Question

TRICK: FIND RADIUS when a function of the form f= p/q is expanded in powers of (x−a), the difference between the nearest 0 and a gives the radius. Ex: f= 1/(x(1−x)) expanded in powers of x− 2/3.

a) 1/3
​b) 2/3
​c) 1/2
d) 3/4
​

Answer 1

The radius of convergence for the function 1/(x(1-x)) expanded at x=2/3 is the distance to the nearest singularity from 2/3, which is 1/3.

Step-by-step explanation:

The question revolves around finding the radius of convergence of a function expanded in a power series around a given point a. Specifically, the function f = p/q is expanded in powers of (x-a), and the radius is the difference between the nearest singularity (where the function is not defined) and the point a. In the example provided, the function f = 1/(x(1-x)) is expanded at x = 2/3. This function has singularities (points where the function is not defined) at x = 0 and x = 1. Since we're expanding around x = 2/3, the nearest singularity is at x = 1. So, the radius of convergence is 1 - 2/3 = 1/3. Therefore, the correct answer is a) 1/3.