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 had forgotten about the Log two entropy approach! 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