4
$\begingroup$

I was wondering about Taylor expansions of functions of the form $x^p$, where p is a real number, about $x = 0$. It seems clear how to do it about any other point, but what happens to the series as I approach 0? What specifically does "break down" in the expansion?

As a simple example, I was looking at $y(x) = (1-x^3)^{1/3}$, where the cube root has the usual property that $(-x)^{1/3} = -x$. When I look at the point $x=1$ and the first derivative there, I get division by zero. The function for $x = 1+\epsilon$ looks like $-3^{1/3} \epsilon^{1/3}$, hence I get $-3^{-2/3} \epsilon^{-2/3}$ for the derivative. How to deal with that?

Thanks for your help. SSF

$\endgroup$
2

0

You must log in to answer this question.

Browse other questions tagged .