Grothard
Dec5-11, 01:11 PM
Let f(z) be a function that is analytic for all |z|≤1, with the exception of z_0, which lies on the circle |z|=1. f(z) has a first order pole at z_0. If Ʃ a_n z^n is the Maclaurin expansion of the function, then z_0 = lim_{n \to \infty} \frac{a_n}{a_{n+1}}.
I have failed to show that this is true. I think the key might lie in the fact that there exist points in the radius of convergence of the Maclaurin series AND in the radius of convergence of the Laurent series about z_0, allowing us to set the two series equal to each other and fix a z value. I can't manage to separate the z_0 term in that equation, though.
Any thoughts? Is this a well-known property?
I have failed to show that this is true. I think the key might lie in the fact that there exist points in the radius of convergence of the Maclaurin series AND in the radius of convergence of the Laurent series about z_0, allowing us to set the two series equal to each other and fix a z value. I can't manage to separate the z_0 term in that equation, though.
Any thoughts? Is this a well-known property?