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Re: Problem with graphics


On 29 Jan 2004, at 11:34, Lautaro Vergara wrote:

> Dear people,
>
> I have found the following problem. I have ploted
>
> f[x_, y_] := Sqrt[(x^2 + y^2)^3]
>
> with Plot3D in order to see how it looks. Then I ploted f[x, 0] (which
> is x^3) between -1 and 1 and I got a symmetric plot. It seems that
> Mathematica plot the function Sqrt[x^6] or (x^2)^(3/2) and evaluate
> first the symmetric term (x^6 or x^2)...
> Does anyone knows what it happens and how can it be fixed?
>
> Thanks in advance!
>
> Lautaro Vergara
>

The problem is that f[x,0]  is not, as you think, x^3 but

In[2]:=
f[x, 0]

Out[2]=
Sqrt[x^6]


This means that

In[6]:=
f[-1,0]

Out[6]=
1

and not -1. This is not Mathematica's falut but the usual convention in 
mathematics. I suppose what you want is:

f[x_, y_] := PowerExpand[Sqrt[(x^2 + y^2)^3]]

which will produce the graph you are asking for. However, I suggest you 
carefully check tha tis really what you need for this contradicts the 
usual conventions and looks rather dubious to me.


Andrzej Kozlowski
Chiba, Japan
http://www.mimuw.edu.pl/~akoz/


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