Re: Q: Recursion on a list

*To*: mathgroup at smc.vnet.net*Subject*: [mg28924] Re: [mg28886] Q: Recursion on a list*From*: BobHanlon at aol.com*Date*: Fri, 18 May 2001 01:13:31 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

After some additional thought, it is more efficient to write the function as f[r_List] := FoldList[Times, 1, 1+r]; Bob Hanlon In a message dated 2001/5/17 7:11:43 AM, writes: >The elements of a list are designated by r[[n]], i.e., with Part[r, n], > >not r[n], which is a function r evaluated for the argument n. > >To generate the complete new list > >f[r_List] := FoldList[#1*(1+#2)&, 1, r]; > >However, to calculate an element of the new list without >calculating and storing the whole new list > >g[r_List, n_Integer?NonNegative] := > Times @@ (1+Take[r, n]); > >r = {r1, r2, r3, r4, r5}; > >f[r] > >{1, r1 + 1, (r1 + 1)*(r2 + 1), > (r1 + 1)*(r2 + 1)*(r3 + 1), > (r1 + 1)*(r2 + 1)*(r3 + 1)* > (r4 + 1), (r1 + 1)*(r2 + 1)* > (r3 + 1)*(r4 + 1)*(r5 + 1)} > >g[r, 4] > >(r1 + 1)*(r2 + 1)*(r3 + 1)*(r4 + 1) > >f[r] == Table[g[r, k], {k, 0, Length[r]}] > >True > >For a large list > >r = Table[2Random[]-1, {10000}]; > >Last[f[r]] == g[r, Length[r]]//Timing > >{0.9166666666665151*Second, True} >