Re: very basic RecurrenceTable puzzle
- To: mathgroup at smc.vnet.net
- Subject: [mg130720] Re: very basic RecurrenceTable puzzle
- From: Bob Hanlon <hanlonr357 at gmail.com>
- Date: Mon, 6 May 2013 04:22:53 -0400 (EDT)
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I don't know the internal workings of RecurrenceTable so I don't know how the attribute affects its behavior, but I noted the difference in the attributes and pointed it out. You could use If like this Clear[st] st[-1] = 0; st[t_?NonNegative] := If[t < 2, st[t - 1], 1]; Table[st[t], {t, -1, 5}] {0, 0, 0, 1, 1, 1, 1} or with memory Clear[st] st[-1] = 0; st[t_?NonNegative] := st[t] = If[t < 2, st[t - 1], 1]; Table[st[t], {t, -1, 5}] {0, 0, 0, 1, 1, 1, 1} Bob Hanlon On Sat, May 4, 2013 at 4:28 PM, Alan G Isaac <alan.isaac at gmail.com> wrote: > On 5/4/2013 8:01 AM, Bob Hanlon wrote: > >> Note that If has attribute HoldRest rather than the HoldAll for Piecewise. >> > > > First, thanks for the suggestion to use Piecewise. > > But ... should I have been able to deduce from that > attribute difference the difference in the output? > If so, might you add a word or two about that? > > Thanks, > Alan Isaac > >
- References:
- very basic RecurrenceTable puzzle
- From: Alan <alan.isaac@gmail.com>
- very basic RecurrenceTable puzzle