difference of square roots approximation

Question

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.

Answer 1

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 $2\sqrt{x}$.

Answer 2

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 $$\sqrt{x + a} - \sqrt{x + b} \approx \frac{a-b}{2\sqrt{x}}$$

Answer 3

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.