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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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Re: [math-fun] Average distance on the unit sphere puzzle
by Dan Asimov 31 May '18

31 May '18

But it's really OK, because we're comparing something like an epsilon neighborhood of the North pole to an epsilon neighborhood of the equator. Even on a 2-sphere the equatorial neighborhood is much bigger. But to sketch a proof of the answer, I used the fact that the average distance can be rigorously calculated as D(n) = 2^n gamma((n+1)/2)^2 / (sqrt(pi) gamma(n+1/2)) and then Stirling's approximation gives lim D(n) = sqrt(2). n—>oo However, this is like the ultimate brute force proof! Is there a far more elegant way to prove this rigorously? —Dan From: Allan Wechsler ----- That is disturbing, isn't it? I still think what I said is true. If you pick any unit vector in N-space as a reference, my intuition is that the dot product of a second, randomly chosen unit vector with the reference vector will follow a distribution more and more tightly centered on 0 as N increases. If this is true, then as Dan points out, for a large enough N, even if you pick a bunch of k different reference vectors, the random vector will tend to be near 90 degrees away from all of them. It may be that N has to increase dramatically to accommodate small increases in k, though. I don't think there is a logical contradiction here. Or my intuition may be completely misguided. Trouble is, there are so many equators ... —Dan ----- ... almost all the hypervolume is near the equator. ----- -----

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Re: [math-fun] Average distance on the unit sphere puzzle
by Dan Asimov 31 May '18

31 May '18

Trouble is, there are so many equators ... —Dan ----- ... almost all the hypervolume is near the equator. -----

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[math-fun] Triangle puzzle
by Dan Asimov 31 May '18

31 May '18

Let * be a point interior to the equilateral triangle T of side 1. Say the lines from the vertices of T through the point * each cut the opposite side in the ratio p:1-p, q:1-q, and r:1-r, respectively, when proceeding in the same sense around the triangle. Puzzle: Express r in terms of p and q. —Dan

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[math-fun] To Allan Re: "I don't find this proof beautiful" quote?
by Dan Asimov 30 May '18

30 May '18

Allan, I just asked essentially the identical question to math-fun a few weeks ago, under a subject line something like "Unsolved number theory problem". There were some interesting responses. —Dan -----Original Message----- >From: Allan Wechsler <acwacw(a)gmail.com> >Sent: May 30, 2018 3:04 PM >To: math-fun <math-fun(a)mailman.xmission.com> >Subject: Re: [math-fun] "I don't find this proof beautiful" quote? > >Is there any progress on the sum-of-three-cubes problem? I think 33 is the >smallest number for which the answer is not known. > >On Wed, May 30, 2018 at 5:29 PM, Mike Stay <metaweta(a)gmail.com> wrote: > >> :D https://cs.uwaterloo.ca/journals/JIS/VOL6/Broughan/broughan25.pdf >> >> On Wed, May 30, 2018 at 2:54 PM, Bill Gosper <billgosper(a)gmail.com> wrote: >> > 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 >> > _______________________________________________ >> > math-fun mailing list >> > math-fun(a)mailman.xmission.com >> > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> >> >> >> -- >> Mike Stay - metaweta(a)gmail.com >> http://www.math.ucr.edu/~mike >> http://reperiendi.wordpress.com >> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> >_______________________________________________ >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] Average distance on the unit sphere puzzle
by Dan Asimov 30 May '18

30 May '18

Let D(n) denote the average distance through R^(n+1) between two points uniformly distributed on the unit n-sphere S^n. E.g., D(1) = 4/π, D(2) = 4/3. Puzzle: Find the limit lim D(n) n—>oo —Dan

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Re: [math-fun] "I don't find this proof beautiful" quote?
by Dan Asimov 30 May '18

30 May '18

My favorite irrationality proof is this which is almost instant if you allow pictures: Suppose there exist integers p, q with (p/q)^2 = 2. Then p^2 = 2 q^2 implies there is a (p, p, q) right triangle by Pythagoras. Draw this (p, p, q) triangle with the hypotenuse as base: Call it ABC with A = right angle. Draw the circle with center at B and radius = p. This cuts the hypotenuse at the point D into lengths p and q-p. Extend a perpendicular from the hypotenuse at D up to point E on side AC. This cuts side AC into pieces of sides q - p and p - (q - p) = 2p - q. Then ADE is a 45-45-90 right triangle with integer sides that are smaller than the original. Which implies an infinite regress of smaller positive integers, contradiction. —Dan 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! -----

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[math-fun] Eliana’s Paradox
by James Propp 30 May '18

30 May '18

My 9-year-old said something yesterday afternoon that I tweeted about, and the tweet blew up like NOTHING I’d ever posted on Twitter before: over 400 likes right now, with more coming in by the minute. So I thought I’d share it with you. Here it is: My kid just told me “I have good news and bad news for you.” It turned out that the good news was that there was no bad news and the bad news was that there was no good news. I’m still trying to figure out how many of these assertions were true. Jim Propp PS: My favorite response was from someone pointing out the relevance of the well-known axiom “No news is good news”.

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