Re: Plot a simple function

• To: mathgroup at smc.vnet.net
• Subject: [mg74898] Re: Plot a simple function
• From: Roger Bagula <rlbagula at sbcglobal.net>
• Date: Tue, 10 Apr 2007 05:20:31 -0400 (EDT)

```Bob Hanlon wrote:

>H[p_] := -p*Log[2, p] -(1-p)*Log[2, 1-p] /; 0 <= p <= 1;
>
>Plot[H[p],{p,0,1},
>    PlotStyle->{AbsoluteThickness[3],Blue},
>    Frame->True,
>    FrameTicks->{Range[0,1,0.5],Range[0,1,0.5],None,None},
>    FrameStyle->White,
>    FrameLabel->{"Pr(X = 1)","H(X)"},
>    AspectRatio->1,
>    TextStyle->{FontSize->18},
>    GridLines->Table[Table[
>          {x,{AbsoluteThickness[2],GrayLevel[0.9]}},
>          {x,0,1,0.1}],{2}],
>    Prolog->{Black,
>        AbsoluteThickness[2],
>        Line[{{0,1.05},{0,0},{1.05,0} }]}];
>
>
>Bob Hanlon
>
>
>
>
>>
>>
>>
>
>
>
>
Bob Hanlon,
You are on top of things as usual, but the question wanted a 3d approach!

Well it appears there is a third method and it isn't the same either!
I got a sign wrong in my approach, too.
But a very nice Zeta[2] like constant.
Can you solve the logistic to Log two as a power or as a linear factor?
H[p_] := -p*Log[2, p] -(1-p)*Log[2, 1-p] /; 0 <= p <= 1;

x0 = t;
y0 = p;
z0 = H[t]*H[p];
ParametricPlot3D[{x0, y0, z0}, {t, 0, 1}, {p, 0, 1}]

Solve:
p[t_] := h /. Solve[H[t] - y[t]^h == 0, h]
g3= Plot[p[x], {x, 0, 1}, PlotRange -> {{0, 1}, {0, 1}}]
pp[t_] = Log[ -t*Log[2, t] - (1 - t)*Log[2, 1 - t]]/(Log[4] + Log[(1 -
t) t])
g4 = Plot[p[x], {x, 0, 1}, PlotRange -> {{0, 1}, {0, 1}}]
Limit[pp[t], t -> 1/2]
1/Log[4]
N[%]
0.7213475204444817

Roger Bagula

```

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