diff is not the problem here, subs is. For example, if you compare the
expressions for diff(x, y) and diff(y, x), they are the same. The problem is
that you get terms like (2*x - 3*y**2)/(x**2 + y**2). IMHO, subs({x:0, y:0})
should not even work here, because you get 0/0. The problem is that once it
replaces one term with 0, it cancels the other.
So I am not even convinced that doing subs in a certain order will always
return a correct result.
Ideally, you should be able to compute a as:
In [3]: a = (x**2*y - y**3*x)/(x**2 + y**2)
In [13]: a = Piecewise((a, And(Ne(x, 0), Ne(y, 0))), (0, False))
And when you take the derivative of a, it will recognize the singularity at 0,
0 (or at least the fact that it is defined differently there), and will compute
the limit for it and return another piecewise.
But that does not work right now, partially because And does not work with
Piecewise yet, and partially because
b = Piecewise((a, Ne(x,0)), (a, Ne(y, 0)), (0, True))
Defines correctly, but b.diff(y) does not compute the limit:
In [37]: limit(a/y, y, 0)
Out[37]: 1
Aaron Meurer
On Apr 7, 2010, at 11:54 PM, chu-ching huang wrote:
> Hi,
>
> I know this problem can be treated correctly as above. The example
> given here is just to say that using "diff" has to be more careful
> since
>
> diff(f,x,y).subs([(x, a), (y,b)])
>
> may be not equal to $lim_{(x,y)\to(a,b)} f(x,y)$.
>
> Thanks,
>
> cch
>
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