Re: question about plot
- To: mathgroup at smc.vnet.net
- Subject: [mg69602] Re: [mg69589] question about plot
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Sun, 17 Sep 2006 06:57:27 -0400 (EDT)
- Reply-to: hanlonr at cox.net
plotDisc[args___] := Module[{p},
p = Plot[args, DisplayFunction -> Identity] //.
(Line[{s___, {x1_, y1_}, {x2_, y2_}, e___}]/;
(Abs[(y2-y1)/(x2-x1)] > 10^3 &&
Sign[y1] != Sign[y2])):>
Sequence[Line[{s, {x1, y1}}], Line[{{x2, y2}, e}]];
Show[p, DisplayFunction -> $DisplayFunction]];
h[x_] := x Tan[x];
plotDisc[h[x], {x, 0, 4Pi}];
Bob Hanlon
---- dimmechan at yahoo.com wrote:
> ***Hello to all.
>
> ***I have a simple function
>
> ln[156]:=Quit
>
> ln[1]:=h[x_]:=x Tan[x]
>
> ***Here its Plot
>
> In[4]:=Plot[h[x],{x,0,Pi}]
>
> ***Plot has incorrectly connected the points at p/2.
>
> ***There are two approaches to deal with this bug.
>
> ***The first is to use a Block structure along with Show
>
> ln[6]:=Block[{$DisplayFunction = Identity}, g1 = Plot[h[x], {x, 0,
> Pi/2}];
> ln[7]:=g2 = Plot[h[x], {x, Pi/2, Pi}]];
> Show[g1, g2, DisplayFunction \[Rule] $DisplayFunction]
>
> ***The other is to define your code. Here is my attempt:
>
> In[12]:=
> plotdisc[f_,a_,b_,x0_,opts___]:=Show[Map[Plot[f,{x,#[[1]],#[[2]]},
>
> DisplayFunction\[Rule]Identity,opts]&,Delete[Partition[{a,x0-10^-8,x0+10^-8,b},\
> 2,1],2]],DisplayFunction\[Rule]$DisplayFunction]
>
> In[13]:=plotdisc[h[x],0,Pi,Pi/2,Axes->False,Frame->True]
>
> ***However my plotdisc[] function has the disadvantage that can deal
> with only one
> discontinuity each time.
>
> ***How can I modify it (or if I can't, I look for another more elegant
> code) so that to deal with
> any number of discontinuities.
>
> ***In other words I look for a function (if mine cannot be modified)
> which for example has the same output with
>
> ln[16]:=Block[{$DisplayFunction = Identity}, gg1 = plotdisc[h[x], 0,
> Pi, Pi/2];
> gg2 = plotdisc[h[x], Pi, 2Pi, 3Pi/2]];
> ln[17]:=Show[gg1, gg2, DisplayFunction \[Rule] $DisplayFunction]
>
>
> ***Thanks in advance for any assistance.
>