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Re: ListPlot: corner problem ??


Drago,

The range of the axes that is shown,  but not the axes origin,  is determine
by PlotRange
Its safer to keep the AxesOrigin

seq = Table[1/n, {n, 4}];;
ListPlot [seq, AxesOrigin -> {0, 0},
  PlotRange -> {{-2, 4}, {-2, 1}}]

Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
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Fax: +44 (0)870 164 0565


Drago Ganic <drago.ganic at in2.hr> wrote in message
news:7o5fq5$rhj at smc.vnet.net...
> Hi there,
>
> When I tried to  plot a sequence ...
>
> seq = Table[1/n, {n, 4}];
> ListPlot [seq, AxesOrigin ->{0,0}]
>
> .. I didn't saw the lower left corner of the picture. Why ?
> I use Mathematica 3.0 and run it from the CD.
>
> I used AxesOrigin because I didn't become the desired picture with:
>     ListPlot[seq]
>
>
> I can get rid of  the "corner-problem" if I use PlotRange:
>
> ListPlot
> [
>  seq,
>  PlotRange->{{0,Length[seq]},{0,1}}
> ]
>
> It's interesting that I don't have to use AxesOrigin now to becamo the
> "desired" picture.
>
>
>
>
>
>



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