3
$\begingroup$

The derivative of $\sqrt{x}$ doesn't have a defined value at x = 0. How then do I find its maclaurin series expansion? Or can it only be approximated with a Taylor series at some value x != 0?

  • 0
    @dragon This might be a crazy thought but take the series around \epsilon>0 and then let it tend to $0$, maybe you get something interesting, maybe you don't.2012-03-27

2 Answers 2

5

It doesn't have a MacLaurin series. It can be expressed as a Taylor series around values of $x\gt0$.

  • 1
    @dragon: The tangent line is the $y$-axis itself. The correspondence between "good" local linear approximations and differentiability breaks down for vertical asymptotes. (The derivative of $\sqrt{x}$, which is $x^{-1/2}$, is undefined at $x=0$ and blows up from the right side.)2012-03-26
4

Since f'(0) does not exist, there is no Maclaurin series at $x=0$, but it does have a Taylor series about any $x>0$. One can use the general binomial theorem to derive the series. For example, $ \begin{align} \sqrt{1+x} &=1+\frac12x-\frac18x^2+\dots+\binom{1/2}{k}x^k+\dots\\ &=1+\sum_{n=1}^\infty\left(-\frac14\right)^{n-1}\binom{2n-2}{n-1}\frac{x^n}{2n} \end{align} $