How can I accurately compute $\sqrt{x + 2} −\sqrt{x}$ when $x$ is large?

Question

How can the values of the function $f(x) = \sqrt{x + 2} −\sqrt{x}$ be computed accurately when $x$ is large?

I have tried using Matllab. I am not able to understand when $x$ will be large.

Answer 1

If directly computing $\sqrt{x+2}-\sqrt{x}$ is giving you problems, you can instead try using

$\sqrt{x+2}-\sqrt{x} = \dfrac{(\sqrt{x+2}-\sqrt{x})(\sqrt{x+2}+\sqrt{x})}{\sqrt{x+2}+\sqrt{x}} = \dfrac{(x+2)-x}{\sqrt{x+2}+\sqrt{x}} = \dfrac{2}{\sqrt{x+2}+\sqrt{x}}$.

For large $x$, this is approximately $\dfrac{2}{\sqrt{x}+\sqrt{x}} = \dfrac{1}{\sqrt{x}}$.

EDIT: I just tried this for $x = 10^{18}$ using MATLAB R2013b.

For $\sqrt{x+2}-\sqrt{x}$ it gives $0$, but for $\dfrac{2}{\sqrt{x+2}+\sqrt{x}}$, it gives $1.0000 \times 10^{-9}$.

Answer 2

Recall that when $t$ is small, $\sqrt{1+t}\approx 1 + t/2$. It follows that, for $x$ large and positive, $$ f(x)= \sqrt{x}(\sqrt{1+2/x}-1)\approx 1/\sqrt{x}, $$ in the sense that $f(x)\sqrt{x}\rightarrow 1$ as $x\rightarrow \infty$.

Answer 3

In fact, an asymptotic expansion of $f(x) = \sqrt{x+a} - \sqrt{x}$ about $x = \infty$ is $$f(x) \approx \frac{a}{2} x^{-1/2} - \frac{a^2}{8} x^{-3/2} + \frac{a^3}{16} x^{-5/2} - \frac{5a^4}{128} x^{-7/2} + \frac{7a^5}{256} x^{-9/2} - \frac{21a^6}{1024} x^{-11/2} + \cdots.$$ We can calculate this via the generalized binomial theorem $$(a+x)^r = \sum_{k=0}^\infty \binom{r}{k} a^k x^{r-k},$$ for $r = 1/2$, so that in particular, $$\begin{align*} \binom{1/2}{k} &= \frac{(1/2)(1/2-1)\cdot \ldots \cdot(1/2-k+1)}{k!} \\ &= \frac{1(-1)(-3)\cdot \ldots \cdot(3-2k)}{2^k k!} \\ &= (-1)^{k-1} \frac{(2k-2)!}{2^{2k-1} k! (k-1)!} \\ &= \frac{(-1)^{k-1}}{2^{2k-1}k } \binom{2k-2}{k-1}, \quad k = 1, 2, \ldots,\end{align*}$$ and $\binom{1/2}{0} = 1$.

Answer 4

This is a variant of heropup's answer; hopefully the useful technique it illustrates justifies the partial duplication.

We may factor $\sqrt{x+2} -\sqrt{x} = \sqrt{x}( \sqrt{1 + 2/x} - 1).$ Now if $x \to \infty,$ then $|2/x| < 1,$ so we can compute $\sqrt{1 + 2/x}$ via the binomial theorem, to get $$\sqrt{1+ 2/x} - 1 = \sum_{n = 1}^{\infty} {{1/2} \choose n } \dfrac{2^n}{x^n} = \sum_{n = 1}^{\infty} (-1)^{n-1}\dfrac{(2n-3)\cdot (2n - 5) \cdots 3 \cdot 1 }{n!} x^{-n},$$ so that $$\sqrt{x+2} - \sqrt{x} = \sum_{n = 1}^{\infty} (-1)^{n-1}\dfrac{(2n-3)\cdot (2n - 5) \cdots 3 \cdot 1 }{n!} x^{-\frac{2n-1}{2}}.$$

Of course this is the same series as in heropup's answer, derived in essentially the same way. But one thing we see, by applying the binomial theorem to $1 + 2/x,$ is that this series converges for $|2/x| < 1,$ i.e. for $x > 2,$ which might be helpful. (And of course one can use the usual theory of Taylor series to estimate the rate of convergence, if necessary.)

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The general technique this illustrates is doing a division by the term that is going to infinity, so that instead one has a term which is going to zero.