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Re: [math-fun] 177 million tons of ice???
by dasimov＠earthlink.net 20 Sep '05

20 Sep '05

Victor writes: << The Times has printed a correction: A front-page article on Saturday about mistakes by the Federal Emergency Management Agency after Hurricane Katrina referred incorrectly to the area being overseen by the agency's federal coordinating officer, William Lokey. It also misstated the amount of ice that had been delivered to the region by Friday. Mr. Lokey oversees Louisiana, not the three-state region. Relief workers delivered 177.6 million pounds, not tons. >> Yes, and not a second too soon. (3 1/3 days after I reported this error to both corrections(a)nytimes.com and to nytnews(a)nytimes.com around midnight Fri/Sat.) --Dan

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[math-fun] 177 million tons of ice???
by dasimov＠earthlink.net 20 Sep '05

20 Sep '05

A front-page Sat., Sept. 17 story on the NY Times website, found at http://snipurl.com/hqxi, claims that FEMA has delivered 177 million tons of ice to Louisiana, Mississippi and Alabama in the aftermath of hurricane Katrina. Is 177 million tons a credible amount? Seems very unlikely to me, but I'd like to know what others think. --Dan

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Re: [math-fun] 177 million tons of ice???
by dasimov＠earthlink.net 16 Sep '05

16 Sep '05

Thanks, Jud and Ed. Jud wrote: << Impossible, as stated. The limit a semi can carry is on the order of 30,000-50,000 pounds. Over 7 million truckloads? >> That's how I convinced myself, too -- est. 8.5 million truckfuls. (As I wrote in an e-mail to the Times, before deciding to ask math-fun in case I was missing something.) --Dan

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[math-fun] RE: A ternary notation
by franktaw＠netscape.net 16 Sep '05

16 Sep '05

"Eric Angelini" <keynews.tv(a)skynet.be> wrote: > Hello SeqFan and Math-Fun, > Consider this (hope this is not old-hat): > > Â 3^0 Â 3^1 Â 3^2 Â 3^3 Â 3^4 Â 3^5 Â 3^6 Â 3^7 ... >= Â 1 Â Â 3 Â Â 9 Â Â 27 Â 81 Â 243 Â 729 Â 2187 ... > >This seq. is known to be very efficient if you >want to weigh integer weights with a two-tray >balance, leaving no "holes" behind; use weights >of 1,3,9,27... units to measure all "natural" >quantities, from 1 to infinity: > >I had the idea, yesterday night, to represent >all natural numbers in the same way -- all I >had to do was to use three symbols : > Â 0 for a power of 3 I don't need > Â 1 for a power of 3 I need > Â 2 as a grahic symbol meaning "minus all the > Â Â rest" This is an old idea, but there is a conceptually much simpler approach to the third digit: it is the digit -1, representing putting just this weight in the other pan. The result is called "balanced ternary" - I first encountered it in Knuth's The Art of Computer Programming (I forget which volume). If you use 2 to represent the -1 digit, you will get: 0, 1, 12, 10, 11, 122, 120, 121, 102, 100, 101, ... This sequence is also not in the OEIS. -- Franklin T. Adams-Watters 16 W. Michigan Ave. Palatine, IL 60067 847-776-7645 __________________________________________________________________ Switch to Netscape Internet Service. As low as $9.95 a month -- Sign up today at http://isp.netscape.com/register Netscape. Just the Net You Need. New! Netscape Toolbar for Internet Explorer Search from anywhere on the Web and block those annoying pop-ups. Download now at http://channels.netscape.com/ns/search/install.jsp

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[math-fun] A ternary notation
by Eric Angelini 15 Sep '05

15 Sep '05

Hello SeqFan and Math-Fun, Consider this (hope this is not old-hat): 3^0 3^1 3^2 3^3 3^4 3^5 3^6 3^7 ... = 1 3 9 27 81 243 729 2187 ... This seq. is known to be very efficient if you want to weigh integer weights with a two-tray balance, leaving no "holes" behind; use weights of 1,3,9,27... units to measure all "natural" quantities, from 1 to infinity: 1 = 1 2 = 3-1 3 = 3 4 = 3+1 5 = 9-(3+1) 6 = 9-3 7 = 9-(3-1) 8 = 9-1 9 = 9 10 = 9+1 11 = 9+(3-1) 12 = 9+3 13 = 9+(3+1) 14 = 27-[9+(3+1)] etc. I had the idea, yesterday night, to represent all natural numbers in the same way -- all I had to do was to use three symbols : 0 for a power of 3 I don't need 1 for a power of 3 I need 2 as a grahic symbol meaning "minus all the rest" ... and use a regular addition rule (which will be explained below in a few seconds). Let's make first a parallel with the binary no- tation : 2^0 2^1 2^2 2^3 2^4 2^5 2^6 2^7 ... = 1 2 4 8 16 32 64 128 ... We use two symbols : 0 for a power of 2 we don't need 1 for a power of 2 we need ... and an easy addition rule. We start from the left, putting 0's and 1's where we want to, and then reverse the sequence of 0's and 1's. To represent 54 (my age) we look for the closest power < or = to 54 and mark it with a "1": 2^0 2^1 2^2 2^3 2^4 2^5 2^6 2^7 ... = 1 2 4 8 16 32 64 128 ... --> 1 54 - 32 = 22; we mark with another "1" the closest power < or = to 22: 2^0 2^1 2^2 2^3 2^4 2^5 2^6 2^7 ... = 1 2 4 8 16 32 64 128 ... --> 1 1 54 - (32+16) = 6, we mark accordingly 2^2: 2^0 2^1 2^2 2^3 2^4 2^5 2^6 2^7 ... = 1 2 4 8 16 32 64 128 ... --> 1 1 1 54 - (32+16+4) = 2, we mark 2^1: 2^0 2^1 2^2 2^3 2^4 2^5 2^6 2^7 ... = 1 2 4 8 16 32 64 128 ... --> 1 1 1 1 To complete the binary representation of 54, we put 0's under all unnecessary powers: 2^0 2^1 2^2 2^3 2^4 2^5 2^6 2^7 ... = 1 2 4 8 16 32 64 128 ... --> 0 1 1 0 1 1 ... last line is reversed to produce 110110, which is indeed the binary expression of 54. We can represent all natural integers with this method, of course: 2^0 2^1 2^2 2^3 2^4 2^5 2^6 2^7 ... = 1 2 4 8 16 32 64 128 ... 1 = 1 => 1 2 = 0 1 (to reverse)=> 10 3 = 1 1 => 11 4 = 0 0 1 => 100 5 = 1 0 1 => 101 6 = 0 1 1 => 110 7 = 1 1 1 => 111 8 = 0 0 0 1 => 1000 9 = 1 0 0 1 => 1001 etc. Let's do the same with the powers of 3; we will mark with a "1" the powers we need, with a "0" the ones we will discard and (novelty!) with a "2" the quantity we will substract (I will use the star symbol (*) to indicate hereunder an operation in base 10): 3^0 3^1 3^2 3^3 3^4 3^5 3^6 3^7 ... = 1 3 9 27 81 243 729 2187 ... 1 = 1 2 = *3-1* = 121 3 = 01 => 10 4 = *3+1* = 11 5 = *9-4* = 1211 6 = *9-3* = 1210 7 = *9-2* = 12121 8 = *9-1* = 1201 9 = 001 => 100 10 = *9+1* = 101 11 = *9+2* = 1121 12 = *9+3* = 110 13 = *9+4* = 111 14 = *27-13* = 12111 15 = *27-12* = 12110 16 = *27-11* = 121121 17 = *27-10* = 12101 18 = *27-9* = 12100 19 = *27-8* = 121201 20 = *27-7* = 1212121 21 = *27-6* = 121210 22 = *27-5* = 121211 23 = *27-4* = 12011 24 = *27-3* = 12010 25 = *27-2* = 120121 26 = *27-1* = 12001 27 = 0001 => 1000 28 = *27+1* = 1001 29 = *27+2* = 10121 30 = *27+3* = 10010 etc. Is this of interest? Is the introduction of a third symbol "economi- cally" efficient? (I think we will use less symbols to represent all first 1000 natural integers with this "ternary notation" than with the usual binary system -- the price might be to high, though)... Speaking of "notation efficiency" (how to re- present in the less "symbol-consuming" way all the natural numbers) -- is there a definitive answer? The "ternary notation" above can certainly be bettered as the "ternary artefacts" 20, 1220, 001 or 002 represent nothing. Best, É. (seq. 0,1,121,10,11,1211,1210,12121,1201,100,... is not in the OEIS)

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RE: [math-fun] A ternary notation
by Pacher Christoph 15 Sep '05

15 Sep '05

Hi, > Hello SeqFan and Math-Fun, > Consider this (hope this is not old-hat): I definitely came across this idea in the past, sorry... I cannot remember right now, where. >Is the introduction of a third symbol "economi- >cally" efficient? (I think we will use less >symbols to represent all first 1000 natural >integers with this "ternary notation" than with >the usual binary system -- the price might be >to high, though)... Speaking in terms of calculations and storage in computers, the binary system has strong technological advantages. It is much easier and more reliable to represent two states than three. E.g. you need only do decide whether there is a certain amount of charge in a capacitor or not. Thus you simply need less chip size (and less power). And I think also the elemental algebraic operations are much simpler (again smaller chip size) in a binary system. >Speaking of "notation efficiency" (how to re- >present in the less "symbol-consuming" way all >the natural numbers) -- is there a definitive >answer? As far as I remember the optimal base would be exp(1)=2.71... . Therefore 3 may be more efficient than two. But two should be the 2nd best choice regarding efficiency. Best regards Christoph

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[math-fun] cutting square into pieces to make equilateral triangle
by N. J. A. Sloane 13 Sep '05

13 Sep '05

I've added three (conjectured) sequences to the OEIS dealing with dissections - see A110000, A110312, A110356 I'm not even sure about the first term of A110000: is it known that there is no way to cut an equilateral triangle into 3 polygonal pieces which can be rearranged to form a square? (Turning over is allowed; 4 pieces is classical - see the entry for references, pictures, etc.) NJAS

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[math-fun] egyptian fractions: equality, comparison, addition, canonicalization
by Schroeppel, Richard 13 Sep '05

13 Sep '05

What would it be like to use egyptian fractions for arithmetic? We would want algorithms for addition, subtraction, multiplication, division, floor, compare, and canonicalization. And maybe converting to decimals. Of course all of these can be done by converting the EF to standard rational num/den representation, doing the arithmetic, and then converting back. But it's an interesting problem to give algorithms that preserve the spirit of the representation as much as possible. For addition, we can take the multiset union of the addends, and then apply some algorithm for getting rid of the 2/n terms. Or we might provide a subalgorithm for adding 1/n to a generic EF. Subtraction and comparison raise similar problems. Providing methods that don't dramatically blow up the number of terms and the largest denominators seems challenging. Rich rcs(a)cs.arizona.edu

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[math-fun] Egyptian fractions question
by Mike Stay 13 Sep '05

13 Sep '05

Given a finite set R of unit fractions, is it possible to represent all rationals as Egyptian fractions without using elements of R? -- Mike Stay metaweta(a)gmail.com http://math.ucr.edu/~mike

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[math-fun] [Fwd: From the Calculus Exam]
by Thane Plambeck 12 Sep '05

12 Sep '05

From http://learningcurves.blogspot.com -------- Original Message -------- Subject: From the Calculus Exam Date: Mon, 12 Sep 2005 17:24:33 GMT From: <Rudbeckia Hirta> Or, "How's that plan to learn nothing in my class working for you, Mr. I-Took-Calculus-in-High-School?" *Question:* Precisely state the Intermediate Value Theorem. *Answers:* In a continuous function, if there are 3 points a, b, and c where a>c, there should be a point b in between a and c so that a>b>c. Given a number a. IVT converts a to the last greatest integer a is divisible by. Between any two points on a graph both L and the limit x -> L exist and can be formed. If 2 numbers are substituted for x at the same distance away from c, then L is the intermediate number. If you have a f(x) and domain (a,b) and the lim x->a f(x) = f(a) and lim x->b f(x) = f(b). If you have one point c between (a,b) then lim x->c f(x) is between f(a) and f(b). Take the intermediate value to find where the function is discontinuous using [[x]]. The IVT states that if a function is not continuous at every number in its domain, then it is not continuous. For every function that is continuous through the given points f(a) and f(b) and has a defined limit, there exists a real number N anywhere between the points f(a) and f(b). On a function f(x) all values of x on the functions domain will be continuous. lim x->a f(x) = L In a closed interval when both f(a) and f(b) exists and are continuous, there exists a constant c between f(a) and f(b) which is also continuous. If a function is continuous on an interval [a,b] there is at least one point c where a < c < b such that there exists f(c)=N. -- Thane Plambeck http://www.plambeck.org/ehome.htm

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