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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• References: <20130504071941.9444A6A2F@smc.vnet.net>

```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
>
>

```

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