# Computing the taylor series of $f(x)=(x+2)^{1/2}$ around $x=2$ up to order 2 terms using binomial series

I can expand $(x+2)^{1/2}$ by taking $y=x+1$ and using the binomial series to find a power series representation of $f$: $$\sqrt{y+1}=\sum_{k=0}^{\infty}{1/2\choose k}y^k=1+\frac{y}{2}+\frac{\frac{1}{2}\frac{-1}{2}y^2}{2!}+O(y^3)=1+\frac{x+1}{2}-\frac{1}{8}(x+1)^2+O(x^3)$$ And grouping terms $$\sqrt{x+2}=(1+1/2-1/8)+\frac{x}{4}-\frac{x^2}{8}+O(x^3)=\frac{11}{8}+\frac{x}{4}-\frac{x^2}{8}+O(x^3)$$

I am not sure how this is supposed to bring me closer to the taylor series representation, and in particular the expansion about $x=2$. Sorry if some of my work was silly, I am a bit confused by the hint and wanted to show what I had done.

• Use $f(x)=f(0)+xf'(0)+(x^2/2!)f''(0)+O(x^3)+..$ instead to get power series in $x$. – Nitin Uniyal Jun 17 '16 at 1:37
• I know how to do that, my question is how to do this using the binomial series – qbert Jun 17 '16 at 1:38

Note that $$(x+2)^{1/2}=(4+(x-2))^{1/2}=2\left(1+\frac{x-2}{4}\right)^{1/2}.$$ Now use what you know about the expansion of $(1+y)^{1/2}$.
• In this case, since we only want to order $2$, and derivatives are easy, I think derivative is faster, and more mechanical. But usually "recycling" known expansions is easier. For example, the Maclaurin series of $\sin(x^2)$ up to the term in $x^{14}$ is easy if we use the series for $\sin t$, and unpleasant if we differentiate. – André Nicolas Jun 17 '16 at 1:37