Re: Analyzing sequences of fractions in Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg93712] Re: Analyzing sequences of fractions in Mathematica*From*: "Steve Luttrell" <steve at _removemefirst_luttrell.org.uk>*Date*: Mon, 24 Nov 2008 04:10:29 -0500 (EST)*References*: <200811200955.EAA20567@smc.vnet.net> <gg62qp$580$1@smc.vnet.net>

You were a little coy about exactly how you did it without using Mathematica 7 (perhaps you do these things every day - I don't), so I had a brief think and here is what I guess you might have done: mysequence = {1/8, 3/16, 11/32, 17/64, 25/128, 59/256, 147/512, 265/1024, 465/2048, 995/4096, 2171/8192, 4161/16384, 7881/32768, 16203/65536, 33571/131072, 65977/262144, 129025/524288, 260979/1048576, 529547/2097152, 1051505/4194304, 2083705/8388608, 4186715/16777216, 8423091/33554432, 16796521/67108864, 33466161/134217728, 67059203/268435456, 134443931/536870912}; Partition the numerator sequence using a sliding window of width n (choose n=6, or whatever span you propose for the recurrence relation). mat = Partition[Numerator[mysequence], 6, 1]; Solve for the vector of weights that linearly maps the first n-1 elements of each window onto its last element - this vector defines a linear recurrence relation. LinearSolve[mat[[All, 1 ;; -2]], mat[[All, -1]]] {-8, 4, 2, -1, 2} QED -- Stephen Luttrell West Malvern, UK "Eric W. Weisstein" <eww at wolfram.com> wrote in message news:gg62qp$580$1 at smc.vnet.net... > Steve Luttrell wrote: >> Is your nice solution obtained using the "Integer Sequence Analysis" >> methods >> that have just been added to Mathematica 7 >> (http://www.wolfram.com/products/mathematica/newin7/content/IntegerSequenceAnalysis/)? > > I actually did it by finding the nice recurrence for the denominator > myself, then using RSolve and simplifying a bit by hand. > > But yes, in V7, you can use FindLinearRecurrence[Numerator[mysequence]] > out of the box. > > In[2]:= mysequence = {1/8, 3/16, 11/32, 17/64, 25/128, 59/256, > 147/512, 265/1024, 465/2048, 995/4096, 2171/8192, 4161/16384, > 7881/32768, 16203/65536, 33571/131072, 65977/262144, 129025/524288, > 260979/1048576, 529547/2097152, 1051505/4194304, 2083705/8388608, > 4186715/16777216, 8423091/33554432, 16796521/67108864, > 33466161/134217728, 67059203/268435456, 134443931/536870912}; > > In[3]:= FindLinearRecurrence[Numerator[mysequence]] > Out[3]= {2, -1, 2, 4, -8} > > and then call RSolve on the linear recurrence generated from that kernel. > > Or you can just use FindSequenceFunction directly: > > In[6]:= FullSimplify[FindSequenceFunction[Numerator[mysequence]][n], > n \[Element] Integers] > > Out[6]= (2^( > 1 - n/2) (17 2^(3 n/2) Sqrt[1 + Sqrt[17]] (13 + 5 Sqrt[17]) + > Sqrt[1 + Sqrt[ > 17]] (85 + > 13 Sqrt[17]) (-I^n (-1 + Sqrt[17])^(n/2) + (1 + Sqrt[17])^( > n/2)) Cos[(n \[Pi])/2] + > Sqrt[2] (-I I^n (-1 + Sqrt[17])^( > n/2) (-17 + 23 Sqrt[17]) - (1 + Sqrt[17])^( > n/2) (323 + 75 Sqrt[17])) Sin[(n \[Pi])/2]))/(17 (1 + Sqrt[ > 17])^(3/2) (9 + Sqrt[17])) > > and with a bit of hand simplification, you get the result I posted. > > So you can see that indeed V7 has some nice new functionality which can > be quite useful for analyzing integer (and even polynomial in some > cases) sequences. I'm sure I will be using it quite often. > > Eric >

**References**:**Re: Analyzing sequences of fractions in Mathematica***From:*"Steve Luttrell" <steve@_removemefirst_luttrell.org.uk>