September 2013 - mathfun@mailman.xmission.com

[math-fun] pythagorean rationals, terminating decimals, and complex analysis
by Warren D Smith 12 Sep '13

12 Sep '13

corrections: As the evil Mike Speciner pointed out, f(z)=e is a counterexample. The correction: Theorem should specify NONCONSTANT analytic functions. Another error was I said the terminating-decimal complex numbers z with |z|, Re z, Im z, all terminating, was a "subring" I should have said "subsubgroup." Re Henry Baker's desires: they are difficult. First, you want a subFIELD while what I said only gave a multiplicative subGROUP (cannot add). Second, even if you had that, you'd still have problems since the sizes of the terminating decimals (or whatever) would grow huge. Third, there already is a notion in the DiffEq world, "symplectic integrators" which are intended to preserve certain constants of motion when simulating Hamiltonian systems, albeit I think in a weaker sense that HB had in mind.

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[math-fun] pythagorean rationals, terminating decimals, and complex analysis
by Warren D Smith 12 Sep '13

12 Sep '13

In my sleep last night, I woke up from a dream having proved a theorem, thus proving that I am smarter when asleep than when awake. THEOREM: Let f(z) be an analytic function. Then the set of z such that Re f(z), Im f(z) and |f(z)| all simultaneously are rational, is dense throughout the region of analyticity. PROOF: As math-fun previously discussed ("pythagorean rationals") the set of points (x,y) on the perimeter of the unit circle x^2+y^2=1 with x,y simultaneously rational, is a dense multiplicative subgroup. Now the circle can be scaled up by any rational scaling factor leading to a dense set of such points in the whole x,y plane. That proves the theorem for f(z)=z. Since f() if analytic is a smooth continuous conformal mapping, locally invertible almost everywhere, the result follows. QED EXTENSION: The theorem also is true if all rationals are demanded to be terminating decimals. To see that note that (0.6, 0.8) lies on the unit circle, and all the integer powers of 0.6+0.8i form a dense subset of that circle-perimeter in the complex plane (a "Weyl sequence" in polar coordinates, using known fact arctan(3/4)/pi is irrational). And the circle can be scaled up by any terminating decimal scale factor to get density in the whole plane, etc. The complex numbers z with Re z, Im z, and |z| all simultaneously rational form a dense multiplicative subgroup of the complex numbers (if terminating decimal, then dense subring). What is we ask for terminating BINARY (not decimal)? Then my proof above does not work (thus proving the superiority of decimal versus binary) but the theorem still holds with a different proof, based on the fact you can scale the integer (x,y) with x^2+y^2=w^2, x,y,w integer, by any (positive or negative) power of 2. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] pythagorean rationals, terminating decimals, and complex analysis
by Henry Baker 12 Sep '13

12 Sep '13

One of the potential applications of this sort of thinking is what I might call 'perfect solutions to differential equations'. In particular, I'd like to figure out how to compute exact solutions to a simple DE -- e.g., simple harmonic motion -- such that the usual 'constants of motion' are truly constant. I'm willing to have delta t non-constant -- i.e., I may have to use relatively chaotic delta t's, but I'd like energy to be _perfectly preserved_ and _angular momentum_ to be _perfectly preserved_. It would also be nice if the denominators didn't grow too fast; I'd like to be ablt to pick delta t such that the denominators grow as slowly as possible (or even better, be bounded). At 08:58 AM 9/12/2013, Warren D Smith wrote: >In my sleep last night, I woke up from a dream having proved a theorem, >thus proving that I am smarter when asleep than when awake. > >THEOREM: >Let f(z) be an analytic function. >Then the set of z such that >Re f(z), Im f(z) and |f(z)| >all simultaneously are rational, is dense throughout the >region of analyticity. > >PROOF: >As math-fun previously discussed ("pythagorean rationals") the set of >points (x,y) on the perimeter of the unit circle x^2+y^2=1 >with x,y simultaneously rational, is a dense multiplicative subgroup. >Now the circle can be scaled up by any rational scaling factor >leading to a dense set of such points in the whole x,y plane. >That proves the theorem for f(z)=z. Since f() if analytic is a smooth >continuous conformal mapping, locally invertible almost everywhere, >the result follows. >QED > >EXTENSION: >The theorem also is true if all rationals are demanded to be >terminating decimals. > >To see that note that (0.6, 0.8) lies on the unit circle, and >all the integer powers of 0.6+0.8i >form a dense subset of that circle-perimeter in the complex plane (a "Weyl >sequence" in polar coordinates, using known fact arctan(3/4)/pi is >irrational). >And the circle can be scaled up by any >terminating decimal scale factor to get density in the whole plane, etc. > >The complex numbers z with Re z, Im z, and |z| all simultaneously rational form >a dense multiplicative subgroup of the complex numbers (if >terminating decimal, then dense subring). > >What is we ask for terminating BINARY (not decimal)? Then my proof >above does not work (thus proving the superiority of decimal versus >binary) but the theorem still >holds with a different proof, based on the fact you can scale the integer (x,y) >with x^2+y^2=w^2, x,y,w integer, by any (positive or negative) power of 2. > >-- >Warren D. Smith >http://RangeVoting.org <-- add your endorsement (by clicking >"endorse" as 1st step)

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[math-fun] rationals hard to approximate by rationals
by Warren D Smith 12 Sep '13

12 Sep '13

From: Allan Wechsler <acwacw(a)gmail.com> Is not B=6 unsolvable? The only reduced rationals with denominator 6 are 1/6 and 5/6. 1/6 has continued fraction 0;6. 5/6 has 0;1,5. Both have terms that exceed 2. Have I misunderstood the question? If I have understood correctly, then 9 and 14 are also unachievable. I have an algorithm for listing these, but I haven't implemented it. I'll go into it in more detail if Warren confirms that I have understood the problem correctly. --WDS: what AW says sounds good to me. Indeed see http://oeis.org/A228804 and http://oeis.org/A228803 for the first 1000 terms. It appears from that the problem is solvable for approximately half the numbers B and unsolvable for the other half. Assuming "approximately half" tends to some limiting constant C with 0<C<1 it would be of some interest to know what C is. It also is interesting to ask about useful subsequences of the integers. For example when B is restricted to the subsequence "powers of 2" we find we have solubility when B=1, 2, 4, 8, 64, 128, 256, but no others up to and including 2048. The other half of the problem, which AW has not yet addressed, is given a number B, is there some fast way to find A's such that A/B has only 1s and 2s in continued fraction, and e.g. what is the answer when B=2^32. (I do have a handwavy untested solution to this problem which in theory a lot faster than the most-naive brute force approach, but there might easily be an idea much better than mine.)

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[math-fun] Simple/dumb field theory question
by Bernie Cosell 12 Sep '13

12 Sep '13

My apologies for asking such a dumb/simple question [I feel like I should have to forfeit my MS for even *needing* to ask], but: I'm working through some texts trying to (a) remember what I once knew about surreal numbers, and (b) get a handle on quaternions (which I never knew). Instead of jumping in and just reading the relevant sections, I thought I'd start at the beginning and try to get some of my very old, not much used, math muscles going again. I was reading the first chapter, on fields, and it was all trivial, easy-going..... until I got to thinking about 0 * a = 0 for all a. That isn't actually mentioned as a theorem in the text [but you needed that for a theorem a few pages in and that's what got me thinking about it] The field axioms it uses are [I mention them only because I think there are different sets of axioms that equivalently define fields but these are also the ones I remember from forever ago]: associativity: (a+b)+c = a+(b+c) (a*b)*c = a*(b*c) commutivity: a+b=b+a; a*b=b*a distributivity: a*(b+c) = (a*b)+(a*c) additive identity 0 + a = a multiplicitive id 1 * a = a additive inverse all b, there exists "-b" such that b + "-b" = 0 multipiciticative inverse: all nonzero b. there exists c such that b*c = 1 And so I've been trying to go from there to prove that 0 * a = 0. I started with: 0 * a = (1 + (-1)) * a = 1 * a + (-1) * a = a + (-1)*a Then I run into an odd roadblock: I don't see exactly how to prove: (-1) * a = -a I even have trouble with something similarly simple/obvious like: -b * -a = -(b*a) I figure I must be missing something unbelievably obvious or maybe this is just the wrong approach [or else I'm approaching senility a LOT faster and scarier than I feared]. I appreciate your forbearance...:o) /Bernie\ -- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--

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Re: [math-fun] tribia
by Bill Gosper 11 Sep '13

11 Sep '13

DanA> When I try to evaluate that exact sum in Mma, it gives me only numbers that are to > 20 decimal places equal to 0. (Adding only 1000 or 10000 terms, asking for 10 digits' precision.) I'd guess only a set of measure 0 of real numbers x would have their Sum[Round[x^n]-x^n,{n,0,∞}] <> 0. But it's easy to believe that many algebraic numbers don't. What's your secret? --Dan On 2013-09-10, at 8:11 PM, Bill Gosper wrote: Let t:=1/3 (1 + (19 - 3 Sqrt[33])^(1/3) + (19 + 3 Sqrt[33])^(1/3)), the tribonacci constant. Then, empirically, Sum[Round[x^n]-x^n,{n,0,∞}] Gaa, I meant Round[t^n]-t^n (Mac OS crashed as I prepared to send.) ==-0.80851211604688125... -2 + ((3 Sqrt[3] - Sqrt[11])^(1/3) + (3 Sqrt[3] + Sqrt[11])^(1/3))/(2^(2/3) Sqrt[3]) In[2]:= NSum[Round[GoldenRatio^k] - GoldenRatio^k, {k, 288}, NSumTerms -> 288, WorkingPrecision -> 99] Out[2]= 0.618033988749894848204586834365638117720309179806089199107257598931385927496544945583310480117151672 In[3]:= (% + 1/2)^2 Out[3]= 1.2500000000000000000000000000000000000000000000007297116525363033564609 5964404522000200488900285933 This can't be new. Anybody? --rwg

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[math-fun] tribia
by Bill Gosper 11 Sep '13

11 Sep '13

In[2]:= Round[#] - Floor[#] &[(1/3* (1 + (19 - 3 Sqrt[33])^(1/3) + (19 + 3 Sqrt[33])^(1/3)))^Range[288]] Out[2]= {1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, \ 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, \ 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, \ 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, \ 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, \ 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, \ 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, \ 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, \ 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, \ 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, \ 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, \ 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, \ 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1} In[3]:= Drop[%, 3] - Drop[%, -3] Out[3]= {-1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, \ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \ 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, \ 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, \ 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, \ 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, \ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \ 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, \ 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, \ 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, \ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, \ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \ 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1} In[4]:= Position[Abs[%], 1] Out[4]= {{1}, {3}, {5}, {18}, {31}, {44}, {57}, {70}, {83}, {93}, \ {106}, {119}, {132}, {145}, {158}, {171}, {184}, {197}, {210}, {223}, \ {236}, {249}, {262}, {275}, {285}} In[5]:= Differences[Flatten[%]] Out[5]= {2, 2, 13, 13, 13, 13, 13, 13, 10, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 10} --rwg (I'd explain the scatology of 288, but it's too gross.)

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[math-fun] Fibonachos
by Henry Baker 11 Sep '13

11 Sep '13

Go to this page & click on "Fibonachoes": https://www.facebook.com/IFeakingLoveScience/photos_stream

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[math-fun] Hunting monster primes: Adam Spencer at TED2013
by Ray Tayek 10 Sep '13

10 Sep '13

http://blog.ted.com/2013/02/28/hunting-monster-primes-adam-spencer-at-ted20… --- co-chair http://ocjug.org/

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[math-fun] Solving for magic squares, addendum
by Bill Gosper 09 Sep '13

09 Sep '13

(A while ago I reported here doing this in Macsyma.) (Barely noteworthy, but undeserving of oblivion.) Find the 3x3 msqs just using Solve. The square: In[2]:= mat = {{a, b, c}, {d, e, f}, {g, h, i}}; The row and diagonal sums: In[6]:= lin = 15 == # & /@ Flatten[{Total /@ mat, Total /@ Transpose[mat], Tr[mat], Tr[Reverse[mat]]}] Out[6]= {15 == a + b + c, 15 == d + e + f, 15 == g + h + i, 15 == a + d + g, 15 == b + e + h, 15 == c + f + i, 15 == a + e + i, 15 == c + e + g} In[7]:= vars = Flatten[mat] Out[7]= {a, b, c, d, e, f, g, h, i} lin is 8 (but really 7) eqns, 9 unknowns. How to impose distinct positive integers? Hint, nonlinearly: In[8]:= cub = Total[vars^3] == Sum[n^3, {n, 9}] Out[8]= a^3 + b^3 + c^3 + d^3 + e^3 + f^3 + g^3 + h^3 + i^3 == 2025 In[9]:= quad = Total[vars^2] == Sum[n^2, {n, 9}] Out[9]= a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + g^2 + h^2 + i^2 == 285 In[10]:= forth = Total[vars^4] == Sum[n^4, {n, 9}] Out[10]= a^4 + b^4 + c^4 + d^4 + e^4 + f^4 + g^4 + h^4 + i^4 == 15333 In[6]:= harm = FullSimplify[1/Total[vars^-1] == 1/Sum[n^-1, {n, 9}]] Out[6]= 1/(1/a + 1/b + 1/c + 1/d + 1/e + 1/f + 1/g + 1/h + 1/i) == 2520/7129 (Note, Solve boggles if this equation is reciprocated.) In[15]:= fac = a*b*c*d*e*f*g*h*i == 9! Out[15]= a b c d e f g h i == 362880 etc. E.g.,we get the eight sol'ns if we impose the cubic and quintic, In[15]:= tim[Solve[Join[lin, {quin, cub}]]] During evaluation of In[15]:= 0.032546 Out[15]= {{b -> 1, d -> 3, f -> 7, h -> 9, a -> 8, c -> 6, g -> 4, i -> 2, e -> 5}, {b -> 1, d -> 7, f -> 3, h -> 9, a -> 6, c -> 8, g -> 2, i -> 4, e -> 5}, {b -> 3, d -> 1, f -> 9, h -> 7, a -> 8, c -> 4, g -> 6, i -> 2, e -> 5}, {b -> 3, d -> 9, f -> 1, h -> 7, a -> 4, c -> 8, g -> 2, i -> 6, e -> 5}, {b -> 7, d -> 1, f -> 9, h -> 3, a -> 6, c -> 2, g -> 8, i -> 4, e -> 5}, {b -> 7, d -> 9, f -> 1, h -> 3, a -> 2, c -> 6, g -> 4, i -> 8, e -> 5}, {b -> 9, d -> 3, f -> 7, h -> 1, a -> 4, c -> 2, g -> 8, i -> 6, e -> 5}, {b -> 9, d -> 7, f -> 3, h -> 1, a -> 2, c -> 4, g -> 6, i -> 8, e -> 5}} tim[Solve[Join[lin, {sex, nine}]]] 3.5 sec gives the above, plus 40 solutions with ghastly surds. MinimalPolynomial boggles. Except e always ->5. tim[Solve[Join[lin, {fac, quad}]]] 0.176237 gives the msq sol'ns plus 8 with docile quartic surds. Earlier, we were puzzled that Solve[Join[lin, {quad,cub}]] remained underconstrained: In[29]:= tim[Solve[Join[lin, {cub, quad}]]] During evaluation of In[29]:= 0.020920 Out[29]= {{b -> i - Sqrt[-15 + 10 i - i^2], d -> i + Sqrt[-15 + 10 i - i^2], f -> 10 - i - Sqrt[-15 + 10 i - i^2], h -> 10 - i + Sqrt[-15 + 10 i - i^2], a -> 10 - i, c -> 5 + Sqrt[-15 + 10 i - i^2], g -> 5 - Sqrt[-15 + 10 i - i^2], e -> 5}, {b -> i + Sqrt[-15 + 10 i - i^2], d -> i - Sqrt[-15 + 10 i - i^2], f -> 10 - i + Sqrt[-15 + 10 i - i^2], h -> 10 - i - Sqrt[-15 + 10 i - i^2], a -> 10 - i, c -> 5 - Sqrt[-15 + 10 i - i^2], g -> 5 + Sqrt[-15 + 10 i - i^2], e -> 5}} (Note two sol'ns vs eight.) An "explanation": In[23]:= tim[GroebnerBasis[Subtract @@ # & /@ Join[lin, {quad}], vars]] During evaluation of In[23]:= 0.005090 Out[23]= {115 - 20 h + h^2 - 30 i + 2 h i + 2 i^2, -15 + g + h + i, -20 + f + h + 2 i, -5 + e, 10 + d - h - 2 i, 5 + c - h - i, -10 + b + h, -10 + a + i} In[24]:= tim[GroebnerBasis[Subtract @@ # & /@ Join[lin, {cub}], vars]] During evaluation of In[24]:= 0.003059 Out[24]= {115 - 20 h + h^2 - 30 i + 2 h i + 2 i^2, -15 + g + h + i, -20 + f + h + 2 i, -5 + e, 10 + d - h - 2 i, 5 + c - h - i, -10 + b + h, -10 + a + i} I.e., cub and quad are completely interchangeable in this problem. NeilB says this is because sum(n^3) = sum(n)^2. Sounds plausible, but I need clarification. For the 4x4 msqs, 4+4+2-1 eqns, 16 unknowns leaves too many nonlinear constraints for current technology, even if we add two more linears for the corners and central 2x2. --rwg

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