Determine the interval of convergence for the Taylor series about the indicated point for the following function [You do not need to find the Taylor series.]

Question

anonymous · Answer

$$\frac{ 1 }{ x ^{3}+1}$$ a=3

anonymous · Answer

this function is undefined at $x=-1$ so the radius of convergence (i think) will be 4

anonymous · Answer

that's what i thought till i saw this$$x^3=cis 5\pi thus x=\frac{ 1 }{ 2 }+ \frac{ \sqrt{3}j }{ 2 } or -1 or \frac{ 1 }{ 2 }- \frac{ \sqrt{3}j }{ 2 } $$

anonymous · Answer

i could be wrong, but clearly this is undefined at $x=-1$ so if you expand about 3 you cannot go more than 4 away

anonymous · Answer

not sure where it all came from

anonymous · Answer

those are the cubed roots of $-1$

anonymous · Answer

since this is a function of real variables, this has nothing to do with the problem
those are the real and complex solutions to $x^3+1=0$

anonymous · Answer

so are you saying that the complex roots of the function should not be included?

anonymous · Answer

no they should not

anonymous · Answer

thx,

anonymous · Answer

yw