I am looking for function in Mathematica, which finds all singularities in an expression. To keep it simple, the variable is only one, say x
.
So given expression 1/x
, it will return $\infty$ and given x*y + 1/(x*y)
it will return $x=0,x=\infty,x=-\infty$. Similar to Maple's singular function. "The singular function will return non-removable as well as removable singularities"
Here are some examples:
singular((c - (a + b + 1)*x)/(x*(1 - x)),x);
{x = 0}, {x = 1}
and
singular(exp(1/x),x);
{x = 0}
and
singular(2*x/((x-1)*(2*x-1)),x);
{x = 1}, {x = 1/2}
The expression will always have the variable $x$ in it. It can be rational or not, and it can be basically any valid expression in x. If no singularities exist, then it returns nothing.
I looked, and not able to find this function in Mathematica.
The question is: What would be the closest thing in Mathematica to the above Maple function? This is not exactly like finding poles, since the expression does not have to be rational polynomials. I know any implementation of this function will end up using Solve
at the end?
Reduce[1/Tan[z] == 0, z]
. That strategy will not work for $\exp(1/x)$ and elliptic functions, to use some examples. $\endgroup$ – J. M.'s ennui♦ Jan 22 '17 at 11:14singular
function works for any expression. $\endgroup$ – Nasser Jan 22 '17 at 11:21