August 2015 - mathfun@mailman.xmission.com

[math-fun] Branch puzzle
by Bill Gosper 20 Aug '15

20 Aug '15

Suppose in the now respectable-looking conjectures K(-(t^(1/4)-1/t^(1/4))^2/4)==t^(1/4) K(1-t), 2 E(-(t^(1/4)-1/t^(1/4))^2/4)==t^(1/4) K(1-t)+E(1-t)/t^(1/4), we replace t by t^4 and then foolishly PowerExpand (t^4)^(1/4) -> t, 1/(t^4)^(1/4) -> 1/t: K(-(t-1/t)^2/4)==t K(1-t^4), 2 E(-(t-1/t)^2/4)==t K(1-t^4)+E(1-t^4)/t . Our identities now hold only in the wedge -π/4 < arg(t) ≤ π/4. But it seems that whenever we have an identity so restricted, simply replacing t by t^(1/4) extends validity back to -π < arg(t) ≤ π, i.e., the whole plane. And indeed this substitution recovers our general identity, undoing our reckless PowerExpand. Puzzle: Suppose we only semifoolishly replace (t^4)^(1/4) -> (t^2)^(1/2), 1/(t^4)^(1/4) -> 1/(t^2)^(1/2). That should work in half the plane. Which half? --rwg

1 0

Re: [math-fun] New(?) elliptic K transformation
by Bill Gosper 20 Aug '15

20 Aug '15

On Mon, Nov 5, 2012 at 11:44 PM, Bill Gosper <billgosper(a)gmail.com> wrote: Purely empirically, for 0<r<1/2, EllipticK[4 E^(I ArcTan[1/2 Sqrt[-4 + 1/(1 - r) + 1/r]]) Sqrt[(1- r) r]] == E^(I ArcTan[Sqrt[r/(1 - r)]]) EllipticK[r] E.g., for r=1/4 EllipticK[Sqrt[3] E^(I*π/6)] == E^(I*π/6) EllipticK[1/4] For r=1/2, it gets the conjugate of the right answer, which is EllipticK[2] == -(-1)^(3/4) EllipticK[1/2] which can be found by listing all eight sign variations of the three square roots. Then we can try complex r: EllipticK[1/2 + I/2] == (-3 + 2 Sqrt[2])^(1/4) EllipticK[4 - 2 Sqrt[2]] which puts the lhs in polar form. [Added 2012-11-16 07:03 Namely, EllipticK[1/2 + I/2] == Sqrt[-1 + Sqrt[2]] E^(I π/4) EllipticK[4 - 2 Sqrt[2]] And for r = I, EllipticK[(8 + 8 I) - 4 Sqrt[-1 + 7 I]] == (2^(1/4) E^(-(I*π/8)) - I E^(I*π/4)) EllipticK[I] ] None of the eight sign guesses works for Re(r)>1/2. If no one finds a proof of this stuff, imagine some future Mathematica just trying everything and choosing the numerically plausible one. --rwg No one seemed to recognize it, so, grotesque though it is, NeilB and I decided it's time to write this up. So Neil bashed on it with Mathematica, and it got a simpler. Then I bashed on it so hard that it got unnervingly ungrotesque: (*) EllipticK[1/4 (2 + (-2 + t)/Sqrt[1 - t])] == (1 - t)^(1/4) EllipticK[t] How could anything this simple be new? It looks like a routine 2F1 quadratic transformation. But which? As before, it makes the strange little results EllipticK[1/4] == E^(-((I \[Pi])/6)) EllipticK[Sqrt[3] E^((I \[Pi])/6)] and EllipticK[I] == (E^((I \[Pi])/16) EllipticK[1/2 - 1/4 Sqrt[7/2 - I/2]])/2^(1/8) Are these familiar? And somehow, (*) has miraculously lost its Re[t]<1/2 restriction:. E.g. for t=2, EllipticK[1/2] == (-1)^(1/4) EllipticK[2], as one might hope. For t=(-1)^(1/6), EllipticK[1/4 (2 + (-2 + (-1)^(1/6))/Sqrt[1 - (-1)^(1/6)])] == (2 - Sqrt[3])^(1/8) E^(-((5 I \[Pi])/48)) EllipticK[E^((I \[Pi])/6)] Angling for a K'/K identity, solve 1/4 (2 + (-2 + t)/Sqrt[1 - t]) == 1 - t : t -> 1/32 (31 + 3 I Sqrt[7]) So, Log[EllipticNomeQ[1/32 (31 - 3 I Sqrt[7])]] == -(1/2) Sqrt[1/2 (-3 I + Sqrt[7])] E^((I \[Pi])/4) \[Pi] along with its formal conjugate. Stop me if you've heard this one. --rwg (Is there a (dlmf?) page of special values of nome? Or just garden varieties:-)

2 3

[math-fun] Math history questions: infinite decimals
by James Propp 17 Aug '15

17 Aug '15

It's well known that Simon Stevin came up with the concept of decimal representation. But did he consider infinite decimals? Who first asserted that 1/3 could be written as .333... (and come to think of it, who came up with "point three bar")? Who first considered decimal expansions of irrational numbers? And who first thought about .999...? Thanks, Jim Propp

3 7

[math-fun] Draft of "The Life of Games"
by James Propp 17 Aug '15

17 Aug '15

I've finished a draft of installment #2 of "Mathematical Enchantments", and I'd welcome comments; visit https://mathenchant.wordpress.com/?p=63&shareadraft=55cbbe0fbd26d to see the current draft. You don't need to use the Wordpress feedback box; you can send me your comments by old-fashioned email if you prefer. PLEASE DON'T SHARE THIS LINK WITH ANYONE. I'll publish the essay on Monday. Thanks! Jim Propp

1 1

[math-fun] This almost feels like cheating.
by Bill Gosper 16 Aug '15

16 Aug '15

"The Chowla–Selberg formula gives a formula for a finite product of values of the eta functions. By combining this with the theory of complex multiplication <https://en.wikipedia.org/wiki/Complex_multiplication>, one can give a formula for the individual absolute values of the eta function as [image: \Im(\tau)|\eta(\tau)|^4 = \frac{\alpha}{4\pi\sqrt{|D|}} \prod_r\Gamma(r/|D|)^{\chi(r)\frac{w}{2h}}] for some algebraic number α." where chi(r) = JacobiSymbol[r,D] and " *−D* is the discriminant <https://en.wikipedia.org/wiki/Discriminant> of an imaginary quadratic field <https://en.wikipedia.org/wiki/Quadratic_field>". That algebraic number can often be expressed in terms of ModularLambda[tau]: DedekindEta[(I Sqrt[11])/2] == 1/(2 \[Pi]^(7/8)) (1/ 11 Gamma[1/22] Gamma[3/22] Gamma[5/22] Gamma[9/22] Gamma[15/ 22])^(1/4) (1 - ModularLambda[I Sqrt[11]])^(1/8) \[Eta][E^(-Sqrt[3] \[Pi])] == ( Gamma[1/3]^(3/2) (3 (1 - InverseEllipticNomeQ[E^(-Sqrt[3] \[Pi])]))^(1/8))/(2 \[Pi]) \[Eta][E^(-Sqrt[5] \[Pi])] == 1/(2 5^(3/8) \[Pi]^(3/4)) Sqrt[Gamma[1/20] Gamma[9/20]] (1/ 2 (5 - Sqrt[5]) (1 - InverseEllipticNomeQ[E^(-Sqrt[5] \[Pi])]))^( 1/8) \[Eta][E^(-(\[Pi]/Sqrt[5]))] == (1/(2 \[Pi]^(3/4))) Sqrt[Gamma[1/20] Gamma[9/20]] (1/ 10 (5 - Sqrt[5]) (1 - InverseEllipticNomeQ[E^(-(\[Pi]/Sqrt[5]))]))^(1/8) It's probably not cheating because it doesn't always work. ( ModularLambda[r] = InverseEllipticNomeQ[E^(I*PI*r)]) --rwg

1 1

Re: [math-fun] An irrational number normal to every radix simultaneously
by Keith F. Lynch 15 Aug '15

15 Aug '15

Warren D Smith <warren.wds(a)gmail.com> wrote: > --Andy, your random number algorithm works, but only with > probability 1. But the FACT that it works with prob=1 proves > EXISTENCE of many normal numbers, ... I'd like to return to what I said two days ago: This also works for defining what's meant by the normality of an infinite list of real numbers that are between 0 and 1. Phrased another way, any large excerpt from such a sequence would have approximately equal distribution across the real number line between 0 and 1, and taken pairwise and graphed against each other would have an approximately equal distribution across the unit square, and taken as triples and graphed against each other would have an approximately equal distribution across the unit cube, etc. Such a sequence would generate a normal number in base B if you choose digits by binning the numbers in the sequence into bins of size 1/B. For instance if your sequence was the non-integer parts of (3/2)^N, whose first few terms are: 0.500000, 0.250000, 0.375000, 0.062500, 0.593750, 0.390625, 0.085938, ... in base 10 you'd simply take the first digit of each, to generate the number: 0.5230530... This would also work for base factorial. You'd just use a bin size of 1/N for the Nth digit. But is any infinite sequence of reals known to be normal in that sense? I used (3/2)^N only for illustrative purposes. The non-integer part of successive powers of X is never a normal sequence, at least not if X is real. Can you see why not? Anyhow, it still wouldn't get you a number that's known to be normal in every base, since this binning technique would get you a *different* normal number in each base.

1 0

[math-fun] "exponential" considered harmful
by Henry Baker 15 Aug '15

15 Aug '15

If I hear another TV pundit talk about "exponential" increase in something, when the vast majority of these examples are *polynomial* (usually only quadratic), I'm going to scream. Today, I heard the following nonsequitur: "Do the math. This {whatever] is increasing exponentially ...". Well, I did the math, and it was increasing quadratically, at best. Can we hook an electrode to some lower extremity of each news person/politician to Pavlov them into understanding the true meaning of "exponential" ?

5 5

[math-fun] An irrational number normal to every radix simultaneously
by Warren D Smith 14 Aug '15

14 Aug '15

> 3. Re: An irrational number normal to every radix simultaneously > (Andy Latto) --Andy, your random number algorithm works, but only with probability 1. But the FACT that it works with prob=1 proves EXISTENCE of many normal numbers, and finite versions of same existence proof prove existence of finite chunks that are used by my algorithm to add a new chunk at a time. It was that sort of existence argument I had in mind to fill (what you complained was a gap, lack of proof) to create a genuine proof. The function F(N) had to grow slowly enough, but it seemed clear to me that if it did such a proof would exist. (Sort of, in the "limit" of "very slow growth" of F, we get the continuum limit which Andy Latto had in mind for his random bit algorithm -- which makes it "obvious" that this attack must work.) Turing also had a proof apparently of this vague ilk in mind, though I have not attempted to read or confirm his details. It might be interesting if ways can be found to make some version of this slick and efficient.

1 0

[math-fun] An irrational number normal to every radix simultaneously
by Warren D Smith 14 Aug '15

14 Aug '15

An irrational number normal to every radix simultaneously It seems to me it is possible to write down a computer program which will output the binary bits of such a number, one by one, with at most polynomial(N) time to output the Nth bit. Essentially, the program keeps track of the counts of digits in radix B (among digits "up to the present location") for all B=2,3,...,F(N), which is order F(N)^2 counts in all. (And note, radix 4 and 8 take care of 2- and 3-tuples in radix 2, etc.) The program chooses the pattern for the next lg(N) bits "greedily" from among the N possibilities to maximize some likelihood measure of "normality up to now." Here F(N) is a sufficiently-slow-growing monotonically-increasing-to-infinity continuous positive-valued function of N, for example lg(N)^(1/4) should be fine. Apparently Alan Turing claimed to have solved essentially this problem, using an approach which sounds vaguely similar, in an unpublished manuscript which was eventually printed in his collected works. He must have solved it before world war II. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

2 1

[math-fun] Aaron Siegel
by James Propp 14 Aug '15

14 Aug '15

Does anyone have a current email address for mathematical game theorist Aaron Siegel? His arXiv contact address at IAS no longer works. Thanks, Jim Propp

2 1