June 2018 - mathfun@mailman.xmission.com

[math-fun] On the existence of Leyland primes with a given number of decimal digits
by Hans Havermann 03 Jun '18

03 Jun '18

I've been searching (some ten months now) for Leyland prime numbers with about 100000 decimal digits. Leyland numbers are of the form x^y + y^x, x,y integers, x>=y>1. Primes of this form, of this size, are infrequent but I've got enough of them now to state that (empirically) the chances of there existing an n-digit Leyland prime for a given n ~ 100000 are about 1 in 75. Currently the largest known Leyland prime has 386434 decimal digits, so I got to wondering what the odds might be for the existence of an n-digit Leyland prime with n ~ 400000. My best guess is that they are about 1 in 95 but my math is somewhat contrived and I'm not at all sure of the result. Anybody want to have a go at this?

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Re: [math-fun] An application of chaos?
by Dan Asimov 02 Jun '18

02 Jun '18

A result in the same vein was that the zeta function zeta(x+iy) = Sum_{1 <= n < oo} 1/n^(x+iy) (defined, e.g., as above for x+iy with x > 1 and by analytic continuation everywhere else, except for x+iy = 1, where it's infinite). is sufficiently crazy (if you go far enough away from the x-axis on the critical strip given by {x+iy | 0 < x < 1}) that zeta(x+iy approximates any preassigned analytic function f(x+iy) *somewhere* to any degree of accuracy: (https://en.wikipedia.org/wiki/Riemann_zeta_function#Universality) —Dan >An interesting but probably useless result. > >https://marginalrevolution.com/marginalrevolution/2018/05/one-parameter-equ… > >Brent > >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] An application of chaos?
by Brent Meeker 02 Jun '18

02 Jun '18

An interesting but probably useless result. https://marginalrevolution.com/marginalrevolution/2018/05/one-parameter-equ… Brent

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Re: [math-fun] "I don't find this proof beautiful" quote?
by Bill Gosper 02 Jun '18

02 Jun '18

On 2018-05-29 09:59, Mike Stay wrote: > My favorite irrationality proof is one I heard from John Baez. > Suppose the cube root of two were not irrational; then there would be > two positive integers p, q such that p/q = ∛2. Multiplying both sides > by q and cubing, we get p³ = 2q³ = q³ + q³, which has no solutions in > the positive integers by Fermat's Last Theorem! I just read this aloud to Rohan, who remarked "You can also use that for ∛9.", barely looking up from his video game. --rwg

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[math-fun] What difference could a few bits of precision make?
by Bill Gosper 02 Jun '18

02 Jun '18

Out[42]= Sec[18190586279576483]/18190586279576483 In[43]:= {N@%, N[%, 22]} Out[43]= {-6.53302256185329*10^-17, 2.154106494388454341277} --rwg Does |Sec@n/n| (integer n) have a supremum, or even a simple max? Hopelessly difficult: The minimal p for which Sec@n/n^p has a limit.

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[math-fun] Complementary Analysis (an idiotic construction)
by françois mendzina essomba2 02 Jun '18

02 Jun '18

I remembered one of my stupid constructions of my tender adolescence, being a pure product of the self-taught tradition. Oh, the beautiful time when one walked with one's feet on the sand and one's head in the stars while singing dramatic poems in honor of the ancient Greeks ... Here is an affine application: f (x) = a * x + b I chose two values of this application (x [1], x [2]) to determine a and b: solve([a*x[1]+b=f(x[1]),a*x[2]+b=f(x[2])], [a,b]); we are getting: a=(f(x[1])-f(x[2]))/(x[1]-x[2]), b=-(f(x[1])*x[2]-x[1]*f(x[2]))/(x[1]-x[2]) If we take two other points of this affine application, namely: (x, x [o]), we will have: solve([a*x+b=f(x), a*x[0]+b=f(x[0])], [a,b]); which gives: a=(f(x)-f(x[0]))/(x-x[0]), b=( x*f(x[0])-f(x)*x[0])/(x-x[0]) Noticing the result of a to which we can associate the derived function f '(x[0])=limit((f(x)-f(x[0]))/(x-x[0]), x, x[0]); At the result of b, I associated a new function that I named complementary function. ' f(x[0])=limit((x*f(x[0])-f(x)*x[0])/(x-x[0]), x, x[0]); I wrote the complementary function adding a "t" to differentiate with the derived function. ' f (x) = dt (f (x)) / dx; partial t (f (x, y) / partial (x); General formula: I came back to my affine application: a * x + b = f (x) by replacing a and b by its values: f(x)=x* (f(x)-f(x[0]))/(x-x[0])+ ( x*f(x[0])-f(x)*x[0])/(x-x[0]); I introduce the limits: limit(f(x) , x, x[0])= limit(x* (f(x)-f(x[0]))/(x-x[0]) , x, x[0])+ limit( ( x*f(x[0])-f(x)*x[0])/(x-x[0]) , x, x[0]); We come to the conclusion: f (x [0]) = ' f (x [0]) + x [0] * f ' (x [0]) Leading to the following formula that seems to me to connect: continuity, differentiability and complementarity. f (x) = ' f (x) + x f ' (x) = dtf (x) / dx + x * df (x) / dx A little silly the construction, it must be recognized, but I liked the complementary geometric operator. What is the point??

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[math-fun] Parametrization of an octant of the sphube puzzle
by Dan Asimov 01 Jun '18

01 Jun '18

By analogy with the squircle we define the sphube (pron.: sfyoob) as the surface Sphu given by Sphu = {(x, y, z) in R^3 | x^4 + y^4 + z^4 = 1}. Puzzle: Find a simple explicit parametrization of the first octant of the sphube: U = {(x, y, z) in [0, oo)^3 | x^4 + y^4 + z^4 = 1} by a continuous bijective map h : Q —> U from the square Q = [0, 1]^2 onto U. —Dan

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