Recent content by BreakaZ

Im not following. Where do I get the interval and radius of convergence from that?

Ive been working on this problem for a while now and this is what I have gotten so far: f(x) = ln(3-x) f(2)= 0 f`(x) = (3-x)^-1 f`(2)= 1 f''(x) = -(3-x)^-2 f''(2)= -1 f'''(x) = 2(3-x)^-3 f'''(2)= 2 taylor series f(x)= 0 + (x-2) - (x-2)^2/2! +...

do I simply plug in the a=2 for n to get the results i want? Ʃ (x^n+1)/(n+1)(3^n+1) where n=0 before, but n=2 now. this would give an interval of ±(81)^1/3 cuberoot of 81 is this right?

so i should make a taylor series of f(a) for f(x)=ln(3-x) and solve it that way? im confused, I am not sure what to do

ƩHomework Statement Determine the radius of convergence and the interval of convergence for the follwing function expanded about the point a=2. f(x)= ln(3-x) Homework Equations ln(1-x) = Ʃ (x^n+1)/n+1 n=0 which has radius of convergence at |x|<1 The Attempt at a Solution...