Comment #22 on issue 1694 by asmeurer: solve has many issues with fractions
http://code.google.com/p/sympy/issues/detail?id=1694
It's called undefined. Mathematicians never run into this problem when
they do things formally because they
define a function by its domain, codomain, and rule, not just implicitly by
the rule as is commonly done. So
f:R\{-1}->R defined as f(x) = (x-1)**2/(x+1) does not have a root at -1
because it is not defined at -1! Because
this is a so-called removable singularity (the limit exists at that point),
you could define another function g:R-
> R as g(x) = f(x) if x != -1 and 0 if x == -1. Because limit(f(x), x, -1)
> == g(-1), g will be a continuous
extension of f and will have a root at -1. Incidentally, in this case, you
could also define g(x) = x+1 and it would
be easy to show that this would define the same function. So, I guess the
moral is that a discontinuity doesn't
have to be an asymptote. It can also be a "hole" in the function or even
something stranger like 0 for sin(1/x).
By the way, we do have a limit() function that works pretty well when the
limit exists. If people would really want
to know where there are removable singularities that would be zeros in the
continuous extension as Chris says,
we could check if the limit of the function at the false zero is equal to 0
and somehow return that separately if
the user wants it. What do you think? Do other CAS's have options like
this that you know of?
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