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

[israel at math.ubc.ca]

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
>
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>
>