September 2011 - mathfun@mailman.xmission.com

Re: [math-fun] How to speak long numbers -- now I'm really curious(-:
by Dan Asimov 27 Sep '11

27 Sep '11

I was taught the same thing. But Wikipedia is a very unreliable reference when it comes to proper word usage. In fact it is almost universally agreed by experts (unfortunately) that either method is correct in the U.S. It's generally considered an issue of style rather than correct spelling. --Dan << I was taught not to use "and" in the integer portion of a number when spelling it out, nor to use commas. A quick web check confirmed you should not use "and" for American spelling, but that you should use it for the hundreds portion in British spelling. Example: "seven hundred twenty-seven" (American) vs. "seven hundred and twenty-seven" (British). Here's a reference: http://en.wikipedia.org/wiki/American_and_British_English_differences#Numbe… >> Sometimes the brain has a mind of its own.

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[math-fun] A puzzle inspired by cartalk
by Victor Miller 26 Sep '11

26 Sep '11

On the program cartalk last week, there was a puzzle of the following form: A list of numbers was given, and one was asked what they had in common (I give the actual puzzler at the end). The answer was that each of these numbers was divisible by the number of letters (excluding spaces) in the standard spelling out of the number in words. This got me to thinking of the following modification: Suppose that we map each positive integer into an integer by the following means: write the number in decimal (no leading zeros). The value of the function will be to total number of letters that one gets by mapping each digit to its English equivalent (e.g. 13 -> length("OneThree") = 8. Or more generally give a fixed map g : {0,1,...,9} -> positive integers, and define f(n) = sum_{d digits in n} g(d). So are there an infinite number of positive integers n such that n is divisible by g(n)? One can generalize this to other bases. If there are an infinite number of them how dense is the set? Victor Here's the original puzzler: When my kids were in school, they, like all the other kids I guess, had to learn their numbers. So each day for homework, they would bring home a list of numbers on a piece of paper, and they were asked to write out the letters that spelled that number, right next to each of them. So the number seven would be there, there'd be a blank space, the kids would have to write S - E - V - E - N. And of course they were also asked which numbers were spelled out by the various combinations of letters, so they'd see S - I - X - T - Y and write Sixty, etc. One day, son number two presented me with a list of numbers and he said, "These numbers are different. There's something special about them." Here are the numbers: Four, Six, Twelve, Thirty, Thirty Three, Thirty Six, Forty, Forty Five, Fifty, Fifty Four, Fifty Six, Sixty, Seventy, Eighty One, Eighty Eight, Ninety, and a Hundred. Now there are no other numbers between one and a hundred inclusive that share this same characteristic. There's something unusual about these numbers that son number two figured out. And I'll give you an additional hint that order does not matter. The best hint is that he determined that these numbers should be on the list perhaps from his homework assignment. What is special about this list of numbers?

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Re: [math-fun] A puzzle inspired by cartalk
by Phil Carmody 26 Sep '11

26 Sep '11

From: Robert Munafo <mrob27(a)gmail.com> > Writing out large numbers in words: > > http://www.grammarbook.com/numbers/numbers.asp Any set of rules that contains the phrase "7 math students" clearly does not apply to the language I know as English. > See examples in Rule 7, and note that the comma is used in > words whenever it is used in digits (but not in 4-digit numbers). > But in any case there is no "and", and the comma when present replaces the > missing "and". Yeah, that's definitely not English as I know it. > Numbers written digitally have a comma "27,101", so the use of the comma in > words is closer to the way it is written in digits. > > "One thousand and one nights" is a contraction of the > original title (in > English) "The Book of the Thousand Nights and One Night" The person using the contracted form was an English speaker who translated poorly and simply used the English rules for spelling 1001. (In the same way that the Frenchman before him translated it into "Les mille et une nuits", and seemingly at least some of the Arabic translations before that.) American rules would not have applied to such a work. That later translations used a more faithful translation is irrelevant, "one thousand and one" means 1001 in English, nothing more. > which implies that > the 1000 nights are to be considered separate from the one night. In the > story, 1000 nights are spend telling tales with cliffhanger endings and > Scheherazade anticipating her execution, and then the final night, in which > she has been pardoned (and will become the queen). It is as if "1000 vases > and one jug" were poorly translated into "1000 and one bottles" which was > then interpreted by the reader as "1001". Apart from the fact that it's "nights" and "night" becoming "nights", rather than "vases" and "jug" becoming "bottle". There was no emphasis at all in the "and one". That was just the French and the English way of spelling the number 1001. If it sounds odd to American ears, that's because you have a different set of rules for spelling numbers, but they are irrelevant to books written in English by Englishmen for an English-speaking audience. Phil

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[math-fun] Bad neutrino joke
by Henry Baker 25 Sep '11

25 Sep '11

Neutrino. Knock knock.

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[math-fun] Composing (T)Chebys
by Bill Gosper 24 Sep '11

24 Sep '11

It is immediate from the cos(n acos(x)) definition that ChebyshevT[n, ChebyshevT[m, x]] == ChebyshevT[n*m, x] which extends to complex m,n. E.g., T[1/n](x) is the inverse of T[n](x). By "cleverly" rewriting sqrt(1-x^2) as sin(one*acos(x)), I got ChebyshevU[n, ChebyshevT[m, x]] == ChebyshevU[-1 + m (1 + n), x]/ChebyshevU[-1 + m, x] when all I needed was d/dx. --rwg Reminder: ChebyshevU[n, I/2]/I^n == Fibonacci[n+1] World's shortest fibula? Er, fib formula?

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Re: [math-fun] Is the asteroid belt a "glider gun" ?
by Henry Baker 23 Sep '11

23 Sep '11

At 12:11 PM 9/22/2011, Bill Gosper wrote: >So if a small asteroid, or that falling satellite, manages to slip in and kill some people, >is Italy going to charge a bunch of astronomers with manslaughter? Or in the U.S., >just a civil tort? We may be about to find out (see below). Perhaps this falling satellite is the reason for the stock market tanking yesterday & today: the sky literally is falling! Antarctica would be far too ironic a resting place for this particular satellite. BTW, how did NASA calculate those 1 in 3200 odds ? http://hosted.ap.org/dynamic/stories/U/US_SCI_FALLING_SATELLITE Sep 22, 5:33 PM EDT Old NASA satellite to tumble to Earth on Friday By ALICIA CHANG AP Science Writer LOS ANGELES (AP) -- While North America appears to be off the hook, scientists are scrambling to pinpoint exactly where and when a dead NASA climate satellite will plummet back to Earth on Friday. The 6-ton, bus-sized satellite is expected to break into more than a hundred pieces as it plunges through the atmosphere, most of it burning up. But if you're hoping for a glimpse, the odds are slim. Most sightings occur by chance because the re-entry path can't be predicted early enough to alert people, said Canadian Ted Molczan, who tracks satellites for a hobby. In all his years of monitoring, Molczan has only witnessed one tumble back to Earth - the 2004 return of a Russian communications satellite. It "looked like a brilliant star with a long glowing tail," he said in an email. The best guess so far is that the 20-year-old Upper Atmosphere Research Satellite will hit sometime on Friday afternoon Eastern time. The latest calculations indicate that it will not be over the United States, Canada and Mexico during that time. Until Thursday, every continent but Antarctic was a potential target. Predicting where and when the freefalling satellite will land is an imprecise science, but officials should be able to narrow it down a few hours ahead. While most of the satellite pieces will disintegrate, 26 large metal chunks - the largest about 300 pounds - are expected to survive, hit and scatter somewhere on the planet. With nearly three-quarters of the world covered in water, chances are that it will be a splashdown. If the re-entry is visible, "it'll look like a long-lived meteor," said Jonathan McDowell of the Harvard-Smithsonian Center for Astrophysics in Cambridge, Mass. Since the dawn of the Space Age, no one has been injured by falling space debris. The only confirmed case of a person being hit by space junk was in 1997 when Lottie Williams of Tulsa, Okla., was grazed in the shoulder by a small bit of debris from a discarded piece of a Delta rocket. The odds of someone somewhere on Earth getting struck by the NASA satellite are 1 in 3,200. But any one person's odds are astronomically lower - 1 in 21 trillion. "You're way more likely to be hit by lightning" than by the satellite, McDowell said. NASA has warned people not to touch any satellite part they might chance upon. There are no hazardous chemicals on board, but people can get hurt by sharp edges, the space agency said. The U.S. tracks the roughly 22,000 pieces of satellites, rockets and other junk orbiting the Earth. Nowadays, the world is more eco-conscious about what it puts up in space. Modern satellites must be designed to disintegrate upon re-entry or have enough fuel to be nudged into a higher orbit or steered into the ocean. The satellite was launched in 1991 aboard the space shuttle Discovery to study the ozone layer, and back then there was no such rule. NASA used up the remaining fuel to put it into a lower orbit in 2005, setting the stage for its uncontrolled return. It will be the biggest NASA spacecraft to fall uncontrolled from the sky in 32 years. It's not unusual for space debris to dive back to Earth. NASA's Orbital Debris Program Office estimates that medium-sized junk falls back once a week. Debris the size of the satellite due back Friday occurs less frequently, about once a year. Harvard's McDowell noted that two massive Russian rocket stages have plunged back this year with little notice. "The only reason this is getting attention is because NASA, as a matter of due diligence, put out a press release," he said. --- Online: Satellite updates: http://www.nasa.gov/uars Aerospace Corp: http://reentrynews.aero.org/1991063b.html --- Follow Alicia Chang's coverage at http://twitter.com/SciWriAlicia

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Re: [math-fun] Is the asteroid belt a "glider gun" ?
by Bill Gosper 22 Sep '11

22 Sep '11

From: "Henry Baker" <hbaker1(a)pipeline.com> FYI -- So the asteroid belt is shooting at us ? http://www.nasa.gov/mission_pages/WISE/news/wise20110919.html Wide-field Infrared Survey Explorer Mission News Origin of Dinosaur-Killing Asteroid Remains a Mystery [...] ------------------------------------------------------------------------- What happened to the old theory that it was a comet (e.g., megaTunguska)? If the chemistry is wrong, how about a retrograde comet killing off most of some asteroid's orbital angular momentum? So if a small asteroid, or that falling satellite, manages to slip in and kill some people, is Italy going to charge a bunch of astronomers with manslaughter? Or in the U.S., just a civil tort? --rwg

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[math-fun] Is the asteroid belt a "glider gun" ?
by Henry Baker 21 Sep '11

21 Sep '11

FYI -- So the asteroid belt is shooting at us ? http://www.nasa.gov/mission_pages/WISE/news/wise20110919.html Wide-field Infrared Survey Explorer Mission News Origin of Dinosaur-Killing Asteroid Remains a Mystery 09.19.11 PASADENA, Calif. -- Observations from NASA's Wide-field Infrared Survey Explorer (WISE) mission indicate the family of asteroids some believed was responsible for the demise of the dinosaurs is not likely the culprit, keeping open the case on one of Earth's greatest mysteries. While scientists are confident a large asteroid crashed into Earth approximately 65 million years ago, leading to the extinction of dinosaurs and some other life forms on our planet, they do not know exactly where the asteroid came from or how it made its way to Earth. A 2007 study using visible-light data from ground-based telescopes first suggested the remnant of a huge asteroid, known as Baptistina, as a possible suspect. According to that theory, Baptistina crashed into another asteroid in the main belt between Mars and Jupiter about 160 million years ago. The collision sent shattered pieces as big as mountains flying. One of those pieces was believed to have impacted Earth, causing the dinosaurs' extinction. Since this scenario was first proposed, evidence developed that the so-called Baptistina family of asteroids was not the responsible party. With the new infrared observations from WISE, astronomers say Baptistina may finally be ruled out. "As a result of the WISE science team's investigation, the demise of the dinosaurs remains in the cold case files," said Lindley Johnson, program executive for the Near Earth Object (NEO) Observation Program at NASA Headquarters in Washington. "The original calculations with visible light estimated the size and reflectivity of the Baptistina family members, leading to estimates of their age, but we now know those estimates were off. With infrared light, WISE was able to get a more accurate estimate, which throws the timing of the Baptistina theory into question." WISE surveyed the entire celestial sky twice in infrared light from January 2010 to February 2011. The asteroid-hunting portion of the mission, called NEOWISE, used the data to catalogue more than 157,000 asteroids in the main belt and discovered more than 33,000 new ones. Visible light reflects off an asteroid. Without knowing how reflective the surface of the asteroid is, it's hard to accurately establish size. Infrared observations allow a more accurate size estimate. They detect infrared light coming from the asteroid itself, which is related to the body's temperature and size. Once the size is known, the object's reflectivity can be re-calculated by combining infrared with visible-light data. The NEOWISE team measured the reflectivity and the size of about 120,000 asteroids in the main belt, including 1,056 members of the Baptistina family. The scientists calculated the original parent Baptistina asteroid actually broke up closer to 80 million years ago, half as long as originally proposed. This calculation was possible because the size and reflectivity of the asteroid family members indicate how much time would have been required to reach their current locations -- larger asteroids would not disperse in their orbits as fast as smaller ones. The results revealed a chunk of the original Baptistina asteroid needed to hit Earth in less time than previously believed, in just about 15 million years, to cause the extinction of the dinosaurs. "This doesn't give the remnants from the collision very much time to move into a resonance spot, and get flung down to Earth 65 million years ago," said Amy Mainzer, a co-author of a new study appearing in the Astrophysical Journal and the principal investigator of NEOWISE at NASA's Jet Propulsion Laboratory (JPL) in Pasadena. Calif. "This process is thought to normally take many tens of millions of years." Resonances are areas in the main belt where gravity nudges from Jupiter and Saturn can act like a pinball machine to fling asteroids out of the main belt and into the region near Earth. The asteroid family that produced the dinosaur-killing asteroid remains at large. Evidence that a 10-kilometer (about 6.2-mile) asteroid impacted Earth 65 million years ago includes a huge, crater-shaped structure in the Gulf of Mexico and rare minerals in the fossil record, which are common in meteorites but seldom found in Earth's crust. In addition to the Baptistina results, the NEOWISE study shows various main belt asteroid families have similar reflective properties. The team hopes to use NEOWISE data to disentangle families that overlap and trace their histories. "We are working on creating an asteroid family tree of sorts," said Joseph Masiero, the lead author of the study. "We are starting to refine our picture of how the asteroids in the main belt smashed together and mixed up." JPL manages and operated WISE for NASA's Science Mission Directorate. The spacecraft was put into hibernation mode after it scanned the entire sky twice, completing its main objectives. The principal investigator, astronomer Edward Wright, is at UCLA. The mission was selected competitively under NASA's Explorers Program managed by the agency's Goddard Space Flight Center in Greenbelt, Md. The science instrument was built by the Space Dynamics Laboratory in Logan. The spacecraft was built by Ball Aerospace & Technologies Corp. in Boulder, Colo. Science operations and data processing take place at the Infrared Processing and Analysis Center at the California Institute of Technology in Pasadena. Caltech manages JPL for NASA. More information is online at http://www.nasa.gov/wise , http://wise.astro.ucla.edu and http://jpl.nasa.gov/wise . Whitney Clavin 818-354-4673 Jet Propulsion Laboratory, Pasadena, Calif. whitney.clavin(a)jpl.nasa.gov Trent J. Perrotto 202-358-0321 Headquarters, Washington trent.j.perrotto(a)nasa.gov

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[math-fun] all solvable sextics are easy!?
by Bill Gosper 20 Sep '11

20 Sep '11

I believe the following ridiculously short function will solve any solvable sextic: Ssexy[q_] := Catch[Block[{rts = #[[1, 2]] & /@ NSolve[q == 0, WorkingPrecision -> 666], foos}, foos = InverseFourier[ rts[[#]]/Sqrt[6]]^6 & /@ (Join[{1}, #] & /@ (1 + Permutations[Range[5]])); Factor[Rationalize[ Expand[Times @@ (# - Rest[foos[[ Ordering[Denominator[Rationalize[Plus @@ #, 9.^-666]] & /@ foos, 1][[1]]]]])]]] /. (a_: 1)*# + b_ :> Throw[Solve[0 == Factor[q, Extension -> Sqrt[-b/a]]]]; Solve[0 == q]]] This is based on the empirical observation that the resolvent quintic, when it exists, will have a linear factor whose root, when square rooted, will split the sextic into cubics. When there is no quintic with rational coeffs, Solve still has a chance of factoring or decomposing. Here's an example of splitting. In[1325]:= MinimalPolynomial[Cos[\[Pi]/5] - Cos[\[Pi]/7], c] Out[1325]= -1 - 8 c + 36 c^2 + 40 c^3 - 128 c^4 - 32 c^5 + 64 c^6 In[1326]:= Ssexy[%] Out[1326]= {{c -> 1/12 (1 - 3 Sqrt[5] + (2 7^(2/3))/(1/2 (1 + 3 I Sqrt[3]))^(1/3) + 2^(2/3) (7 (1 + 3 I Sqrt[3]))^(1/3))}, {c -> 1/12 (1 + 3 Sqrt[5] + (2 7^(2/3))/(1/2 (1 + 3 I Sqrt[3]))^(1/3) + 2^(2/3) (7 (1 + 3 I Sqrt[3]))^(1/3))}, {c -> -((7^(2/3) (1 + I Sqrt[3]))/(6 2^(2/3) (1 + 3 I Sqrt[3])^(1/3))) - 1/12 (1 - I Sqrt[3]) (7/2 (1 + 3 I Sqrt[3]))^(1/3) + 1/12 (1 - 3 Sqrt[5])}, {c -> -((7^(2/3) (1 - I Sqrt[3]))/( 6 2^(2/3) (1 + 3 I Sqrt[3])^(1/3))) - 1/12 (1 + I Sqrt[3]) (7/2 (1 + 3 I Sqrt[3]))^(1/3) +1/12 (1 - 3 Sqrt[5])}, {c -> -((7^(2/3) (1 + I Sqrt[3]))/(6 2^(2/3) (1 + 3 I Sqrt[3])^(1/3))) - 1/12 (1 - I Sqrt[3]) (7/2 (1 + 3 I Sqrt[3]))^(1/3) + 1/12 (1 + 3 Sqrt[5])}, {c -> -((7^(2/3) (1 - I Sqrt[3]))/(6 2^(2/3) (1 + 3 I Sqrt[3])^(1/3))) - 1/12 (1 + I Sqrt[3]) (7/2 (1 + 3 I Sqrt[3]))^(1/3) +1/12 (1 + 3 Sqrt[5])}} In[1332]:= FindIntegerNullVector[ Append[#[[1, 2]] & /@ %1326, Cos[\[Pi]/5] - Cos[\[Pi]/7]]] Out[1332]= {0, 0, 0, 0, 1, 0, -1} So Cos[\[Pi]/5] - Cos[\[Pi]/7] is the penultimate solution. --rwg

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[math-fun] 1750, Big and Small, iterate
by Bill Gosper 19 Sep '11

19 Sep '11

I'm not a big fan of results peculiar to the decimal radix, but this reminds me of a Fortran compiler that Peter Samson wrote in PDP-6 machine language the late 60s, based on an absolutely inscrutable paper on generative grammars by Edgar T. Irons. Since the task of commenting the code would have been much larger than the coding itself, there was only one comment in the entire listing, where Peter needed the constant 1000. The line was ...: 1750 ;RIP JSB (The assembler default radix was octal. You had to add a decimal point if you wanted decimal. And Peter was a huge Bach fan.) --rwg ---------------------------- Original Message ---------------------------- From: "Eric Angelini" <Eric.Angelini(a)kntv.be> Date: Mon, September 19, 2011 2:20 pm -------------------------------------------------------------------------- Hello Math-fun,, a light (iterating) idea (too light, I'm afraid)... Take an integer N (N = 1750 here) Make the sum of N's digits and call it D (D = 13 here) Now build the integers B (big) and S (small) like this: - B is the biggest integer you can write using one digit of N and all the digits of D (here 7 and 1,3 --> B = 731) - S is the smallest integer you can write using one digit of N and all the digits of D (here 0 and 1,3 --> S = 103) Now iterate, computing B, then S, then B, then S, etc. Let's see what happens, starting with 1 (the line under the sequence shows D (digit sum) then B (big), then D, then S (small), then D, then B, then D, then S, ... 1---11---12---32---25---75---125---85---135---95---145 1 B 2 S 3 B 5 S 7 B 12 S 8 B 13 S 9 B 14 S 145---510---16---76---136---610---17---87 ... S 10 B 6 S 7 B 13 S 10 B 7 S 8 B ... Where do we get? (loops, fixed points, flights, etc.) Best, �. (and what about iterating only B's? Or S's?)

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