Re: newbie: programmatic sequence of plots?

*To*: mathgroup at smc.vnet.net*Subject*: [mg96921] Re: newbie: programmatic sequence of plots?*From*: Tom Roche <tlroche at gmail.com>*Date*: Fri, 27 Feb 2009 06:13:36 -0500 (EST)*References*: <go0j4b$n0o$1@smc.vnet.net> <go31og$fga$1@smc.vnet.net>

Sjoerd C. de Vries Wed, 25 Feb 2009 09:05:27 +0000 (UTC) > Loops themselves only have a null return value. Doh! I was assuming ListPlot would have a "side effect." Albert Retey Wed, 25 Feb 2009 09:06:32 +0000 (UTC) > return a list of these plots additionaly or instead of just > printing them, so that you can use ListAnimation or TabView or > whatever you like for displaying them. And the way to do that is "m.g." Wed, 25 Feb 2009 09:07:28 +0000 (UTC) > Clear[a, y, y0, nRecurr, yinit, yinc, yfinal]; > a = 3.5; > nRecurr = 50; > yinit = 0.1; > yinc = 0.01; > yfinal = 0.2; > out = {}; (* modification here *) > For[y0 = yinit, y0 <= yfinal, y0 += yinc, > AppendTo[out, (* modification here *) > ListPlot[ > RecurrenceTable[{y[n + 1] == a y[n] (1 - y[n]), y[0] == y0}, > y, {n, 1, nRecurr}]]]; > ]; > out (* modification here *) So I modified that to do For[y0 = yinit, y0 <= yfinal, y0 += yinc, AppendTo[out, ListPlot[ RecurrenceTable[{y[n + 1] == a y[n] (1 - y[n]), y[0] == y0}, y, {n, 1, nRecurr}], Joined -> True ] ]; ]; ListAnimate[out] which is even better than printing a list of static plots. Thanks all, Tom Roche <Tom_Roche at pobox.com>