application Note #3.
As seen in the calculations above, a 4W unit running at full rated power rarely needs any supplemental heat sinking. However, a 20W, plastic-encased “A” Series running continuously at full power will
dissipate approximately 3.5W, assuming 85% efficiency (which is a rough approximation used only for this example; do not use this number for your own calculations).
Without any supplemental heat sinking and with one surface of the power supply mounted onto a PCB, the case temperature will rise 26°C, an unacceptable temperature increase in applications with more
than a 40°C ambient temperature. The design of 20W “A” Series, as well as all other UV modular HVPSs, facilitates easy interfacing to a heat-dissipation medium. The top surface of the 20W,
plastic-case “A” Series is actually a 0.062”-thick aluminum plate, which will efficiently conduct heat to an external heat-sink medium, such as a chassis wall or other customer-supplied heat sink.
This aluminum plate also helps regulate internal power supply hot spots, which occur because of the heat dissipations of different components within the power supply itself (as mentioned previously).
A 4A24-P20-H (a 4kV, 20W supply) delivering 2500V at 3mA (much less than the rated 20W out), draws .46A from its 24V supply. Solving for Pheat,
Pheat = PDCin – PDCout = ((24V)(0.46A) – (2.5kV)(3mA))
= 3.54W.
Therefore, this power supply will dissipate 3.54 watts of heat when operated, according to the operating conditions (68% efficiency due to the light load). For comparison purposes, a 4A24-P20-H
running at full output voltage has been analyzed in the appendix of this Application Note, providing a typical 89% efficiency example.
For the previously mentioned power supply, flush-mounted to a PCB, with the -H (heat sink) option, and with all sides except the mounting side exposed to free air, calculate the temperature rise of
the power supply's case.
First, the exposed surface area of the standard 20W “A” Series must be calculated. From simple geometry (dimensions from “A” Series datasheet), the SA of the sides exposed to free air (neglecting the
mounting and the heat sink sides) can be calculated as
SA1 = 2(lh + wh) = 2((3.7in)(0.77in) + (1.5in)(0.77in))
= 8.01in^2.
The -H heat sink adds approximately another 13in^2 of surface area, so the total exposed still-air surface area is
SAtot = SA1 + SAhs = (8.01in^2) + (13in^2)
= 21.01in^2.
It is again assumed that 1W will raise 1in^2 of surface area approximately 100°C (in still air, which we will assume here for simplicity).
Since we are dissipating 3.54W, we can calculate ∆T as
∆T = (Pheat / SAtot)(100°C • in^2 / W) = (3.54W / 21.01 in^2)(100°C • in^2 / W)
= 16.85°C.
If the enclosure within which the UV HVPS is placed has a continuous operating temperature of 40°C, we will have a final UV HVPS average case temperature of 56.9°C. (Again, due to the differing heat
dissipations of certain components within the power supply, certain regions of the unit will be warmer than others.) Of course, this value is below the recommended maximum “A” Series operating
temperature of 65°C.
It is important to consider that all of our calculations have assumed the power supply is supplying a continuous amount of power. Low duty-cycle or pulsed operation of a UV power supply may dissipate
a considerably smaller amount of average heat than the calculations here have yielded. In these situations, the HVPS may not need supplemental heat sinking. For example, a power supply operating at a
10% duty cycle will dissipate a small fraction of the continuous full-output dissipated heat (and hence can be operated without supplemental heat sinking).
Note, should these units be operated at less than full-rated output power, their operating efficiency will be less than their potential maximum efficiency. UltraVolt power supplies are designed to
operate most efficiently when delivering their full-rated output power.
In summary, when dealing with an HVPS unit that does not have any supplemental heat sinking (or has only a -H heat sink), a few basic rules can be followed to simplify calculations. When calculating
the surface area of the unit, ensure only sides that have access to free, open air are included in the calculations. If a side is in close proximity to a surface, such as a high-voltage power supply
mounted with sockets onto a PCB, a fraction of the actual surface area of the hindered side(s) should be included in the calculations. If a -H heat sink is used, be sure that its side is summed into
the equation for surface area only once (for that particular side, the 13in^2 surface area of the -H heat sink is the surface exposed to air).
Units with user-supplied heat sinks (and other cooling methods)
Through testing, an UltraVolt 6C24-P125 is determined to draw 5.9A from its 24V supply while delivering 6kV into its rated load.
Under the above operating conditions, determine whether a fully enclosed metal container of dimensions 12” x 6” x 4” (288in^2 total surface area) will sufficiently cool this HVPS if it is mounted to
the wall inside the chassis. (Only the exterior of the container is exposed to the outside free air.) All sides of the metal container are exposed to free air. Assume the container is a good thermal
conductor and has a low thermal-resistance interface between the HVPS and the container.
Pheat = PDCin – PDCout = ((24V)(5.9A)) – 125W
= 16.6W (88% efficiency).
Here, since the HVPS is mounted inside the enclosed container, the surface of the HVPS will not directly dissipate its heat to the outside air. Instead, it will transfer its thermal energy to the
chassis. The only surface that can dissipate heat to the outside air is the chassis' surface. The cooling provided by the HVPS's surface inside this container is negligible compared to the cooling
provided by the heat transfer from the HVPS to the chassis wall. We approximate the dissipating surface area to be 12” x 12” x 2” = 288in^2.
Since we are dissipating 16.6W,
∆T = (Pheat / SAtot)(100°C • in^2 / W) = (16.6W / 288in^2)(100°C • in^2 / W)
= 5.7°C.
A chassis temperature increase of 5.7°C should be quite acceptable under nearly all circumstances. So in conclusion, the HVPS would operate quite satisfactorily from a thermal standpoint.
In the above example, we made some assumptions that will now be discussed. First, it was assumed that a low thermal-resistance interface was used between the HVPS and the chassis wall. This implies
the HVPS can transfer its heat to the heat sink - the chassis in the above example - without hindrance. In practice, this can be accomplished using an excellent heat conduction
medium, such as thermal elastomer (0.010” or 0.020” are common thicknesses), double-sided thermal tape, or thermal grease. This heat-conduction medium is then placed between the HVPS and the heat
sink, and the HVPS is securely mounted to the heat sink using the proper bracket and/or mounting fasteners. These fasteners are then torqued sufficiently to guarantee the correct contact pressure on
the heat-conduction medium. For example, the #8 studs on a 60W unit should be torqued to 8ft • lb of torque to ensure the proper ‘cold flow’ of thermal-elastomer heat-conduction medium. Remember, in
order to keep the thermal resistance low between the HVPS and its heat sink (allowing the greatest transfer of heat), the HVPS should have as much surface area as possible facing the smooth heat-sink
medium (via the heatconduction medium).
Second, it was assumed the heat-sink medium — the chassis wall in the above example — was a good conductor of heat. This assumption simplifies calculations by allowing us to utilize the entire
open-air, exposed surface area of the heat sink as a dissipation medium (since the heat to be dissipated is easily conducted to the exposed surface area of the dissipation medium). Not surprisingly,
heat sinks are usually made from metals known for their excellent heat conduction properties (such as aluminum).
When cooling an HVPS, many design variables enter into the equation and can determine how actual implementation is carried out. Airflow is one very important variable, and is, unfortunately, too
complex to fully analyze here. Generally speaking, forced-air cooling from a simple fan can drastically reduce the size of a required heat sink, since the flowing air removes the dissipated heat more
efficiently (by transferring the heat to a larger volume of air and not relying solely on convection). The ambient temperature of the HVPS's surroundings is also a very important variable. A 15°C
temperature increase may be acceptable if the ambient is 25°C, but may be totally unacceptable if the ambient temperature is 60°C (due to other devices within the same chassis dissipating a large
amount of heat, for example). The mounting method is another crucial design variable.
If the HVPS is mounted to a large chassis wall, then it is unlikely a supplemental heat sink will be required.
Oddly enough, one of the most important considerations when it comes to cooling design is to allow for a large margin of safety. Even common dust coating a heat sink can drastically reduce its
cooling effectiveness. Obstructed or dust-coated fan vents can seriously reduce the cooling efficiency of the fans themselves. The importance of air-vent filters and regular cleaning becomes obvious.
Also, load changes (due to slight design changes) can require more power to be delivered from the HVPS and, hence, change its actual heat dissipation. In summary, include a large margin of safety in
your heat calculations to ensure your high-voltage power supply cools sufficiently under all conditions. This can be achieved simply by assuming the power supply will operate in an ambient
temperature higher than previously calculated.
Although one can carry out thermal calculations using thermal junction analysis and actual thermal resistance values, we have not opted for that method here. While they will yield more accurate
results, the methods described here will sufficiently approximate the cooling requirements for your HVPS, given an adequate safety margin. Complex cooling requirements, such as very limited space
restrictions with forced-air cooling may require a more in-depth analysis than proposed here. In these cases, we recommend you consult both a text on thermal management and UltraVolt's Customer
Service Department. Remember to keep your UltraVolt HVPS cool, and it will reward you with a long, trouble-free life.
│ Surface Area of UltraVolt Models │
│ UltraVolt Model │ Total Surface Area (in square inches) │
│ “A” / with -F / with -C / with -F-C │ 19.11 / 25.26 / 28 / 37.69 │
│ “10A” / with -C │ 20.46 / 29.2 │
│ “15A” / with -C │ 25.26 / 35.4 │
│ “20A” / with -C │ 31.5 / 43.2 │
│ “25A or 30A” / with -C │ 39.93 / 57 │
│ “C” / with -C │ 19.11 / 28 │
│ 60W/125W “C” │ 54.28 │
│ 250W “C” │ 98.88 │
│ “8C - 15C” │ 98.88 │
│ “20C - 25C” │ 98.88 │
A 4A24-P20-H (a 4kV, 20W supply) delivers 4kV at 5mA (the rated 20W out), and draws .94A from its 24V supply, operating at 89% efficiency. The heat energy dissipated is
Pheat = PDCin – PDCout = ((24V)(0.94A)) – ((4kV)(5mA))
= 2.56W.
This power supply will dissipate 2.56W of heat when operated as described above.
For the previously mentioned power supply, flush-mounted to a PCB, with the -H (heat sink) option, and with all sides except the mounting side exposed to free air, calculate the temperature rise of
the power supply's case.
First, the exposed surface area of the standard 20W “A” Series must be calculated. From simple geometry (dimensions from the “A” Series data sheet), the SA of the sides exposed to free air
(neglecting the mounting and the heat sink sides) can be calculated as
SA1 = 2(lh + wh) = 2((3.7in)(0.77in) + (1.5in)(0.77in))
= 8.01in^2.
The -H heat sink adds approximately another 13in^2 of surface area, so the total exposed still-air surface area is
SAtot = SA1 + SAhs = (8.01in^2) + (13in^2)
= 21.01in^2.
It is again assumed that 1W will raise 1in^2 of surface area approximately 100°C (in still air, which we will assume here for simplicity).
Since we are dissipating 2.56W, we can calculate ∆T as
∆T = (Pheat / SAtot)(100°C • in^2 / W) = (2.56W / 21.01in^2)(100°C • in^2 / W)
= 12.2°C.
Assuming the enclosure within which the UV HVPS is placed has a continuous operating temperature of 40°C, we have a final UV HVPS average case temperature of 52.2°C. This value is comfortably below
the recommended maximum “A” Series operating temperature of 65°C.
Rev. D1
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