September 2005 - mathfun@mailman.xmission.com

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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[math-fun] the q Filbert matrix [long]
by R. William Gosper 11 Sep '05

11 Sep '05

Many of you may remember the spectacularly ill-conditioned "Filbert" matrix pointed out to me by Peter ("Feathers") Groot in the late 60s: E.g., (c89) (genmatrix(lambda([i,j],1/fib(i+j+9)),3),%% = float(%%)) [ 1 1 1 ] [ -- --- --- ] [ 89 144 233 ] [ ] [ 0.01124 0.00694 0.00429 ] [ 1 1 1 ] [ ] (d89) [ --- --- --- ] = [ 0.00694 0.00429 0.00265 ] [ 144 233 377 ] [ ] [ ] [ 0.00429 0.00265 0.00164 ] [ 1 1 1 ] [ --- --- --- ] [ 233 377 610 ] Contrary to appearance, this is good to six figures: (c90) 1/%; [ 89 144 233 ] [ 89.0 144.0 233.0 ] [ ] [ ] (d90) [ 144 233 377 ] = [ 144.0 233.0 377.0 ] [ ] [ ] [ 233 377 610 ] [ 233.0 377.0 610.0 ] But (c91) float(map('invert,d89)) [ 1.0019e+11 2.62304e+11 - 6.86715e+11 ] [ ] (d91) [ 2.62304e+11 6.86691e+11 - 1.79782e+12 ] = [ ] [ - 6.86715e+11 - 1.79782e+12 4.70676e+12 ] [ 5.01351e+7 1.44257e+8 - 3.64668e+8 ] [ ] [ 1.3994e+8 3.67339e+8 - 9.60738e+8 ] [ ] [ - 3.57682e+8 - 9.72036e+8 2.50923e+9 ] The disagreement is even *more* dramatic if the floating numbers on the rhs of d89 are first converted to exact rationals, and the inversion performed exactly! The reason is that the matrix is far closer than 10^-6 to being singular, and truncation to six figures pushes it way over the brink. The true inverse is symmetric and integer. I tried substituting a couple of candidates for q-Fibonaccis to see if inverting made huge q-polynomials. It is tempting to define qib(0):=0, qib(1):=1, n - 2 qib(n) = qib(n - 1) + q qib(n - 2)), for then n 1 floor(- - -) 2 2 ==== 2 \ k qib(n) = > qbn(n - k - 1, k) q / ==== k = 0 where qbn is the q-binomial coefficient (q; q) n qbn(n, k) = -------------------. (q; q) (q; q) k n - k Then, thru qib(11) 0 1 1 1 + q 2 1 + q + q 2 3 4 1 + q + q + q + q 2 3 4 5 6 1 + q + q + q + 2 q + q + q 2 3 4 5 6 7 8 9 1 + q + q + q + 2 q + 2 q + 2 q + q + q + q 4 6 2 3 4 5 6 (1 + q + q ) (1 + q + q + q + q + q + q ) 4 2 3 4 5 6 7 8 9 10 (1 + q ) (1 + q + q + q + q + q + 2 q + 2 q + 2 q + 2 q + q 11 12 + q + q ) 2 3 4 2 3 4 (1 - q + q - q + q ) (1 + q + q + q + q ) 4 5 6 7 8 9 10 11 12 (1 + q + q + q + q + q + q + q + q + q + q ) 2 3 4 5 6 7 8 9 10 1 + q + q + q + 2 q + 2 q + 3 q + 3 q + 4 q + 5 q + 5 q 11 12 13 14 15 16 17 18 + 5 q + 6 q + 6 q + 6 q + 6 q + 6 q + 5 q + 5 q 19 20 21 22 23 24 25 + 4 q + 4 q + 3 q + 2 q + q + q + q and qib(oo) is defined and equal to inf ==== \ A003114(n) 1 > q = ----------------------------------- / oo ==== /===\ n = 0 | | 5 n + 1 5 n + 4 | | (1 - q ) (1 - q ) | | n = 0 oo 2 ==== n \ q = > -------, / (q; q) ==== n n = 0 the famous Rogers-Ramanujan function. But this qib definition produces messy denominators in the Filbert matrix, and worse, fails to distribute over gcd, f(gcd(n,k)) = gcd(f(n),f(k)), unlike f:= vanilla fib. Defining instead qfb[0] := 0, qfb[1] := 1, qfb[n] := qfb[n-1] + q qfb[n-2] (A011973) answers these objections: 0 0 1 1 2 1 3 1 + q 4 1 + 2 q 2 5 1 + 3 q + q 6 (1 + q) (1 + 3 q) 2 3 7 1 + 5 q + 6 q + q 2 8 (1 + 2 q) (1 + 4 q + 2 q ) 2 3 9 (1 + q) (1 + 6 q + 9 q + q ) 2 2 10 (1 + 3 q + q ) (1 + 5 q + 5 q ) 2 3 4 5 11 1 + 9 q + 28 q + 35 q + 15 q + q 2 12 (1 + q) (1 + 2 q) (1 + 3 q) (1 + 4 q + q ) 2 3 4 5 6 13 1 + 11 q + 45 q + 84 q + 70 q + 21 q + q 2 3 2 3 14 (1 + 5 q + 6 q + q ) (1 + 7 q + 14 q + 7 q ) Note that the degree of qfb[n] is floor((n-1)/2), and qfb[2 p] = qfb[p] times a polynomial of equal degree which is 1 mod p. We also have the pleasant addition formula [ qfb qfb ] [ qfb qfb ] [ k + 1 k ] [ 1 0 ] [ n + 1 n ] [ ] . [ ] . [ ] = [ qfb qfb ] [ 0 q ] [ qfb qfb ] [ k k - 1 ] [ n n - 1 ] [ qfb qfb ] [ n + k + 1 n + k ] [ ]. [ qfb qfb ] [ n + k n + k - 1 ] Note that this contradicts the assertion F(n+m,x) = F(m,x)*F(n,x) + F(m-1,x)*F(n-1,x) in the A011973 article. And the q-Filbert inverse contains satisfyingly huge polynomials divided by mere monomial powers of q. (It looks possible to guess the general form, which will probably require q-"superfactorials", i.e. q-extensions of 1^1 2^2 3^3 ... n^n or 1! 2! 3! ... n!.) Finally, analogous to the Stirling polynomials, one can form an interesting sequence by dividing out the predictable (by gcd thm) factors of qfb[n]: 1 1 2 1 3 1 + q 4 1 + 2 q 2 5 1 + 3 q + q 6 1 + 3 q 2 3 7 1 + 5 q + 6 q + q 2 8 1 + 4 q + 2 q 2 3 9 1 + 6 q + 9 q + q 2 10 1 + 5 q + 5 q 2 3 4 5 11 1 + 9 q + 28 q + 35 q + 15 q + q 2 12 1 + 4 q + q 2 3 4 5 6 13 1 + 11 q + 45 q + 84 q + 70 q + 21 q + q 2 3 14 1 + 7 q + 14 q + 7 q 2 3 4 15 1 + 9 q + 26 q + 24 q + q 2 3 4 16 1 + 8 q + 20 q + 16 q + 2 q 2 3 4 5 6 7 8 17 1 + 15 q + 91 q + 286 q + 495 q + 462 q + 210 q + 36 q + q 2 3 18 1 + 6 q + 9 q + 3 q 2 3 4 5 6 7 19 1 + 17 q + 120 q + 455 q + 1001 q + 1287 q + 924 q + 330 q 8 9 + 45 q + q 2 3 4 20 1 + 8 q + 19 q + 12 q + q 2 3 4 5 6 21 1 + 13 q + 64 q + 146 q + 148 q + 48 q + q 2 3 4 5 22 1 + 11 q + 44 q + 77 q + 55 q + 11 q 2 3 4 5 6 7 23 1 + 21 q + 190 q + 969 q + 3060 q + 6188 q + 8008 q + 6435 q 8 9 10 11 + 3003 q + 715 q + 66 q + q 2 3 4 24 1 + 8 q + 20 q + 16 q + q 2 3 4 5 6 7 25 1 + 20 q + 170 q + 800 q + 2275 q + 4003 q + 4280 q + 2605 q 8 9 10 + 775 q + 75 q + q . . . 2 3 4 30 1 + 7 q + 14 q + 8 q + q --rwg PS: 4 2 %i %pi k ==== ---------- \ 1/5 5 > sqrt(8 - 7 %e ) / ==== k = 0 1 1/5 3/5 1 1/5 2/5 2 (------- + 1) 7 (7 + 3) (1 - -------) (7 + 2) 7 sqrt(5) sqrt(5) 2 = ------------------------------- + ----------------------------- + - sqrt(5) sqrt(5) 5 ~ 14.1420530099223

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RE: [math-fun] Rosser provability
by Andy Latto 09 Sep '05

09 Sep '05

> > > S is a Rosser proof of the formula A iff > S is a proof of A and all proofs S' with > smaller Godel code than S don't prove not(A). > (2) Is there a normal provabile but not Rosser > provabile formula A? No if the Peano axioms are consistent; yes if they are inconsistent. That is, the statement 'Every provable formula is Rosser provable" is equivalent to the statement "The Peano axioms are consistent". Both are true (or at least provable in ZF, if you believe ZF is consistent), but neither is provable from the Peano axioms. Andy Latto andy.latto(a)pobox.com

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[math-fun] radical denesting puzzle
by Richard Schroeppel 09 Sep '05

09 Sep '05

rwg writes ... 4/5 3/5 2/5 1/5 1/5 2 7 - 7 - 2 7 + 6 7 + 2 sqrt(8 - 7 ) = -----------------------------------, 5 This set off a "can't be right" buzzer for me. Since it's an algebraic identity, it should also be valid for the complex 5th roots of 7. Suppose we add up all five conjugates. As we run through the 5 fifth roots of 7, the LHS radicand will have real parts between 6 and 10, and we can expect the sum of all five sqrts to be roughly 5*sqrt8, about 14. But in summing the RHS, all the 7^(k/5) terms will cancel, since the sum of the five fifth roots of 7 is 0. This leaves only the 2/5 term, times 5, = 2. Thus, an apparent contradiction, 14 = 2. So I checked the equation numerically, and it's correct in double precision, about 16 decimal places. PRIME>(sqrt (- 8 (expt 7 1/5))) 2.5542566116984893 PRIME>(/ (+ (* 2 (expt 7 4/5)) (* -1 (expt 7 3/5)) (* -2 (expt 7 2/5)) (* 6 (expt 7 1/5)) 2) 5) 2.5542566116984893 My best guess as to what's going on is that the signs of the sqrts vary as different conjugates are selected. But this creates two new puzzles: (a) There are only a few combinations of sign assignments that will still cancel the imaginary parts when summing the LHS, and it's mysterious that the real parts mostly cancel too. (b) And, what's the rule or rationale for selecting the LHS sqrt signs? Normally one expects equations with different (relatively prime) degree roots to permit separate choices of the roots. A likely related curiosity: 8^5 - 7 = 32761 = 181^2. rwg> Two days ago I found several one-parameter continua (too cumbrous to display) of solutions to sqrt(a^(1/5)+-b^(1/5)) = <five terms>, e.g., They must be huge. Rich rcs(a)cs.arizona.edu

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[math-fun] denesting wins & losses
by R. William Gosper 08 Sep '05

08 Sep '05

The answer to the denesting of 1/5 1/5 sqrt(4 - 3 ) I mentioned last week is 3/5 4/5 3/5 2/5 2/5 4/5 1/5 1/5 - 2 3 + 3 + 2 2 3 - 2 3 + 2 ---------------------------------------------------. 5 Two days ago I found several one-parameter continua (too cumbrous to display) of solutions to sqrt(a^(1/5)+-b^(1/5)) = <five terms>, e.g., 1/5 sqrt(11 12 - 7) = sqrt(5) 3/5 4/5 1/5 3/5 4/5 2/5 2/5 1/5 - 2 3 + 2 3 - 2 3 + 4 2 3 + -------------------------------------------------, sqrt(5) 4/5 3/5 2/5 1/5 1/5 2 7 - 7 - 2 7 + 6 7 + 2 sqrt(8 - 7 ) = -----------------------------------, 5 1/5 2/5 sqrt(41 29 + 33 2 ) = 3/5 4/5 3/5 2/5 2/5 4/5 1/5 1/5 2 29 - 3 29 + 2 2 29 + 7 2 29 + 19 2 ------------------------------------------------------------, 5 4/5 1/5 sqrt(808 3 79 - 4551) = 1/5 4/5 2/5 3/5 3/5 2/5 4/5 1/5 6 3 79 - 2 3 79 + 14 3 79 - 38 3 79 - 87 ----------------------------------------------------------------. 5 However, assuming there are only finitely many rational r and d satisfying 4 3 2 2 2 0 = 2 d (2 d - 1) r + 2 d (2 d - 4 d - 1) r - (5 d - 4 d - 7) r - 2 and five close cousins, and that I have found them all, then there are no denestings of the form sqrt(a^(1/6)+-b^(1/6)) = <six real terms>, a, b positive rationals. My nearest miss: 1/6 1/3 5/6 sqrt(56 55 %i - 51) = (5 11 %i - 2 sqrt(11) %i 2/3 1/6 1/6 2/3 5/6 1/3 + 5 5 11 %i - 5 11 + 5 11 + 7 sqrt(5))/3. I'm almost ready to conjecture that there are no denestings of the form sqrt(a^(1/n)+-b^(1/n)) = <six or more real terms> for any n (with a, b positive rationals). (In spite of my sqrt(2-2^(1/7))= <five terms>.)

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[math-fun] Array with primes
by Eric Angelini 07 Sep '05

07 Sep '05

[Eric A.]: > 55 doesn't belong to "our" array -- but is in A105441. ... no, sorry, Michael, you are right, A105441 is (except "1") the sequence with all array missing integers... ... A093642 has no 55 -- thus is *not* the sequence we were looking at... Best, É. + A.

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