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In two of my physics courses in the past week, I've come across an approximation for the difference of two square roots for large radicands: $\sqrt{x+a}-\sqrt{x+b}\approx\frac{a-b}{2\sqrt x}$ for large $x$.

I have no idea where this comes from. Wolfram|Alpha gives me the simplest answer by handing me this as the first term of the expansion around $x=\infty$, but how is that calculated? I know how normal Taylor Series expansions are formulated, but I don't get how to run this calculation.

Just trying to get the approximation alone, the closest I've gotten is with the logic: $$ \frac{\sqrt{x+b+(a-b)}-\sqrt{x+b}}{a-b}\approx\frac{d}{dx}\sqrt{x+b} \\ \sqrt{x+a}-\sqrt{x+b}\approx\frac{a-b}{2\sqrt{x+b}} $$ But, as you can tell, the radicand in the approximation is incorrect. (Not to mention the approximation there is that $a\approx b$, not large $x$.) I can try to say that for large $x$, $\sqrt{x+b}\approx\sqrt{x}$, but that doesn't explain why the version with just $x$ as the radicand is more accurate. (Which is true.) Plus, I also want to know how to get the higher-order terms in the expansion.

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The approximation that you gave is in fact often more accurate. Multiply $\sqrt{x+a}-\sqrt{x+b}$ by $\dfrac{\sqrt{x+a}+\sqrt{x+b}}{\sqrt{x+a}+\sqrt{x+b}}$. We get $$\sqrt{x+a}-\sqrt{x+b}=\frac{a-b}{\sqrt{x+a}+\sqrt{x+b}}.\tag{$1$}$$ This is not an approximation, it is exact.

Moreover, when $x$ is very large in comparison to $|a|$ and $|b|$, the right-hand side of $(1)$ is computationally much better than the left-hand side. To see this, imagine we are doing the calculation on a calculator, or in floating point on a computer. If $x$ is very large, like $10^{10}$, adding $a$ to it on a calculator does not change the value, because of the limited precision. This means that our result may have a large relative error.

If $x$ is very large in comparison with $|a|$ or $|b|$, there is little relative error made if we replace $\sqrt{x+a}+\sqrt{x+b}$ in $(1)$ by $2\sqrt{x}$.

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Factor out $x$ to get $$\sqrt{x}\left(\sqrt{1 + \frac{a}{x}} - \sqrt{1 + \frac{b}{x}}\right)$$ Now expand each square root in a Binomial series $$=\sqrt{x}\left(1 + \frac{a}{2x} - 1 - \frac{b}{2x} + \mathcal{O}(x^{-2})\right)$$ For large $x$ we can neglect the terms of order $x^{-2}$ and smaller to make the approximation $$\sqrt{x + a} - \sqrt{x + b} \approx \frac{a-b}{2\sqrt{x}}$$

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Let $f(x)=\sqrt{x}$. Then, for $x>0$, we have $f(x+h) \approx f(x) + f'(x)h$. Hence $f(x+a)-f(x+b) = f(x+a)-f(x) - (f(x+b)-f(x)) \approx f'(x)(a-b)$. Since $f'(x) = \frac{1}{2\sqrt{x}}$, we have the desired approximation: $\sqrt{x+a}-\sqrt{x+b} \approx \frac{1}{2\sqrt{x}}(a-b)$.

The approximation comes from the Taylor theorem, and can be extended to higher orders.

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