# Taylor series and radius of convergence: $\sqrt{x}$ with centre $x = 16$?

I've been struggling with this question for a while now and getting nowhere with it. Could someone please help me out?

Assuming that the function has a power series expansion about the given point, find its Taylor series and the radius of convergence:

$f(x) = x^{1/2}$ with centre $x=16$.

• Do you know the general procedure to construct a Taylor series for a given function? Alternatively, you could look up the definition on Wikipedia. For the radius of convergence, you might want to use the ratio test. Nov 27 '15 at 1:22

A Taylor series about $x=a$ is given by
$$f(x) = \sum_{n=0}^\infty \frac{ f^{(n)}(a)}{n!} (x-a)^n$$
where $f^{(n)}(a)$ is the $n$h derivative of $f$ evaluated at $x=a$. The $0th$ derivative is just the function itself. To actually find the series you can try to compute the first several terms given from the series above and see if you discover a pattern. I'll get you started:
$$f(x) = \sum_{n=0}^\infty \frac{ f^{(n)}(16)}{n!} (x-16)^n$$ $$f(x) = \frac{f(16)}{0!} + \frac{f'(16)}{1!}(x-16) + \frac{f''(16)}{2!}(x-16)^2 + \cdots$$