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[math-fun] How can Mathematica not have Mediant?
by Bill Gosper 03 Nov '19

03 Nov '19

To get all "best" rational approximations to a real, instead of continued fractions, use Mediant: Mediant[a_] := a; Mediant[a__] := Divide @@ (Total /@ {Numerator[{a}], Denominator[{a}]}) (Just sum of numerators over sum of denominators.) Then we can, e.g., In[62]:= mediate[π, 10] Out[62]= {0, 1, 2, 3} | {22/7, 19/6, 16/5, 13/4, 10/3, 7/2, 4, 1/0} where the fractions are in numerical order and improve toward the "|". In[60]:= mediate[√2, 6] Out[60]= {0, 1, 4/3, 7/5} | {10/7, 3/2, 2, 1/0} In[41]:= mediate[√2, 12] Out[41]= {0, 1, 4/3, 7/5, 24/17, 41/29, 140/99} | {99/70, 58/41, 17/12, 10/7, 3/2, 2, 1/0} (I confess to being unable to convince Mathematica to say 1/0 vs ComplexInfinity.) Here's mediate: Unprotect[Numerator, Denominator]; Numerator[ComplexInfinity] = 1; Denominator[ComplexInfinity] = 0; Off[Power::infy] (* 1/0 is an important rational number! *) mediate[x_?NumericQ, ct_?NumericQ /; ct > 0, lo_ : {0}, hi_ : {ComplexInfinity}] := Piecewise[{{mediate [x, ct - 1, Append[lo, #], hi], # <= x}}, mediate[x, ct - 1, lo, Prepend[hi, #]]] &@ Mediant[lo[[-1]], hi[[1]]] /; ct >= 1 mediate[_, _, lo_, hi_] := lo | hi (Julian, is there a cool way to avoid repeating "mediate [x, ct - 1" with Sequence?) Pushing harder on π, In[39]:= mediate[π, 26] Out[39]= {0, 1, 2, 3, 25/8, 47/15, 69/22, 91/29, 113/36, 135/43, 157/50, 179/57, 201/64, 223/71, 245/78, 267/85, 289/92, 311/99, 333/106} | {355/113, 22/7, 19/6, 16/5, 13/4, 10/3, 7/2, 4, 1/0} so we see that 22/7 and 355/113 are both overestimates. And we see the drawback of not using continued fractions. The "15" in π = {3,7,15,1,292,...} caused 15 consecutive slowly converging underestimates. Soon, we shall face 292 more! To see how easy it is to perform mediate by hand, arithmetic with boxes <http://gosper.org/rectarith12.pdf> . —rwg

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[math-fun] Fwd: Unlikely looking Sin[π/22]
by Bill Gosper 01 Nov '19

01 Nov '19

More puzzles for FullSimplify: Sin[π/22] == 1/10 + √11 (Cos[1/5 ArcTan[5/Sqrt[877 + 1958/√5]]] - Cos[π/5 - 1/5 ArcTan[5/109 Sqrt[4385 + 1958 √5]]])/5 == 1/10 + 2/5 √11 Sin[1/10 (π- ArcTan[5/199 Sqrt[71045 - 26378 √5]])] Sin[ 1/10 (π - ArcTan[5/199 Sqrt[71045 + 26378√5]])] "Explanation": Sin[π/22] satisfies a (solvable) quintic. But isn't π/22 preferable to ArcTan[...]/5? Not if you want the radicals. Whatever happened to Mathematica's quintic solver? I have something that seems to solve (the solvable) sextics and septics as well. —rwg

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[math-fun] Orphan numbers
by Éric Angelini 31 Oct '19

31 Oct '19

Hello Math-Fun, Let's call K the kid of integer J. K is the concatenation of the successive absolute differences of J's digits: For J = 2019, we have K = 218. Note that 22019 has the same "218" kid because we don't accept the possible leading 0s of K. Orphan numbers are integers with no father, like 919 or 68955. Is a list of such orphan numbers already in the OEIS? Not sure... Best, É.

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[math-fun] Orphan numbers typo
by Éric Angelini 31 Oct '19

31 Oct '19

68955 has at least one father: 719050 But 68956 has none -- sorry. Best, É.

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[math-fun] q-numbers are actually fun.
by Bill Gosper 29 Oct '19

29 Oct '19

(Kids: I highly recommend grokking this. Ask me anything.) Years ago I wrote WRI a package of q-utilities, e.g. QPochhammer, QFactorial, QBinomial, ... . One of the items in the spec was QNumber, which is just, e.g., In[120]:= qNumber[4, -q^2] Out[120]= 1 - q^2 + q^4 - q^6 I.e., replace the number n by n terms of the geometric series with the specified term ratio. It seemed sort of useless and trivial. I was a bit relieved when they didn't install it. I was wrong. Let's play. Exercise: Expand In[121]:= qNumber[5, q] qNumber[5, -q] Out[121]= (1 - q + q^2 - q^3 + q^4) (1 + q + q^2 + q^3 + q^4) in your head! SPOILER: StringReverse@"]2^q,5[rebmuNq" Why? Nice Mathematica exercise: Inverse qNumber. E.g., In[125]:= unqNumber /@ %121 Out[125]= Inactive[qNumber][5, -q] Inactive[qNumber][5, q] Try it! Here's a way, if you give up: unqNumber[p_] := Module[{L = List @@ Together@p}, Inactive[qNumber][Length@L, L[[2]]] /; PolynomialQ[p, Variables[p][[1]]]&&SameQ @@ (Drop[L, 1]/Drop[L, -1])] If you study the following factorizations, you should formulate enough theorems to mentally factorize any q-number whose q➝1 you can mentally factorize. True or False: the factors are always q-numbers. 0 0 1 1 2 1+q 3 1+q+q^2 2^2 (1+q) (1+q^2) 5 1+q+q^2+q^3+q^4 2 3 (1+q) (1-q+q^2) (1+q+q^2) 7 1+q+q^2+q^3+q^4+q^5+q^6 2^3 (1+q) (1+q^2) (1+q^4) 3^2 (1+q+q^2) (1+q^3+q^6) 2 5 (1+q) (1-q+q^2-q^3+q^4) (1+q+q^2+q^3+q^4) 11 1+q+q^2+q^3+q^4+q^5+q^6+q^7+q^8+q^9+q^10 2^2 3 (1+q) (1+q^2) (1-q+q^2) (1+q+q^2) (1-q^2+q^4) 13 1+q+q^2+q^3+q^4+q^5+q^6+q^7+q^8+q^9+q^10+q^11+q^12 2 7 (1+q) (1-q+q^2-q^3+q^4-q^5+q^6) (1+q+q^2+q^3+q^4+q^5+q^6) 3 5 (1+q+q^2) (1+q+q^2+q^3+q^4) (1-q+q^3-q^4+q^5-q^7+q^8) 2^4 (1+q) (1+q^2) (1+q^4) (1+q^8) 17 1+q+q^2+q^3+q^4+q^5+q^6+q^7+q^8+q^9+q^10+q^11+q^12+q^13+q^14+q^15+q^16 2 3^2 (1+q) (1-q+q^2) (1+q+q^2) (1-q^3+q^6) (1+q^3+q^6) 19 1+q+q^2+q^3+q^4+q^5+q^6+q^7+q^8+q^9+q^10+q^11+q^12+q^13+q^14+q^15+q^16+q^17+q^18 2^2 5 (1+q) (1+q^2) (1-q+q^2-q^3+q^4) (1+q+q^2+q^3+q^4) (1-q^2+q^4-q^6+q^8) 3 7 (1+q+q^2) (1+q+q^2+q^3+q^4+q^5+q^6) (1-q+q^3-q^4+q^6-q^8+q^9-q^11+q^12) 2 11 (1+q) (1-q+q^2-q^3+q^4-q^5+q^6-q^7+q^8-q^9+q^10) (1+q+q^2+q^3+q^4+q^5+q^6+q^7+q^8+q^9+q^10) 23 1+q+q^2+q^3+q^4+q^5+q^6+q^7+q^8+q^9+q^10+q^11+q^12+q^13+q^14+q^15+q^16+q^17+q^18+q^19+q^20+q^21+q^22 2^3 3 (1+q) (1+q^2) (1-q+q^2) (1+q+q^2) (1+q^4) (1-q^2+q^4) (1-q^4+q^8) 5^2 (1+q+q^2+q^3+q^4) (1+q^5+q^10+q^15+q^20) 2 13 (1+q) (1-q+q^2-q^3+q^4-q^5+q^6-q^7+q^8-q^9+q^10-q^11+q^12) (1+q+q^2+q^3+q^4+q^5+q^6+q^7+q^8+q^9+q^10+q^11+q^12) 3^3 (1+q+q^2) (1+q^3+q^6) (1+q^9+q^18) —rwg

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[math-fun] primoids. An elementary primes lesson ends in a bit of a surprise.
by Bill Gosper 27 Oct '19

27 Oct '19

primoids <http://gosper.org/primoids.pdf> —rwg

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[math-fun] Chunk & Sum (minimal rule)
by Éric Angelini 26 Oct '19

26 Oct '19

Hello Allan, I have found -- thanks to a remark made to me by Frank Stevenson in a private mail -- the "minimal pieces" rule I was trying to explain. Alice always wants to check if there is a sum that is sound (according to the first rule -- increasing chunks) _and_ inferior to the sum claimed by Bob -- when considering Bob's arrangement. This is sometimes difficult to see. Best, and thanks to Frank, É.

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[math-fun] "Self-describing" integer tuples.
by Allan Wechsler 26 Oct '19

26 Oct '19

I am sure this has been discussed here before, so if somebody just has a pointer to any previous discussion, let me know. Suppose we have a finite sequence of non-negative integers. We can create a new sequence that is a "census" of the first. For example, the census of (1,4,1,4,2,1,3) is (3,1,1,2,1,3,2,4) -- that is, three 1's, one 2, one 3, and two 4's. I am interested in any tuple that is its own census. A simple example is (2,2); a more complicated example is (2,1,3,2,2,3,1,K). The K can be replaced with any number bigger than 3. I found another class of examples of the form (K,1,3,2,2,3,2,K,1,A,1,B...) where K,A,B... are distinct from each other and bigger than 3, and there are exactly K-1 terms in the A,B... set. Let's declare that the choice of the value of singletons doesn't matter. I have an intuition that the number of types of self-censusing sequences is quite small -- maybe only the three types I just mentioned? Or did I miss a few? I would not be surprised if I did, but I would be surprised if there were a _lot_ more types.

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[math-fun] The "bottom row" of q-Pascal's triangle
by Bill Gosper 26 Oct '19

26 Oct '19

makes sense because successive rows approach a limit, which is experimentally Inactive[Table][1/QPochhammer[q,q,k],{k,0,∞}], and shouldn't be hard to prove. For finite segments, you should just be able to say Table[QBinomial[∞,k],{k,0,...}], but Mathematica returns nonsense. —rwg

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[math-fun] Mathematician Solves 48-Year-Old Problem, Finds New Way To Multiply
by Ray Tayek 23 Oct '19

23 Oct '19

https://science.slashdot.org/story/19/10/20/164235/mathematician-solves-48-… An anonymous reader quotes Popular Mechanics: /An assistant professor from the University of New South Wales Sydney in Australia has developed a new method for multiplying giant numbers together that's more efficient <https://www.popularmechanics.com/science/a29514208/faster-way-multiply/> than the "long multiplication" so many are taught at an early age. "More technically, we have proved a 1971 conjecture of Schönhage and Strassen about the complexity of integer multiplication," associate professor David Harvey says in this video <https://youtu.be/2w-lDOQfI5s>... Schönhage and Strassen predicted that an algorithm multiplying /n/-digit numbers using /n/ * log(/n/) basic operations should exist, Harvey says. His paper is the first known proof that it does... The /[original 1971]/ Schönhage-Strassen method <https://en.wikipedia.org/wiki/Sch%C3%B6nhage%E2%80%93Strassen_algorithm> is very fast, Harvey says. If a computer were to use the squared method taught in school on a problem where two numbers had a billion digits each, it would take months. A computer using the Schönhage-Strassen method could do so in 30 seconds. But if the numbers keep rising into the trillions and beyond, the algorithm developed by Harvey and collaborator Joris van der Hoeven at École Polytechnique in France could find solutions faster than the 1971 Schönhage-Strassen algorithm. "It means you can do all sorts of arithmetic more efficiently, for example division and square roots," he says. "You could also calculate digits of pi more efficiently than before. It even has applications to problems involving huge prime numbers./ "The question is, how deep does /n/ have to be for this algorithm to actually be faster than the previous algorithms?" the assistant professor says in the video. "The answer is we don't know. "It could be billions of digits. It could be trillions. It could be much bigger than that. We really have no idea at this point." -- Honesty is a very expensive gift. So, don't expect it from cheap people - Warren Buffett http://tayek.com/

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