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.
Best Answer
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 $2\sqrt{x}$.