Taylor expansion for $sqrt{x+2}$

I'm enrolled in Coursera's calculus with a single variable and am trying to solve one of the homework problems.

In lecture, it was stated that to expand $\sqrt x$ about $x=a$, you would have:

$$\sqrt{x} = \sqrt{a} + {1 \over 2 \sqrt{a}}(x-a)- {1\over 8 \sqrt{a^3}}(x-a)^2 + H.O.T$$

The homework hint says you can us the Binomial series to find the Taylor series expansion for expressions with non-integer powers.

Wikipedia says the Binomial series expands to
$$(x +1)^{ \alpha }= \sum \limits_{k=0}^{\infty} {\alpha \choose k} x^k$$

$${\alpha\choose{k}} = \frac{\alpha \cdot (\alpha - 1) \cdot (\alpha - 2) \cdot \dots \cdot (\alpha - k + 1)}{k!}$$

My first question is where the term $$a^{1/2 - k}$$ comes from, given the Binomial series formula.

My second question is how to properly evaluate the series about a particular value other than zero.

The homework problem asks me to compute the Taylor series for $$f(x) = \sqrt{x+2}$$ about $x=2$. I also tried to use substitution with $h=x+2$, $x=h-2$ and then compute the Taylor series expansion about h=0 using the definition of Taylor series formula with

$$\sum_{n=0} {{f^{(n)}\over n!}(x-a)^n}$$

$$f(h) = \sqrt{h-2}$$

But with $f(h=0)$, I get imaginary numbers.

Answer

Hint: I am assuming you want to expand $\sqrt{x}$ about $x=2$.

If you want to use the "formula" for the Taylor expansion, you need the derivatives of $\sqrt{x}$ at $x=2$. These derivatives are well-behaved, and you can find an explicit formula for the $n$-th derivative at $x=2$.

If you are allowed to quote the general Binomial Theorem, note that
$$\sqrt{x}=\sqrt{2+(x-2)}=\sqrt{2}\left(1+\frac{x-2}{2}\right)^{1/2}.$$

Then we are looking at $(1+t)^{1/2}$ for $t=\frac{x-2}{2}$.