- #1
MMS
- 148
- 4
Hi guys,
Is it right to say that a simple pole (pole of order 1) is a removable singularity (and vice versa)?
Is it right to say that a simple pole (pole of order 1) is a removable singularity (and vice versa)?
No. A pole of order 1 is not removable. Think about the integral of 1/z on the unit circle.MMS said:Hi guys,
Is it right to say that a simple pole (pole of order 1) is a removable singularity (and vice versa)?
A removable singularity is a type of singularity that can be removed or "filled in" to make the function continuous. It occurs when the limit of a function at a certain point exists, but the value of the function at that point does not.
A simple pole is a type of singularity that cannot be removed or made continuous. It occurs when the limit of a function at a certain point does not exist. In other words, a removable singularity can be fixed, while a simple pole cannot.
A removable singularity is typically caused by a factor of (x-a) in the denominator of a function, where 'a' is the point where the function is undefined. This can be seen in functions such as f(x) = (x-2)/(x-2), where x = 2 is a removable singularity.
Yes, a removable singularity can be graphed. It will appear as a hole or gap in the graph at the point where the singularity occurs. This is because the function is undefined at that point, but can still be evaluated as the limit exists.
To determine if a singularity is removable or not, you can evaluate the limit of the function at the point in question. If the limit exists, but the function is undefined at that point, then it is a removable singularity. If the limit does not exist, then it is a simple pole.