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I'm trying to figure out how to get this result via taylor expansion as $x,y \rightarrow \infty$:

$f(x) = \sqrt{(x-1)y} = \sqrt{xy} - \frac{1}{2}\sqrt{\frac{y}{x}} + ...$

I've been told (on yahoo answers), that since you can't take taylor expansions at infinity, you do it on intervals (what does that mean?). Specifically, I was told:

You can write

f(x) = f(a) + (x-a) f'(a) + error term


f(a) = f(x) -(x-a)f'(x) + error term

Here your 2 ends are X and X - 1

I'm not sure what that means =P.

I am looking for a step-by-step taylor series calculation up to the first 2 terms. I've never approximated a function "at infinity" nor am I familiar with the generalized formula for taylor expansion.

So, any help would be great.


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To take a Taylor expansion of $f$ "at infinity", write the Taylor expansion of $f(\frac{1}{x})$ at $0$. – Joel Cohen Jul 29 '12 at 20:27
Yes. It's actually a Laurent series. – Robert Israel Jul 29 '12 at 20:33
... or it would be if $f(1/x)$ was analytic at $0$. In this case it's a Puiseux series. – Robert Israel Jul 29 '12 at 20:35
up vote 2 down vote accepted

In your case, you can just ignore the common factor $\sqrt{y}$ and concentrate on $\sqrt{x-1}$. To take the Taylor expansion at infinity is to replace $x$ by $1/u$ and take the Taylor expansion of at $u=0$ instead. Naïvely: $$\sqrt{x-1}=\sqrt{\frac1u-1}$$ which is of course singular at $u=0$, so you take out a troublesome factor: $$\sqrt{x-1}=\frac{\sqrt{1-u}}{\sqrt{u}}.$$ Now you expand the numerator as $\sqrt{1-u}=1-\frac12u+\cdots$, replace $u$ by $1/x$, and put the pieces together. The approximation will be good when $\lvert u\rvert$ is small, i.e., when $\lvert x\rvert$ is large.

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