[seqfan] Re: The Virtues of X_{n+1}=(4+3*X_{n})/(3+2*X_{n})

Brad Klee bradklee at gmail.com
Sun Sep 17 22:36:29 CEST 2017


Robert,

Thanks, I will try to figure it out. It's worth mentioning that the matrix
approach leads nicely to more proofs:

I. Iterator Matrices, and element sequences for M^n
================================
M_2     |  {{1,2},{1,1}}  | A001333, A000129
M_3     | {{2,3},{1,2}}   | A001075, A001353
M_5     | {{2,5},{1,2}}   | A001007, A001076
M_phi  | {{2,1},{1,1}}   | A001519, A001906

All are invertible, and the elements of powers of these matrices apparently
belong to just few OEIS entries ( nice coverage, nothing missing ! ) .
Appropriate initial conditions can be chosen, then by Dirichlet's
irrationality criteria, we arrive at the following:

II. Quadratic Irrationality Critera, and Decimal Expansions
=======================================================
xi = sqrt(2)   |      dxi * (1 - dxi *(6 - 4 *xi))         <   1/Q^2   |
[-dxi^2] =  0.343 . . .
xi = sqrt(3)   |      dxi * (1 - dxi *(2 - xi))              <   1/Q^2   |
[-dxi^2] =  0.267 . . .
xi = sqrt(5)   |      dxi * (1 - dxi *4*(9 - 4 *xi))     <   1/Q^2   |
[-dxi^2] = 0.222 . . .
xi = phi         |      dxi * (1 - dxi *(2 - xi))              <   1/Q^2
|   [-dxi^2] = 0.381 . . .

For these few examples, the maximum coefficient of [-dxi^2] is for
xi=phi=golden ratio. This seems to recapitulate the idea that the golden
ratio is the most irrational number.

It's also worth mentioning the geometry.

M_2 and M_phi are the inflation matrices for the Ammann-Beenker and the
Penrose aperiodic tilings, respectively. Then we can picture the convergent
fractions as the ratio of different tile types in successively larger
supertiles, which are produced by iteration of the geometric replacement
rules. Many pictures are already available, especially at the tilings
encyclopedia [1].

I don't think that these criteria specifically are in Baake's "Aperiodic
Order Vol. I"; however, on p.37 there is a nice table about cyclotomic
fields with phi(n) L.E. 4. It seems like this table is probably important
to understanding why the cases discussed here have inflation & deflation
matrices over the integers, i.e. {{Z,Z}{Z,Z}}.

As for total generality, the absence of an integer-only-inverse does throw
a wrench into the machinery, because factors of F can--in some examples
do--cancel with denominators in the inverse transform. Then the induction
to prove coprime property becomes inoperable. Again my intuition &
experiments are telling me that there is some sufficient condition, perhaps
odd denominators, which will work even when the inverse leads to a
fractional matrix.

Okay... the speculation and other Faux Pas are regrettable, but at least I
am printing off a few numbers. If you are still wondering about Dirichlet's
criterion, I checked a few references, and the "pigeonhole principle" is
presented on page 1 of Ivan Niven, "Diophantine Approximations", John Wiley
& Sons, 1963. Easy enough to understand, but again, I would have like to
see a drawing, maybe with actual pigeons, ha!

Cheers,

Brad

http://tilings.math.uni-bielefeld.de/




On Sun, Sep 17, 2017 at 12:51 PM, <israel at math.ubc.ca> wrote:

> You have [P_{n+1}, Q_{n+1}]^T = M [P_n, Q_n] where
>
>    [ 3 4 ]
> M = [ 2 3 ]
>
> This matrix has characteristic polynomial lambda^2 - 6 lambda + 1
> so yes, P_n = A005319 and Q_n = A001541.
>
> Cheers,
> Robert
>
>
> On Sep 17 2017, Brad Klee wrote:
>
> Hey Seqfans,
>>
>> It's Friday night somewhere,
>> maybe Saturday morning,
>> Let's do more iteration,
>> an easy-enough proof !
>>
>> X_{n+1}=(4+3*X_{n})/(3+2*X_{n})
>> ======================
>> 4/3, 24/17, 140/99, 816/577, 4756/3363, 27720/19601, 161564/114243,
>> 941664/665857, 5488420/3880899
>>
>> List of Iterator Virtues
>> =============
>> * The rational iterator admits a fixed point at (+/-) sqrt(2), by setting
>> X_{n+1} = X_n and solving.
>>
>> * With X = P/Q, the rational iterator decomposes to integer-only form :
>> ( P_{n+1} , Q_{n+1} )=( 3*P_n + 4*Q_n , 2*P_n + 3*Q_n )
>>
>> * By induction: If P_0 even, and Q_0 odd, then all P_n even all Q_n odd.
>> Vice versa also true.
>>
>> * ( P_0 , Q_0 ) = ( 4 , 3 ) is a good enough initial condition, better
>> than
>> ( 0 , 1 ) in the sense that 0 does not provide any semblance of an
>> approximation to sqrt(2), ha! The iterated sequence seems to recover (
>> A005319, A001541 ); though, my reading of the entries did not rigorously
>> clarify this question.
>>
>> * The transformation is invertible:
>> ( P_n , Q_n ) = (3*P_{n+1} - 4*Q_{n+1},-2*P_{n+1} + 3*Q_{n+1} )
>>
>> * All P_n coprime to Q_n, when P_0 coprime to Q_0. I stumbled over this
>> initially, making a stupid symmetry argument. Actually it's obvious after
>> calculating the inverse transformation. Let's work out the details
>> anyways.  Proof: Assume otherwise. With B coprime to C, and F a
>> hypothesized common factor, (P_{n+1},Q_{n+1})=F*( B , C ) is true if and
>> only if ( P_n , Q_n ) = F*(3*B - 4*C,-2*B + 3*C ), so ( P_n , Q_n ) also
>> have a common factor F. This sets up an induction with base case (4,3),
>> two
>> coprime integers.
>>
>> * Starting with (4/3), it is always possible to write: delta_n = xi - P_n
>> /
>> Q_n, with xi = sqrt(2), and \delta_n > \delta_{n+1} > 0. That is, the
>> iterates approach increasingly close to sqrt(2). Or equivalently, X_{n+1}
>> -
>> X_n > 0 whenever sqrt(2)>X>0, and 4/3 = 1.3... < sqrt(2) = 1.4...
>>
>> * Applying the iteration, delta_{n+1} < (1/Q_{n+1})^2 leads to equivalent
>> condition delta_n*(1 + delta_n*(-6 + 4*xi)) < 1/Q_n^2. If delta_n <
>> (1/Q_n)^2 the condition at n+1 is satisfied by a narrow margin, for:
>> -6+4*xi = -0.3431457505... < 0 ( decimal expansion not in OEIS, ha! ).
>> This
>> instantiates yet another induction, with base case xi-4/3 = 0.08... < 1/9
>> =
>> 0.11... True.
>>
>> * So, the iterates, also irreducible X_n = P_n/Q_n , satisfy Dirichlet's
>> irrationality criterion ( cf. Hardy & Wright, Introduction to the theory
>> of
>> numbers, Ch. XI, Eq. 11.1.1 ). Then, when Hardy and Wright ask: "...how
>> /rapidly/ can we approximate to xi?" We are free to answer: The iterates
>> approach exponentially close to xi, for the recursion is derived as a
>> special case of the elliptic curve addition law on the Jacobi Quartic!
>> More precisely, in the limit around sqrt(2), the invariant differential
>> becomes dt = -dx/(x-sqrt(2)) , and the sequence X_n chooses discrete
>> points
>> from the continuous time evolution x(t) at equal time intervals X_n =
>> x(n*t_0).
>>
>> Caveat Emptor . . . I'm not much of an author, in math, even in physics,
>> nor in poetry. If you think I'm wrong, don't be too quiet.
>>
>> Loose Ends
>> =======
>> * Does this iterator of many virtues produce A005319 / A001541 ?
>>
>> * Did Jacobi or Weierstrass ever argue along these lines ?
>>
>> * Does anyone else care about the picture with the parabolas intersecting
>> at (+/-) sqrt(2) ?
>>
>> * Assuming the proof is already written, will we ever find it in the
>> library?
>>
>> * And there is another secret identity, tending to infinity, and having to
>> do with Ramanujan's work.
>>
>> I think these questions
>> and the last assertion,
>> will wait for another night,
>> maybe another morning.
>>
>> Cheers,
>>
>> Brad
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>>
>>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>


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