Find Taylor series of $\sqrt{x}$ centered at $x=4$ and the order 3 Find Taylor series of $\sqrt x$, about $x=4$ and the order 3
I've tried a few timesm but I keep getting a result that does not comply with the answer. Following are the steps I've taken, hopefully I can get a pointer to the factor I'm missing in some of the denominators. I hope all the steps are clear.
Find derivatives of $\sqrt x$
If $f(x) = \sqrt x$, then:


*

*$f'(x) = \frac{1}{2}x^{-\frac{1}{2}}$

*$f''(x) = -\frac{1}{4}x^{-\frac{3}{2}}$

*$f'''(x) = \frac{3}{8}x^{-\frac{5}{2}}$
Write out the Taylor polynomial
$P_3(x) = \sqrt 4 + \frac{1}{2} 4^{-\frac{1}{2}}(x - 4) -  \frac{1}{4} 4^{-\frac{3}{2}}(x - 4)^2 + \frac{3}{8} 4^{-\frac{5}{2}}(x - 4)^3$
$P_3(x) = 2 + (\frac{1}{2}) (\frac{1}{\sqrt 4})(x - 4) - (\frac{1}{4}) (\frac{1}{4^{\frac{3}{2}}})(x - 4)^2 + (\frac{3}{8})(\frac{1}{x ^{\frac{5}{2}}})(x - 4)^3$
$P_3(x) = 2 + \frac{1}{4}(x - 4) - \frac{1}{4} \frac{1}{4^{2/2} 4^{1/2}}(x - 4)^2 + \frac{3}{8}\frac{1}{4^{4/2}} \frac{1}{4^{1/2}}(x - 4)^3$
$P_3(x) = 2 + \frac{1}{4}(x - 4) - \frac{1}{32}(x - 4)^2 + \frac{3}{256}(x - 4)^3$
But according to the book the series develops like: 
$$P_3(x) = 2 + \frac{1}{4}(x - 4) - \frac{1}{64}(x - 4)^2 + \frac{3}{1536}(x - 4)^3$$
So I'm missing some (increasing) factor, I just can't seem to find it. Any hints?

A Taylor polynomial of degree n centered at $x_0$ is defined as follows:
$$
p_n(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\\
\frac{f'''(x_0)}{3!}(x-x_0)^3+\cdots+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$$
You forgot to include the factorials in your Taylor polynomial.
