Limiting behavior of $x + \sqrt{x} - \sqrt{x + \sqrt{x}}$

Question

As $x$ and $y$ vary through the real numbers, how do the interval families $\left[ x, x + \sqrt{x} \right]$ and $\left[ y - \sqrt{y}, y \right]$ differ?

Not by much, it turns out: consider the two maps: $f : x \mapsto x + \sqrt{x}$, $g : y \mapsto y - \sqrt y$. The composition $g \circ f$ rapidly converges to the the identity minus a constant:

$$\lim_{x \rightarrow \infty} x - (g \circ f)(x) = \lim_{x \rightarrow \infty} x - \left( (x + \sqrt{x}) - \sqrt{x + \sqrt x)} \right) = 0.5.$$

Yet this fact is false for any exponent greater than 0.5. Indeed, for any fixed $\varepsilon > 0$:

$$\lim_{x \rightarrow \infty} = x - \left( (x + x^{0.5 + \varepsilon}) - (x + x^{0.5 + \varepsilon})^{0.5 + \varepsilon} \right) = \infty.$$

How can I understand this sudden change in behavior???

EDIT: another unusual observation. what if we add ceiling functions to both, so that $\hat{f} : \lceil x \mapsto x + \sqrt{x} \rceil $, $\lceil g : y \mapsto y - \sqrt y \rceil$. Then $g \circ f$ appears to oscillate between $-1$ and 0, though possibly converges to 0. I'd like to prove or disprove: does it always take the values either $-1$ or $0$? Is it "eventually" 0? Thanks.

Answer 1

First, $$x-\big(x+\sqrt{x}-\sqrt{x+\sqrt{x}}\big)=\sqrt{x}-\sqrt{x+\sqrt{x}},$$ so I'm just going to work with that. $$\sqrt{x}-\sqrt{x+\sqrt{x}}=\sqrt{x}-\sqrt{x}\sqrt{1+x^{-1/2}}.$$ When $x>1$, we can use the binomial series $$\sqrt{1+x^{-1/2}}=1+\frac{1}{2}x^{-1/2}-\frac{1}{8}x^{-1}+\frac{1}{16}x^{-3/2}\cdots$$ Then we get $$\sqrt{x}-\sqrt{x}\sqrt{1+x^{-1/2}}=\sqrt{x}-\sqrt{x}-\frac{1}{2}+\frac{1}{8}x^{-1/2}-\frac{1}{16}x^{-1}\cdots.$$ The $\sqrt{x}$ terms cancel, and the terms of order $x^{-1/2}$ and smaller vanish as $x$ gets large, leaving only $-0.5$.

Now, in the other case: $$x^{0.5+\epsilon}-(x+x^{0.5+\epsilon})^{0.5+\epsilon}=x^{0.5+\epsilon}-x^{0.5+\epsilon}(1+x^{-0.5+\epsilon})^{0.5+\epsilon}.$$ For $\epsilon<0.5$ and $x>1$, we can again use the binomial series $$(1+x^{-0.5+\epsilon})^{0.5+\epsilon}=1+(0.5+\epsilon)x^{-0.5+\epsilon}+\cdots$$ This time, however, we get $$x^{0.5+\epsilon}-x^{0.5+\epsilon}(1+x^{-0.5+\epsilon})^{0.5+\epsilon}=x^{0.5+\epsilon}-x^{0.5+\epsilon}+(0.5+\epsilon)x^{2\epsilon}+\cdots.$$ The higher order terms in the series will vanish for large $x$ (under the conditions we put on $\epsilon$) as you can check, but the $x^{2\epsilon}$ will blow up.

Answer 2

Let $a=\frac12+\epsilon$, $$ \begin{align} x-g\circ f(x) &=\left(x+x^a\right)^a-x^a\tag{1a}\\ &=x^a\left(\left(1+x^{a-1}\right)^a-1\right)\tag{1b}\\ &=x^a\left(ax^{a-1}+O\!\left(x^{2a-2}\right)\right)\tag{1c}\\ &=ax^{2a-1}+O\!\left(x^{3a-2}\right)\tag{1d}\\ &=\left(\tfrac12+\epsilon\right)x^{2\epsilon}+O\!\left(x^{3\epsilon-\frac12}\right)\tag{1e} \end{align} $$ Explanation:
$\text{(1a):}$ apply $f$ and $g$
$\text{(1b):}$ factor $x^a$ out front
$\text{(1c):}$ apply the Binomial Theorem
$\text{(1d):}$ distribute the $x^a$ back in
$\text{(1e):}$ substitute $a=\frac12+\epsilon$

For $\epsilon\lt\frac16$, the $O\!\left(x^{3\epsilon-\frac12}\right)$ term vanishes as $x\to\infty$. If $\epsilon\gt0$, the $x^{2\epsilon}$ term tends to $\infty$; whereas, if $\epsilon\lt0$, the $x^{2\epsilon}$ term tends to $0$. Of course, if $\epsilon=0$, $\left(\tfrac12+\epsilon\right)x^{2\epsilon}=\frac12$.