<s>[INST] 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.
 [/INST] This is what I have now:
Evaluating derivatives for $\sqrt{x+2}$ at x=2, I get:
$ f(0) = \sqrt{2+ (2)} = 2 $
$ f'(0) = {1\over2} {1 \over \sqrt{2+(2)} }= 1/4$
$ f''(0) = -{1\over 4} { 1 \over (\sqrt{ 2+(2)})^{3} }= -1/32$
The taylor series expansion for $ \sqrt{x+2} \space \space at \space \space x=2 :$
$2 + {1\over 4\cdot 1!} (x-2) - {1 \over 32 \cdot 2! }+...$
Which I think this is correct, but when I try to use the general Binomial theorem, I run into trouble.
$\sqrt{2+(x-2)}=\sqrt{2}\left(1+\frac{x-2}{2}\right)^{1/2}$
$(1+t)^{1/2}$ where $t=\frac{x-2}{2}$
$1 { {1/2} \choose 0} + \sqrt{2}{ 1/ 2 \choose 1}(1+t)+ 2^{3/2}{ 1/ 2 \choose 2}(1+t)^2 = $
$1 + \sqrt{2}\space{ {1\over 2} \over  1!}(1+t)+ 2^{3/2}\space{ {1\over 2} \cdot {-1 \over 2} \over  2!}(1+t)^2 + ...$
$1 + \sqrt{2}\space{ {1\over 2} \over  1!}(1+\frac{x-2}{2})+ 2^{3/2}\space{ {1\over 2} \cdot {-1 \over 2} \over  2!}(1+\frac{x-2}{2})^2 + ...$
I assume that ${ {1/2} \choose 0} = 1$, since there is only one way to choose 0 things, but I'm not sure if that is correct.
</s>