Thread View: sci.math.num-analysis
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Started by Paul Bruckman
Sat, 05 Sep 1998 00:00
Extending continued fractions
Author: Paul Bruckman
Date: Sat, 05 Sep 1998 00:00
Date: Sat, 05 Sep 1998 00:00
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Here is a problem that has been gnawing at me for years. I am giving up on it and I'm going to share it with the rest of you. Consider the recurrence un = an un-1 + bn un-2 , where we are given initial values u0 and u1 and sequences of an's and bn's . We all know that if pn and qn are solutions of the recurrence, with p0 = 1, p1 = a1, q0 = 0, q1 = b1 (and also p-1 = 0, q-1 = 1), then the fractions pn/qn are the convergents of the continued fraction : a1+ b1/a2+ b2/a3+... , which may be terminated at any predetermined point. There is also a matrix analogy to this process, namely if we multiply the 2 X 2 matrices ( ai 1 ) (pn pn-1 ) ( bi 0 ) from i = 1 to n, we get the matrix (qn qn-1 ) . This process can be extended, at least with the matrices and the recurrence. For 3 dimensions, we would have the recurrence un = an un-1 + bn un-2 + cn un-3 , with an, bn and cn given. If pn, qn and rn are solutions of this recurrence, with p-2 = p-1 = 0, p0 = 1 ; q-2 = 0 , q-1 = 1 , q0 = 0 ; r-2 = 1, r-1 = r-0 = 0 (which implies p1 = a1 , q1 = b1, r1 = c1 ), we can also generate the sequences pn, qn and rn as the entries in the following matrix product : (ai 1 0 ) (pn pn-1 pn-2 ) (bi 0 1 ) from i = 1 to n , which yields the matrix (qn qn-1 qn-2 ) (ci 0 0 ) (rn rn-1 rn-2 ) Now comes the $64 million question: what objects comparable to continued fractions can be conjured from the situation for 3 dimensions? Same question for higher dimensions. The extension to such higher dimensions is obvious, as far as the recurrence and the matrix product are concerned. But not so with the continued fraction algorithm. The difficulty seems to be that a fraction is inherently a binary operation operating on two integers, while in the case of 3 or higher dimensions as indicated above, there does not seem to be a "ternary" (or higher) process involving an extension of such fractions. Any and all assistance with this research problem would be appreciated. I am receptive to co-authoring a paper, if this should lead to a productive result. Thanks for your time. Paul Bruckman
Re: Extending continued fractions
Author: MCKAY john
Date: Sun, 06 Sep 1998 00:00
Date: Sun, 06 Sep 1998 00:00
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In article <6srm44$678$1@news-1.news.gte.net> aardwolf@gte.net writes: >Here is a problem that has been gnawing at me for years. I am giving up >on it and I'm going to share it with the rest of you. > >Consider the recurrence un = an un-1 + bn un-2 , where we are given >initial values >u0 and u1 and sequences of an's and bn's . We all know that if pn and >qn are solutions of the recurrence, with p0 = 1, p1 = a1, q0 = 0, q1 = >b1 (and also p-1 = 0, q-1 = 1), then the fractions pn/qn are the >convergents of the continued fraction : > >a1+ b1/a2+ b2/a3+... , which may be terminated at any predetermined >point. > >There is also a matrix analogy to this process, namely if we multiply >the 2 X 2 matrices > >( ai 1 ) (pn pn-1 ) >( bi 0 ) from i = 1 to n, we get the matrix (qn qn-1 ) . > >This process can be extended, at least with the matrices and the >recurrence. For >3 dimensions, we would have the recurrence un = an un-1 + bn un-2 + cn >un-3 , with >an, bn and cn given. If pn, qn and rn are solutions of this recurrence, >with >p-2 = p-1 = 0, p0 = 1 ; q-2 = 0 , q-1 = 1 , q0 = 0 ; r-2 = 1, r-1 = r-0 >= 0 >(which implies p1 = a1 , q1 = b1, r1 = c1 ), we can also generate the >sequences >pn, qn and rn as the entries in the following matrix product : > >(ai 1 0 ) (pn pn-1 pn-2 ) >(bi 0 1 ) from i = 1 to n , which yields the matrix (qn qn-1 qn-2 ) >(ci 0 0 ) (rn rn-1 rn-2 ) > >Now comes the $64 million question: what objects comparable to >continued fractions >can be conjured from the situation for 3 dimensions? Same question for >higher dimensions. The extension to such higher dimensions is obvious, >as far as the recurrence and the matrix product are concerned. But not >so with the continued fraction algorithm. >The difficulty seems to be that a fraction is inherently a binary >operation operating >on two integers, while in the case of 3 or higher dimensions as >indicated above, there >does not seem to be a "ternary" (or higher) process involving an >extension of such >fractions. > >Any and all assistance with this research problem would be appreciated. >I am receptive >to co-authoring a paper, if this should lead to a productive result. >Thanks for your >time. Paul Bruckman > Your question is a good one. I can only refer you to what I know in terms of people: First we have lattice reduction algorithms (best-known is LLL = Lenstra, Lenstra, Lovasz) - here I would recommend looking at the work of H.R.P Ferguson. A person who has axiomatized continued fractions is Rabin at Harvard. A student of V.I. Arnol'd (at Paris 9) is writing a thesis on a general continued fraction algorithm using ideas from topology and geometry. Continued fractions and their generalizations remain a fecund area for research. Good luck! John McKay -- But leave the wise to wrangle, and with me the quarrel of the universe let be; and, in some corner of the hubbub couched, make game of that which makes as much of thee.
Re: Extending continued fractions
Author: G. A. Edgar
Date: Sun, 06 Sep 1998 00:00
Date: Sun, 06 Sep 1998 00:00
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In article <6srm44$678$1@news-1.news.gte.net>, aardwolf@gte.net wrote: > > Now comes the $64 million question: what objects comparable to > continued fractions > can be conjured from the situation for 3 dimensions? A good question. A 3-dimensional analogs of continued fractions are studied in the branch of mathematics called "the geometry of numbers". One source is H. Hancock, "The Development of the Minkowski Geometry of Numbers" (2 volumes) re-printed by Dover. -- Gerald A. Edgar edgar@math.ohio-state.edu
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