Re: Re: How can I get the previous Fibonacci calculated

*To*: mathgroup at smc.vnet.net*Subject*: [mg100397] Re: [mg100312] Re: How can I get the previous Fibonacci calculated*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Tue, 2 Jun 2009 06:47:54 -0400 (EDT)*Reply-to*: hanlonr at cox.net

Fibonacci number is not restricted to Integer arguments Fibonacci[Pi] // N 2.11703 Fibonacci[Pi] == Fibonacci[Pi - 1] + Fibonacci[Pi - 2] // FullSimplify True prevFib[x_] := Module[{m, n, x2}, x2 = SetPrecision[x, 120]; If[x > 10^15, N[x/GoldenRatio], m = (n /. FindRoot[Fibonacci[n] - x2 == 0, {n, 100}, WorkingPrecision -> 100]) - 1; If[Abs[m - Round[m]] < 10^-10, Fibonacci[Round[m]], N[Fibonacci[m]]]]] Table[x = 10^n*Pi; prevFib[Fibonacci[x]] == Fibonacci[x - 1], {n, 0, 2}] {True,True,True} Fibonacci is not even restricted to real arguments, although prevFib will not handle complex arguments. Fibonacci[Pi + E*I] // N -45.13963322822449 - 246.04041318893496*I Fibonacci[Pi + E*I] == Fibonacci[Pi + E*I - 1] + Fibonacci[Pi + E*I - 2] // FullSimplify True Bob Hanlon ---- "David W. Cantrell" <DWCantrell at sigmaxi.net> wrote: ============= Bob Hanlon <hanlonr at cox.net> wrote: > prevFib[x_] := Module[{m, n, x2}, > x2 = SetPrecision[x, 100]; > If[x > 10^15, > N[x/GoldenRatio], > m = (n /. FindRoot[Fibonacci[n] - x2 == 0, {n, 100}, > WorkingPrecision -> 100]) - 1; > If[Abs[m - Round[m]] < 10^-10, > Fibonacci[Round[m]], > N[Fibonacci[m]]]]] > > prevFib[13] > > 8 > > prevFib[Fibonacci[50]] == Fibonacci[49] > > True > > prevFib[Fibonacci[10000]] == Fibonacci[9999] > > True > > prevFib[Fibonacci[20000]] == Fibonacci[19999] > > True > > prevFib[234.34*^10] > > 1.4483008492365037*^12 But that's not a Fibonacci number! I suggest In[7]:= n = 234.34*^10; inv = Round[Log[GoldenRatio, Sqrt[(Sqrt[5]*n + Sqrt[5*n^2 - 4])* (Sqrt[5]*n + Sqrt[5*n^2 + 4])]/2]]; Fibonacci[inv - 1] Out[7]= 1548008755920 I am guessing that the OP actually wants to find the Fibonacci number which immediately precedes a given _Fibonacci number_. If that guess is correct, that is, if n is actually supposed to be a Fibonacci number, then my inv will always work correctly for n >= 2. But CAUTION: If n is not a Fibonacci number, then my method may give an incorrect answer. In the example, 234.34*^10 is not a Fibonacci number and so there was no guarantee that my method would give the correct result, although it happened to do so. David W. Cantrell > ---- ratulloch at gmail.com wrote: > > ============= > Greetings > > How can I get a previous Fibonacci calculated number in a specific > range. I can calculate the previous number of phi see below > > phi1 =(1+sqrt(5))/2 > > 13/phi1 > > I tested it with 13 and got 8.034441857 .....so that works > > but how can I get only the previous numbers of Fibonacci in a > specified range from 20 to 20000 if I input a large number like > 234.34e10 > > tia sal2 -- Bob Hanlon