July 2012 - mathfun@mailman.xmission.com

Re: [math-fun] Mathematical Reviews and Zentralblatt fur Mathematik
by Richard Guy 30 Jul '12

30 Jul '12

Dear seqfans & mathfunsters, I'm being dispossessed and would like to find a good home for Math Reviews (from Vol.1, 1940, up to a few years ago) and Zentralblatt (ditto, from 1931). If anyone knows of one, please let Yanmei know [email address above]. I can cover transportation costs. But perhaps no-one uses books anymore? I'll be away from email until Monday. I'll be in Madison --- perhaps I'll bump into some of you. Best to all, R.

1 0

[math-fun] Erdos conjecture on arithmetic progressions
by Charles Greathouse 30 Jul '12

30 Jul '12

Erdos conjectured that if the reciprocal sum of a sequence of positive integers diverges, then it contains k-arithmetic progressions for any k. It is known* that this implies the existence of constants N_k such that if the sum of the reciprocals of a sequence is greater than N_k, the sequence contains a k-arithmetic progression. Of all subsets of positive integers not containing a 3-arithmetic progression, what's the highest reciprocal sum you can find? The greedy method gets over 3.0078. * Gerver 1977, Brown & Freedman 1987 in a stronger form Charles Greathouse Analyst/Programmer Case Western Reserve University

6 6

[math-fun] NYTimes: Is Algebra Necessary?
by Henry Baker 29 Jul '12

29 Jul '12

FYI -- I'm not fond of the way algebra is taught in schools, but I think it would be insane not to teach it at all. The comments on this article are interesting. http://www.nytimes.com/2012/07/29/opinion/sunday/is-algebra-necessary.html Opinion Is Algebra Necessary? Adam Hayes By ANDREW HACKER Published: July 28, 2012 A TYPICAL American school day finds some six million high school students and two million college freshmen struggling with algebra. In both high school and college, all too many students are expected to fail. Why do we subject American students to this ordeal? IÂve found myself moving toward the strong view that we shouldnÂt. My question extends beyond algebra and applies more broadly to the usual mathematics sequence, from geometry through calculus. State regents and legislators Â and much of the public Â take it as self-evident that every young person should be made to master polynomial functions and parametric equations. There are many defenses of algebra and the virtue of learning it. Most of them sound reasonable on first hearing; many of them I once accepted. But the more I examine them, the clearer it seems that they are largely or wholly wrong Â unsupported by research or evidence, or based on wishful logic. (IÂm not talking about quantitative skills, critical for informed citizenship and personal finance, but a very different ballgame.) This debate matters. Making mathematics mandatory prevents us from discovering and developing young talent. In the interest of maintaining rigor, weÂre actually depleting our pool of brainpower. I say this as a writer and social scientist whose work relies heavily on the use of numbers. My aim is not to spare students from a difficult subject, but to call attention to the real problems we are causing by misdirecting precious resources. The toll mathematics takes begins early. To our nationÂs shame, one in four ninth graders fail to finish high school. In South Carolina, 34 percent fell away in 2008-9, according to national data released last year; for Nevada, it was 45 percent. Most of the educators IÂve talked with cite algebra as the major academic reason. Shirley Bagwell, a longtime Tennessee teacher, warns that Âto expect all students to master algebra will cause more students to drop out.Â For those who stay in school, there are often Âexit exams,Â almost all of which contain an algebra component. In Oklahoma, 33 percent failed to pass last year, as did 35 percent in West Virginia. Algebra is an onerous stumbling block for all kinds of students: disadvantaged and affluent, black and white. In New Mexico, 43 percent of white students fell below Âproficient,Â along with 39 percent in Tennessee. Even well-endowed schools have otherwise talented students who are impeded by algebra, to say nothing of calculus and trigonometry. CaliforniaÂs two university systems, for instance, consider applications only from students who have taken three years of mathematics and in that way exclude many applicants who might excel in fields like art or history. Community college students face an equally prohibitive mathematics wall. A study of two-year schools found that fewer than a quarter of their entrants passed the algebra classes they were required to take. ÂThere are students taking these courses three, four, five times,Â says Barbara Bonham of Appalachian State University. While some ultimately pass, she adds, Âmany drop out.Â Another dropout statistic should cause equal chagrin. Of all who embark on higher education, only 58 percent end up with bachelorÂs degrees. The main impediment to graduation: freshman math. The City University of New York, where I have taught since 1971, found that 57 percent of its students didnÂt pass its mandated algebra course. The depressing conclusion of a faculty report: Âfailing math at all levels affects retention more than any other academic factor.Â A national sample of transcripts found mathematics had twice as many FÂs and DÂs compared as other subjects. Nor will just passing grades suffice. Many colleges seek to raise their status by setting a high mathematics bar. Hence, they look for 700 on the math section of the SAT, a height attained in 2009 by only 9 percent of men and 4 percent of women. And itÂs not just Ivy League colleges that do this: at schools like Vanderbilt, Rice and Washington University in St. Louis, applicants had best be legacies or athletes if they have scored less than 700 on their math SATs. ItÂs true that students in Finland, South Korea and Canada score better on mathematics tests. But itÂs their perseverance, not their classroom algebra, that fits them for demanding jobs. Nor is it clear that the math we learn in the classroom has any relation to the quantitative reasoning we need on the job. John P. Smith III, an educational psychologist at Michigan State University who has studied math education, has found that Âmathematical reasoning in workplaces differs markedly from the algorithms taught in school.Â Even in jobs that rely on so-called STEM credentials Â science, technology, engineering, math Â considerable training occurs after hiring, including the kinds of computations that will be required. Toyota, for example, recently chose to locate a plant in a remote Mississippi county, even though its schools are far from stellar. It works with a nearby community college, which has tailored classes in Âmachine tool mathematics.Â That sort of collaboration has long undergirded German apprenticeship programs. I fully concur that high-tech knowledge is needed to sustain an advanced industrial economy. But weÂre deluding ourselves if we believe the solution is largely academic. A skeptic might argue that, even if our current mathematics education discourages large numbers of students, math itself isnÂt to blame. IsnÂt this discipline a critical part of education, providing quantitative tools and honing conceptual abilities that are indispensable Â especially in our high tech age? In fact, we hear it argued that we have a shortage of graduates with STEM credentials. Of course, people should learn basic numerical skills: decimals, ratios and estimating, sharpened by a good grounding in arithmetic. But a definitive analysis by the Georgetown Center on Education and the Workforce forecasts that in the decade ahead a mere 5 percent of entry-level workers will need to be proficient in algebra or above. And if there is a shortage of STEM graduates, an equally crucial issue is how many available positions there are for men and women with these skills. A January 2012 analysis from the Georgetown center found 7.5 percent unemployment for engineering graduates and 8.2 percent among computer scientists. Peter Braunfeld of the University of Illinois tells his students, ÂOur civilization would collapse without mathematics.Â HeÂs absolutely right. Algebraic algorithms underpin animated movies, investment strategies and airline ticket prices. And we need people to understand how those things work and to advance our frontiers. Quantitative literacy clearly is useful in weighing all manner of public policies, from the Affordable Care Act, to the costs and benefits of environmental regulation, to the impact of climate change. Being able to detect and identify ideology at work behind the numbers is of obvious use. Ours is fast becoming a statistical age, which raises the bar for informed citizenship. What is needed is not textbook formulas but greater understanding of where various numbers come from, and what they actually convey. What of the claim that mathematics sharpens our minds and makes us more intellectually adept as individuals and a citizen body? ItÂs true that mathematics requires mental exertion. But thereÂs no evidence that being able to prove (xÂ² + yÂ²)Â² = (xÂ² - yÂ²)Â² + (2xy)Â² leads to more credible political opinions or social analysis. Many of those who struggled through a traditional math regimen feel that doing so annealed their character. This may or may not speak to the fact that institutions and occupations often install prerequisites just to look rigorous Â hardly a rational justification for maintaining so many mathematics mandates. Certification programs for veterinary technicians require algebra, although none of the graduates IÂve met have ever used it in diagnosing or treating their patients. Medical schools like Harvard and Johns Hopkins demand calculus of all their applicants, even if it doesnÂt figure in the clinical curriculum, let alone in subsequent practice. Mathematics is used as a hoop, a badge, a totem to impress outsiders and elevate a professionÂs status. ItÂs not hard to understand why Caltech and M.I.T. want everyone to be proficient in mathematics. But itÂs not easy to see why potential poets and philosophers face a lofty mathematics bar. Demanding algebra across the board actually skews a student body, not necessarily for the better. I WANT to end on a positive note. Mathematics, both pure and applied, is integral to our civilization, whether the realm is aesthetic or electronic. But for most adults, it is more feared or revered than understood. ItÂs clear that requiring algebra for everyone has not increased our appreciation of a calling someone once called Âthe poetry of the universe.Â (How many college graduates remember what FermatÂs dilemma was all about?) Instead of investing so much of our academic energy in a subject that blocks further attainment for much of our population, I propose that we start thinking about alternatives. Thus mathematics teachers at every level could create exciting courses in what I call Âcitizen statistics.Â This would not be a backdoor version of algebra, as in the Advanced Placement syllabus. Nor would it focus on equations used by scholars when they write for one another. Instead, it would familiarize students with the kinds of numbers that describe and delineate our personal and public lives. It could, for example, teach students how the Consumer Price Index is computed, what is included and how each item in the index is weighted Â and include discussion about which items should be included and what weights they should be given. This need not involve dumbing down. Researching the reliability of numbers can be as demanding as geometry. More and more colleges are requiring courses in Âquantitative reasoning.Â In fact, we should be starting that in kindergarten. I hope that mathematics departments can also create courses in the history and philosophy of their discipline, as well as its applications in early cultures. Why not mathematics in art and music Â even poetry Â along with its role in assorted sciences? The aim would be to treat mathematics as a liberal art, making it as accessible and welcoming as sculpture or ballet. If we rethink how the discipline is conceived, word will get around and math enrollments are bound to rise. It can only help. Of the 1.7 million bachelorÂs degrees awarded in 2010, only 15,396 Â less than 1 percent Â were in mathematics. IÂve observed a host of high school and college classes, from Michigan to Mississippi, and have been impressed by conscientious teaching and dutiful students. IÂll grant that with an outpouring of resources, we could reclaim many dropouts and help them get through quadratic equations. But that would misuse teaching talent and student effort. It would be far better to reduce, not expand, the mathematics we ask young people to imbibe. (That said, I do not advocate vocational tracks for students considered, almost always unfairly, as less studious.) Yes, young people should learn to read and write and do long division, whether they want to or not. But there is no reason to force them to grasp vectorial angles and discontinuous functions. Think of math as a huge boulder we make everyone pull, without assessing what all this pain achieves. So why require it, without alternatives or exceptions? Thus far I havenÂt found a compelling answer.

3 2

Re: [math-fun] peculiar 2x2 products
by Bill Gosper 23 Jul '12

23 Jul '12

HGB>Speaking information theoretically, if the matrix elements are integers, wouldn't they _have_ to blow up in order to converge to a real number with supposedly infinite information? Certainly, but the matrix below has rational function entries. Unfortunately, in your copy, the oo and /= characters got turned into âˆž and â‰ . At 12:01 AM 7/23/2012, Bill Gosper wrote:>We're used to evaluating an infinite continued fraction as a limit of ratios of elements of a running product of 2x2 matrices, while the actual elements of the product blow up.>>E.g., for 4 arctan(1)>>In[1153]:= FoldList[Dot, {{0, 4}, {1, 0}}, Table[{{2*k - 1, k^2}, {1, 0}}, {k, 9}]]>>Out[1153]= {{{0, 4}, {1, 0}}, {{4, 0}, {1, 1}}, {{12, 16}, {4, 4}}, {{76, 108}, {24, 36}}, {{640, 1216}, {204, 384}}, {{6976, 16000}, {2220, 5100}}, {{92736, 251136}, {29520, 79920}}, {{1456704, 4544064}, {463680, 1446480}}, {{26394624, 93229056}, {8401680, 29675520}}, {{541937664, 2137964544}, {172504080, 680536080}}}>>In[1154]:= N[Divide @@ # & /@ Rest[%]]>>Out[1154]= {{4., 0.}, {3., 4.}, {3.16667, 3.}, {3.13725, 3.16667}, {3.14234, 3.13725}, {3.14146, 3.14234}, {3.14161, 3.14146}, {3.14159, 3.14161}, {3.14159, 3.14159}}>>But here's an infinite matrix product where the elements converge individually:>>Product[MatrixForm[{{(3*a^2 - 3*a*j + j^2)/(j*(-1 - 3*a + b + j)), a^3/(j*(-1 - 3*a + b + j))},> {1/j, 1}}], {j, b, âˆž}] ==> MatrixForm[{{"?", (a^3*HypergeometricPFQ[{1 + a, 1 + a, 1 + a}, {1 + b, 1 + b}, 1])/b^2},> {"?", HypergeometricPFQ[{a, a, a}, {b, b}, 1]}}]>>(Product means matrix product, not Mathematica's product.)>>The ?s converge, too, but I don't know to what.>>Note that the determinant â‰ 1, but approaches 1 in a very particular way.>>It's not obvious to me how to scale a matrix to make its running product converge elementwise.>>I actually have an elementwise convergent 2x2 product for the general 3F2[1], Duh, you can compute *any* sum as prod({rn,an};{0,1}). My 2x2 computes 3F2[a,b,c;d,e] and 3F2[a+1,b+1,c+1;d+1,e+1] in separate elements. It is desirable to compute such contiguous pairs, as *all* other contiguous 3F2[1] (or 2F1[z]) may be computed w/o additional infinite series. [...] --rwg

1 0

Re: [math-fun] peculiar 2x2 products
by Henry Baker 23 Jul '12

23 Jul '12

Speaking information theoretically, if the matrix elements are integers, wouldn't they _have_ to blow up in order to converge to a real number with supposedly infinite information? At 12:01 AM 7/23/2012, Bill Gosper wrote: >We're used to evaluating an infinite continued fraction as a limit of ratios of elements of a running product of 2x2 matrices, while the actual elements of the product blow up. > >E.g., for 4 arctan(1) > >In[1153]:= FoldList[Dot, {{0, 4}, {1, 0}}, Table[{{2*k - 1, k^2}, {1, 0}}, {k, 9}]] > >Out[1153]= {{{0, 4}, {1, 0}}, {{4, 0}, {1, 1}}, {{12, 16}, {4, 4}}, {{76, 108}, {24, 36}}, {{640, 1216}, {204, 384}}, {{6976, 16000}, {2220, 5100}}, {{92736, 251136}, {29520, 79920}}, {{1456704, 4544064}, {463680, 1446480}}, {{26394624, 93229056}, {8401680, 29675520}}, {{541937664, 2137964544}, {172504080, 680536080}}} > >In[1154]:= N[Divide @@ # & /@ Rest[%]] > >Out[1154]= {{4., 0.}, {3., 4.}, {3.16667, 3.}, {3.13725, 3.16667}, {3.14234, 3.13725}, {3.14146, 3.14234}, {3.14161, 3.14146}, {3.14159, 3.14161}, {3.14159, 3.14159}} > >But here's an infinite matrix product where the elements converge individually: > >Product[MatrixForm[{{(3*a^2 - 3*a*j + j^2)/(j*(-1 - 3*a + b + j)), a^3/(j*(-1 - 3*a + b + j))}, > {1/j, 1}}], {j, b, â}] == > MatrixForm[{{"?", (a^3*HypergeometricPFQ[{1 + a, 1 + a, 1 + a}, {1 + b, 1 + b}, 1])/b^2}, > {"?", HypergeometricPFQ[{a, a, a}, {b, b}, 1]}}] > >(Product means matrix product, not Mathematica's product.) > >The ?s converge, too, but I don't know to what. > >Note that the determinant Ã¢Â 1, but approaches 1 in a very particular way. > >It's not obvious to me how to scale a matrix to make its running product converge elementwise. > >I actually have an elementwise convergent 2x2 product for the general 3F2[1], but idiot Mathematica 8.04 takes (literally?) forever to convert a simple sum of Pochhammers to a 3F2. > >--rwg

1 0

[math-fun] Coxeter group terminology
by Mike Stay 22 Jul '12

22 Jul '12

Is there a term for the operation of adding another generator to a Coxeter group? It has to be an involution, of course, but shouldn't satisfy any other relations---i.e. in the Coxeter matrix presentation, the off-diagonal elements of the new row and column should be infinity. This turns up in the context of continued fractions. The infinite dihedral group acts on the real line translating by 2 and negating. If you do this operation, you get the (inf, inf, inf) hyperbolic triangle group, which acts on the upper half-plane in the same way plus geometric inversion, which is a hyperbolic reflection. If the three generators are s,t,u, then the continued fraction [a, b, c, ...] means the element ... u s^c u t^b u s^a. It works abstractly on any Coxeter group, but since not all Coxeter groups have linear representations, it doesn't always have a nice geometric meaning. -- Mike Stay - metaweta(a)gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com

2 2

Re: [math-fun] what is known about the squareful Fibonacci numbers?
by Robert Munafo 22 Jul '12

22 Jul '12

So I'm guessing we have at least three sequences under discussion (or at least confusion): "Squareful numbers are numbers for which at least one prime factor exponent is [exactly] 2" which gives A038109: 4, 9, 12, 18, 20, 25, 28, 36, 44, 45, 49, 50, 52, 60, ... "squarefull numbers [are] numbers for which each prime factor exponent is at least 2" which gives A001694: 1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 72, 81, 100, ... Mike Speciner's "at least one prime factor exponent is at least 2" which gives A013929: 4, 8, 9, 12, 16, 18, 20, 24, 25, 27, 28, 32, 36, 40, ... On 7/16/12, Mike Speciner <ms(a)alum.mit.edu> wrote: > Under that definition, 8 would not be squareful. From the example in the > referenced webpage, I would guess it's supposed to say "at least one > prime factor exponent is at least 2" > On 7/16/2012 9:37 AM, Wouter Meeussen wrote: >> http://oeis.org/wiki/Squareful_numbers >> >> "Squareful numbers are numbers for which at least one prime factor >> exponent is 2, thus are not squarefree numbers, not to be confused >> with squarefull numbers, numbers for which each prime factor exponent >> is at least 2. " >> -----Original Message----- From: Robert Munafo >> Richard Guy probably uses "UPINT" [to refer] to "Unsolved Problems In >> Number Theory", a book that perhaps defines "powerful number" in a >> similar way to the following: >> >> An integer m such that if p | m, then p^2 | m, is called a powerful >> number. -- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com

2 1

Re: [math-fun] Invariant metrics
by Dan Asimov 22 Jul '12

22 Jul '12

Yes, that's one way to see it. (But it takes as known something very close to the answer.) A direct way is to note that the family of all lines parallel to a given line forms a line itself perpendicular to them. After you rotate such a parallel family 180 degrees you're back to the original line. So the perpendicular lines form a Moebius band, the space of all lines in the plane. --Dan Adam wrote: << I wrote: << . . . It's fun to verify (and well-known) that topologically, this space [of all lines in the plane] is the (open) Moebius band. . . . >> Indeed. The projective lines (ax + by + cz = 0), where (a,b,c) is not (0,0,0) and (la,lb,lc) = (a,b,c) form a projective plane P^2, which is a consequence of projective duality. . . . >> ________________________________________________________________________________________ It goes without saying that .

1 0

[math-fun] Wechsler's Disk problem solved via linear program (clarification)
by Warren Smith 22 Jul '12

22 Jul '12

>> --Given a set of points P and another Q in the plane, >> the question of whether a disk D exists with P inside D but Q outside D, >> is a linear programming problem, for each point xy >> we demand >> a*x*x+b*x*y+a*y*y+c*x+d*y+e > 0 >> (or <0) > > + b*x*y ??? > > I don't think so; the general equation of a circle is: > > A(x^2 + y^2) + Bx + Cy + D = 0 > > which forms a projective 3-space of possible circles. --sorry, typo, I should have omitted the b*x*y term by making b=0. To correct+clarify: If you do a*x*x+b*x*y+c*y*y+d*x+e*y+f then by the linear programming approach you can decide separability of point set P from Q by any conic section (i.e. quadratic algebraic curve). Indeed in any fixed #dimensions D for any fixed-degree polynomial(x1,x2,...,xN) you can decide separability of two point sets via the algebraic surface defined by such a polynomial, via linear programming in the same kind of way. So anyway, back to the circle case, a*(x-x0)^2 + a*(y-y0)^2 - 1 = a*x*x+a*y*y-2*a*x0*x-2*a*y0*y + x0^2 + y0^2 - 1 (left hand side >0 or <0 if x,y outside/inside circle with center x0,y0 and radius^2=1/a) can be rewritten as a*x*x + a*y*y + b*x + c*y + d so the question of whether a separating circle exists is just a linear program in (a,b,c,d) to tell if (a,b,c,d) exist which cause the above expression to be >0 for points in one set and <0 for points in the other. And by solving this linear program you can find such a circle, indeed you can find the "best" such circle as I'd described. (You could also demand a>0 as an additional inequality in the linear program; this demand should be unnecessary but might lead to a speedup.) > It looks like > you have a 4-parameter family of conic sections instead. But, yes, > your idea does generalise to determining whether there exists an > algebraic curve (within a particular pencil) which separates the > sets P and Q. --yes. But as I said above, it also works just for circles. > Returning to my original algorithm (just for discs), it is sufficient to > check the interior points on the perimeter of the polyomino (for > the same reason it is sufficient to check the exterior points > adjacent to them). This reduces the running time from O(n^2) to > O(n) for confirming a disc polyomino -- asymptotically, the best > possible. --the linear programming method will also run in optimal O(n) time. Specifically, it is known that solving a linear program with V variables and N inequalities can be done in O(N) time if V is fixed. I think this was first shown by Nimrod Megiddo 20-30 years ago but there are now many known algorithms to do it, the nicest being randomized algorithms, including ones by K.Clarkson and by R.Seidel.

1 0

Re: [math-fun] Wechsler disk problem
by Dan Asimov 22 Jul '12

22 Jul '12

I'd say that since a circle in R^2 is determined by its center and radius, the space of circles is naturally isometric to R^2 x (0,oo). For the general equation below to give a circle, it must hold that B^2 + C^2 > 4AD. So the space of coefficients (mod giving the same circle) isn't quite P^3. --Dan << . . . the general equation of a circle is: A(x^2 + y^2) + Bx + Cy + D = 0 which forms a projective 3-space of possible circles. >> ________________________________________________________________________________________ It goes without saying that .

2 3