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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.

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

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  • $\begingroup$ got it. Do you think this is a faster way to tackle taylor expansions of this form? Is it better to try and avoid computing derivatives? $\endgroup$ – qbert Jun 17 '16 at 1:31
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    $\begingroup$ 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. $\endgroup$ – André Nicolas Jun 17 '16 at 1:37
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Jun 17 '16 at 1:39

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