Re: Re: functions

*To*: mathgroup at smc.vnet.net*Subject*: [mg45735] Re: [mg45525] Re: [mg45481] functions*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Thu, 22 Jan 2004 03:37:36 -0500 (EST)*References*: <200401100500.AAA02409@smc.vnet.net> <200401120715.CAA29358@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Selwyn Hollis wrote: > > This is a pretty wild idea, but how about the following as a criterion > for deeming a single-precision floating-point number between 0 and 1 to > be "irrational": It's irrational if its base-10 form contains all 10 > digits, more precisely, if > > IrrationalQ[x_] := Length[Union[IntegerDigits[Round[10^16*x]]]] == 10 > > returns True. Then > > grph = Plot[If[IrrationalQ[x], x, 0], {x, 0, 1}, PlotPoints -> 100] > > draws a graph that, with a little imagination, suggests what the graph > of > If[x is irrational, x, 0] could look like. > > Still better is > > Show[Graphics[{AbsolutePointSize[1.5], Point[#]} & /@ > (grph/.Graphics[{{Line[z_]}}, ___] -> z)]] > > This is all nonsense, of course, but fun nonetheless. > > Experiments show that IrrationalQ[Random[]] returns True with > probability roughly 0.07. (Anybody have a clue why?) > > ----- > Selwyn Hollis > http://www.math.armstrong.edu/faculty/hollis > [...] As to that experimental probability... If you ignore the case where the leading digit is zero and disappears, the probability is modelled as follows. Let p[j] denote the probability that the first j digits contain all 10 digits (so obviously p[j]=0 for j<10). We have p[10] = Product[j/10, {j,1,9}]; Now p[11] = p[10] + probability that we need exactly 11 digits to get all 10 possibilities. This means exactly one of the first ten digits replicates an earlier digit. This arises if the second digit replicates the first (probability 1/10), or the third replicates the second or first (probability 2/10), etc. In Mathematica this would be given by Sum[k[1]/10,{k[1],9}] * p[10] We need exactly 12 digits if exactly two of the first eleven digits replicate earlier digits. This is given by Sum[k[1]/10*k[2]/10, {k[1],9}, {k[2],k[1],9}] * p[10] In general we have: productsum[j_Integer] := Module[ {k, vars, prod, iters}, vars = Array[k,j-10]; k[0] = 1; prod = Apply[Times,vars]/10^(j-10); iters = Map[{#,#[[0]][#[[1]]-1],9}&,vars]; Sum[prod,Evaluate[Apply[Sequence,iters]]] ] (I'm sure this is not the cleanest way to code this). p[j_Integer /; j>10] := p[j-1] + productsum[j]*p[10] In[59]:= InputForm[p16 = p[16]] Out[59]//InputForm= 10985907933/156250000000 In[60]:= N[p16] Out[60]= 0.0703098 This pretty well replicates your experimental result. By the way, those random decimals are double precision floats, not single. Daniel Lichtblau Wolfram Research

**References**:**functions***From:*lorenzo.keegan@handbag.com

**Re: functions***From:*Selwyn Hollis <sh2.7183@misspelled.erthlink.net>