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[math-fun] Puzzle re maximum perimeter of triangle inscribed in ellipse
by Dan Asimov 07 Sep '07

07 Sep '07

I recently heard an interesting question via the grapevine: Given an ellipse E in the plane, consider any point A of E. Then two other points B,C of E can be found so that the perimeter of the triangle ABC is maximized. Denote this maximum perimeter by P(A). Prove or disprove: P(A) is necessarily independent of A. --Dan

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Re: OOPS, Re: [math-fun] Algebra I (RiP)
by Henry Baker 07 Sep '07

07 Sep '07

At 09:59 AM 8/23/2007, Dan Asimov wrote: >This should read: > >This is because newspaper articles et al. are well known for >royally screwing up clarity in such matters. (They will commonly >use a phrase like "150% more" whether they mean 3/2 as much or >5/2 as much.) This reminds me of a (possibly apocryphal) tale of a Congressional hearing on some very difficult technology. The scientist said that they were able to get some quantity to 10^10, but the goal required 10^20. Whereupon the Congressman said "then we're halfway there, already!".

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[math-fun] Algebra I (RiP)
by Bill Gosper 07 Sep '07

07 Sep '07

I said >PPS, my former student aced his tests but skipped his homework, and is now repeating Algebra I. As soon as I can photocopy a few pages, I'll share some of the stupendous howlers I found on a quick perusal of his new text. Amazingly, it's *not* Prentice-Hall ... Key Curriculum Press, Discovering Algebra, An Ivestigative Approach, by Murdock, Kamischke, & Kamischke. Along with a committee filling two columns on page ii, including five Multicultural and Equity Reviewers, two Social Sciences and Humanities reviewers, and, get this, three Scientific Content Reviewers, and one Accuracy Checkers (sic!). Seven hundred pages of irrelevant color pictures (e.g., Elvis impersonators) with inane captions (e.g., "Dead skin cells are one of the components of dust."). The people I know who are good at algebra had books with *zero* pictures, and few, if any, black and white graphs. Those books weighed maybe five times less, and contained way more than twice the algebra. (I learned it from my grandmother's.) If I were an aspiring algebra student, I would be deeply offended by this book for diluting the subject beyond recognition, as if algebra were so difficult and distasteful as to require huge doses of eye and brain candy, and for forcing me to lug to school hundreds of stupid pictures far less interesting than an equally heavy stack of magazines of my own choosing. And I came here to learn algebra, not to be cured of racism and sexism. And as a teacher faced with this book, I would invite any "student" who wished he was elsewhere to place his favorite magazine inside his open book. The pictures would be at least as relevant. I'll bet that even those who get an A in the course will be completely unable to apply algebra to any problem that appears to differ significantly from those "investigated" in the book. Howlers: In a section devoted to wind chill, p227, photo: snow blowing across highway across a barren plain. Caption: High wind speeds in Saskatchewan, Canada, drop temperatures below freezing. Just to make sure you understand, p228, photo: orange tents and heavily clad people on an expanse of snow. Caption: The wind chill factor drops temperatures well below zero in Antarctica. Wow, I had no idea how many of those poor Katrina victims must have frozen to death. According to this table, a temperature of 5F with a 20mph wind gives a chill of -31. So the temperature of an object at 0F should decrease rapidly. Think of the starvation we could avert by giving each victim a cheap refrigerator--a fan in a box! And those stupid citrus growers actually think they can avert frost damage with giant blowers! And imagine the wind chill at 40 below and 1000mph. It must be dang near minus infinity. And yet those idiot Europeans built the Concorde out of titanium because they thought aluminum would melt! Geez I'm glad I stayed in school. Oops, here in the main text it says wind chill is only how cold it *feels*. How do you measure how cold it feels? Does snow in Saskatchewan stay frozen because the wind makes it feel cold? In a section on radioactive decay, photo: human skull. Caption: Archaeologists can approximate the age of artifacts with carbon dating. This process uses the rate of radioactive decay of cabon-14. Carbon is found in all living things, so the amount left in a bone, for example, is an indicator of the bone's age. This is a plastic casting of a skull found in 1997 in Richland Washington. Carbon dating has dated this skull as 9200 years old. It's less frightening now that I know it's plastic. I wonder if the Scientific Content Reviewers and the Accuracy Checker used their real names. This skull, by the way, is of the infamous Kennewick Man, http://www.harbornet.com/folks/theedrich/hive/Kennewic.htm . Apparently, someone slipped it past the Multicultural and Equity Reviewers. In the fractals section, p19(!): As the Koch curve develops [For some reason, the book studiously avoids the name Snowflake], segment length decreases as the number of segments increases. As you draw higher stages, the length of individual segments approaches, or gets closer and closer to, what number? What number does the number of line segments approach? ZAP! Without even knowing it, the student has been infected with the poisonous notion that infinity is a number. And the poisonous notion that this line segment recursion actually defines which points are on the fractal. There is, of course, no mention of dimension. As a student, I would wonder how Mandelbrot and Sierpinski got so much credit for "discovering" designs that people have doubtless doodled since the invention of writing. The authors could have promised a discussion of dimension in a later chapter, although it might have required a few thousand more intervening pages of photographs. E.g., antique photo of seated woman. Caption: Maria Mitchell (1818-1889) was the first professional woman astronomer in the United States. Relevance: nil. But the Multicultural and Equity Reviewers must have detected a patch of inequity. I would think that even female students would begin to doubt the competence of women, if they require such heavy-handed propaganda. Side by side on p15, Caption 1: Because a coastline, like a fractal curve, is winding and irregular, it is not possible to measure its length accurately. The Koch curve can be used as a model of a coastline. [...] Photo: Aerial shot of a pair of beautifully smooth, crescent beaches! Then, to perfect the confusion, Caption 2: At later stages, the Koch curve looks smoother and smoother.[!] But if you magnify a section at a later stage, it is just as jagged as at Stage 1. So much for self-similarity. Photo: a minority adolescent holding a magnifying glass over a Snowflake approximation to reveal a part of the initial hexagram (Heaven forbid that I say Star of David), but with HUGE, FAT line segments! Line segments are the bane to understanding fractals. A properly drawn line segment is invisible! (Except as a color boundary.) p72, problem 9: The graph at right is a hexagon [True] whose vertices are seven[!] ordered pairs. Two of the points are (3,0) and (1.5,2.6). The hexagon is centered at the origin. a. What are the coordinates of the other points? HUH?? Assuming regular, even though it isn't? There are three kinds of mathematicians: ... To sum up, p295, Caption: Wall Drug is a landmark in South Dakota. The store's fame began during the 1930s--the Great Depression--when it offered free ice water to travelers. --rwg (Confidential to you-know-who-you-are: There is only one b in Fibonacci.) --------------------------------- Take the Internet to Go: Yahoo!Go puts the Internet in your pocket: mail, news, photos & more.

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[math-fun] pi-3, 1/pi
by Bill Gosper 07 Sep '07

07 Sep '07

I said >I know some kids who go to fancy private schools--maybe they'll let me sneak a peek at their texts. I just peeked at a *public* (Palo Alto, vs Los Altos) 7th grader's Algebra 1 text, and it's ten times better. The pictures are relevant, there is no crediting an affirmative action committee, and the only braino I found was the caption of a photo of Saturn claiming that the reason it has an average surface temperature is that it gives off more heat than the sun provides.(!) The book factors polynomials and derives the quadratic formula. I assume it's "honors algebra" instead of "dishonored algebra", but there's nothing on the respective covers suggesting anything except they're both for California standards. I guess this is to protect the self-esteem of the poor slobs carrying around the dishonorable versions. (I assume there's an honors course at Los Altos that my young friend wasn't offered.) Enough of that. A Bruce Berndt question about 1/pi series prompted me to try reconstructing my derivation (in the ancient ftp://publications.ai.mit.edu/ai-publications/pdf/AIM-304.pdf) of Ramanujan's dyadic rational inf 2 k 3 ==== (42 k + 5) ( ) 1 \ k -- = > ----------------- pi / 12 k + 4 ==== 2 k = 0 F[1/64] series directly from Gauss's 2F1[1]. But I spazzed and got inf 2 i ==== (2 i + 1) (6 i + 7) (7 i + 9) ( ) \ i pi - 3 = 8 > ----------------------------------- / 2 2 4 i + 4 2 ==== (2 i + 2) (4 i + 5) ( ) i = 0 2 i + 2 instead. As usual, pFq notation is ridiculous, (c300) make1hyper(%) (d300) %pi - 3 = 3 3 3 13 16 7 7 9 9 7 9 1 7 hyper_f ([1, 1, -, -, -, --, --], [-, -, -, -, -, -], --) 7, 6 2 2 2 6 7 4 4 4 4 6 7 64 -------------------------------------------------------------- 50 (c301) dfloat(%) (d301) 0.14159265358979d0 = 0.14159265358979d0 but at least the parameters are rational. I'm almost sure I found such in the 70s, but failed several times to rediscover it. Note how economical is the equivalent matrix product: (c302) factor(d170) [ 3 ] [ k (2 k + 1) (6 k + 1) (7 k + 2) ] [ ----------------------- ------------------- ] (d302) [ 2 2 450 ] [ 2 (4 k + 3) (4 k + 5) ] [ ] [ 0 1 ] (c303) prud(%,k,1,9.0d0) [ 2.60286523404292d-19 0.14159265358979d0 ] (d303) [ ] [ 0.0d0 1.0d0 ] I'm afraid this comes a few months too late for another young friend who needed a pi-3 series for some kind of programming contest. Perhaps he should join this list, if we're done with prom(iscuity.-) Note that if we squeeze out one more term, we get an exact series for pi - 3.14: [ 3 ] [ k (2 k + 1) (6 k + 1) (7 k + 2) ] [ ----------------------- ------------------- ] (d304) [ 2 2 132300 ] [ 2 (4 k + 3) (4 k + 5) ] [ ] [ 0 1 ] (c305) prud(%,k,2,9.0d0) [ 7.6524237880862d-17 0.00159265358979d0 ] (d305) [ ] [ 0.0d0 1.0d0 ] Here is the real indictment of pFq notation. From 2F1[a,-a;1] = sin(pi a)/(pi a) I derived the more general F[a|1/64], which puts a cubic(a,k) in the UR of d302, but intractable, needless surds in the pFq form. Correcting my derivation spaz and retrying, inf 3 2 i + 1 3 ==== (i + 1) (14 i - 3) ( ) 1 \ i -- = 3 > ------------------------------. pi / 2 12 i + 2 ==== (2 i - 3) (2 i - 1) 2 i = 0 It's 1/pi, dyadic, and F[1/64]. But it's weird. Try again. inf 2 k - 1 3 ==== (42 k + 5) ( ) 1 \ k 35 -- = > --------------------- - --. pi / 12 k + 1 16 ==== 2 k = 0 Finally. This is equivalent to Ramanujan's. (Nice exercise.) Now I can apply the exact same process to Heine's series instead of Gauss's and get a q-extension (maybe Mizan knows a prettier one), 3 2 oo 2 6k 2k+1 3 6k+1 6k+1 6k+3 ==== (q; q ) q ((q + 1) (1 - q ) - q (1 - q )) \ k > ----------------------------------------------------------- / 3 3 ==== 4 4 2 k = 0 (q ; q ) (- q; q ) k k + 1 2 2 (q; q ) 1/4 2 oo q (1 - q ) = ---------- = -------------. 2 pi(q) 2 2 (q ; q ) oo Test: (c479) taylor(makeprod(lhs(%)),q,0,24) 2 3 4 5 6 7 8 9 (d479)/T/ 1 - 2 q + 3 q - 6 q + 11 q - 18 q + 28 q - 44 q + 69 q - 104 q 10 11 12 13 14 15 16 + 152 q - 222 q + 323 q - 460 q + 645 q - 902 q + 1254 q 17 18 19 20 21 22 23 - 1722 q + 2343 q - 3174 q + 4278 q - 5722 q + 7601 q - 10056 q 24 2 3 4 5 6 7 + 13250 q + . . . = 1 - 2 q + 3 q - 6 q + 11 q - 18 q + 28 q - 44 q 8 9 10 11 12 13 14 15 + 69 q - 104 q + 152 q - 222 q + 323 q - 460 q + 645 q - 902 q 16 17 18 19 20 21 22 + 1254 q - 1722 q + 2343 q - 3174 q + 4278 q - 5722 q + 7601 q 23 24 - 10056 q + 13250 q + . . . (c480) 1/lhs(%) 2 3 4 6 7 9 10 11 12 (d480)/T/ 1 + 2 q + q + 2 q + 2 q + 3 q + 2 q + 2 q + 2 q + 2 q + q 13 15 16 18 20 21 22 24 + 2 q + 2 q + 4 q + 2 q + q + 4 q + 2 q + 2 q + . . . which will obey nonlinear, three term relations among pi(q), pi(q^a), and pi(q^b), for rational a and b. (Actually, we want Pi(q):=pi(q)/(1-q^2), which changes q^(6k^2) to q^(6(k^2-1/24)) in our series.) --rwg --------------------------------- Building a website is a piece of cake. Yahoo! Small Business gives you all the tools to get online.

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[math-fun] Constructivist logic
by Jason Holt 07 Sep '07

07 Sep '07

I recently went through my approximately 3-year cycle of wondering what the significance of the Halting problem is all about and wrote up my thoughts here: http://credentiality2.blogspot.com/index.html (I include the link for completeness, but it's not the point of this message, and I can't claim that it's in any way worth the time it takes to read it.) My old CS theory prof tells me I take offense to the notion of uncomputability because I prefer the constructivist school of logic: http://en.wikipedia.org/wiki/Constructivism_(mathematics) I poked around wikipedia a bit and have to agree that it does appeal to me. In particular, I like the idea that the reals are computable and perhaps thus not 200% more infinite than the naturals as I had been lead to believe, but rather 100% less different in cardinality, there being no more fewer reals than naturals than there are naturals and integers. Excuse me, I think I need to sit down for a moment. Anyway, it seems that Constructionism has not yet converted all the heathen masses, and I'm curious what this mass of heathens thinks of it. In particular, I'm curious what it means for diagonalization proofs, which have always seemed suspicious to me, doing magical things that only work out at infinity.

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[math-fun] Violating transitivity
by Guy Haworth 07 Sep '07

07 Sep '07

Isn't there something in Arrow's work on Voting Systems that shows violations of transitivity. I think there was at least a Scientific American article on Arrow's results - way back. Guy Haworth

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[math-fun] Causation and correlation
by Mike Stay 05 Sep '07

05 Sep '07

http://ghill.customer.netspace.net.au/correlation-causation.html -- Mike Stay metaweta(a)gmail.com http://math.ucr.edu/~mike

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[math-fun] Melzak reprinted !!
by Eugene Salamin 05 Sep '07

05 Sep '07

Dover has reprinted Melzak's "Companion to Concrete Mathematics". http://store.doverpublications.com/0486457818.html Gene ____________________________________________________________________________________Ready for the edge of your seat? Check out tonight's top picks on Yahoo! TV. http://tv.yahoo.com/

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[math-fun] A060900 formula
by David Wilson 05 Sep '07

05 Sep '07

A060900 is the number of unrestricted walks with n edges starting at the origin on the 3/8 square lattice. There is a recursive formula there, though I don't remember putting it there, I could have. Given the nature of the sequence and the number of matching elements, I'm sure the formula is correct, but I admit inability to prove it. However, I have been told the formula is conjectural, and if there is not a proof, it should be marked as such. It is just seems aggravating that I can't come up with one. Anyway, thought the seqfans might want to take a look at it, I know my betters are out there. Similarly, can we come up with a formula for A060897 or A060898? Can we confirm the formula for A060898?

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[math-fun] gefundenes Fressen 2
by wouter meeussen 04 Sep '07

04 Sep '07

consider the table of necklace polynomials, n beads, c<=n colours, substitute colour k by Exp[2 Pi I k/c], Table[NecklacePolynomial[n,Exp[2 I Pi Range[c]/c],Cyclic]//FullSimplify,{n,13},{c,n}] and find: {1}, {1,1}, {1,0,2}, {1,2,0,2}, {1,0,0,0,4}, {1,2,4,0,0,2}, {1,0,0,0,0,0,6}, {1,4,0,6,0,0,0,4}, {1,0,8,0,0,0,0,0,6}, {1,4,0,0,12,0,0,0,0,4}, {1,0,0,0,0,0,0,0,0,0,10}, {1,8,16,12,0,8,0,0,0,0,0,4}, {1,0,0,0,0,0,0,0,0,0,0,0,12}, and that might very well be: if c|n then Sum(d|(n/c), phi(c d) d^(n/c/d) /n) else 0 or, Table[If[MemberQ[Divisors[n],c], Fold[ #1 + EulerPhi[c #2]c^(n/c/#2)/(n) &, 0, Divisors[n/c]],0 ],{n,20},{c,n}] which is nice too, Wouter. earlier: ----- Original Message ----- From: "wouter meeussen" <wouter.meeussen(a)pandora.be> To: "math-fun" <math-fun(a)mailman.xmission.com> Sent: Saturday, July 28, 2007 9:57 PM Subject: [math-fun] gefundenes Fressen (a cheap find) > Imagine a necklace with n beads of up to n colours, > say n=3 beads made of colours x[1], x[2] and x[3], > then these are counted by: > > Table[NecklacePolynomial[n,Array[x,n],Cyclic],{n,3,3}] > {x[1]^3 + x[1]^2*x[2] + x[1]*x[2]^2 + x[2]^3 + x[1]^2*x[3] + > 2*x[1]*x[2]*x[3] + > x[2]^2*x[3] + x[1]*x[3]^2 + x[2]*x[3]^2 + x[3]^3} > > now, change the colours into complex roots of 1 > x[k] -> E^(2I Pi k/n) > so that they do a 'complex cancellation' on the unit circle, > and, hey presto, the whole caboodle collapses to EulerPhi[n]: > > EulerPhi[n] == NecklacePolynomial[n,E^(2I Pi Range[n]/n),Cyclic] > > which is nice... > > Wouter.

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