       Re: Problem with a diagram

• To: mathgroup at smc.vnet.net
• Subject: [mg110839] Re: Problem with a diagram
• From: "David Park" <djmpark at comcast.net>
• Date: Thu, 8 Jul 2010 03:15:24 -0400 (EDT)

```One thing I don't understand is why users so often put an extensive
calculation inside a plot statement, and then if it goes wrong they never
think of taking it out and looking at it.

In this case you could just use Evaluate on the first argument in the
ListPlot. But why not use...

plotpoints =
Flatten[Table[
Transpose[{Table[alpha, {129}],
NestList[f, Nest[f, 0.5, 500], 128]}], {alpha, 0, 0.75, 0.001}],
1]

Now you can see that you obtain a very long list of points suitable for
plotting. Then, make the plot with the pre-computed points.

ListPlot[plotpoints, PlotStyle -> PointSize[0.001], Axes -> False,
Frame -> True]

Among other advantages, if you wanted to change the PlotStyle or other
aspects of the plot, you wouldn't have to recomputed all the points each
time.

David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/

From: Screech [mailto:dildovpizdo at gmail.com]

I am constructing a bifurcation diagram for interval 0.15 <= alpha <= 0.75
but it always go into loop and can't find an error. I used word alpha for
greek symbol.

recursive equation is:
In:= f[x_]:=(1-alpha)x+((alpha*0.8)/0.25)-((alpha*ArcTan[4x])/0.25)
In:= f[x]

I used:

ListPlot[
Flatten[Table[
Transpose[{
Table [alpha, {129}],
NestList[f, Nest[f,0.5,500],128]
}],
{alpha,0,0.75,0.001}
],1],
PlotStyle->PointSize[0.001],Axes->False,Frame->True];