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I've been reading through a thesis and trying to rederive all of the equations used. However, I've come across an expansion that I'm unsure on. The following is a simplified form of two consecutive equations in the thesis, where $x$ is assumed to be close to 0:

\begin{equation} h=\frac{x^2}{a}+\mathcal{O}(x^3) \end{equation} By inverting this relationship, it follows \begin{equation} x=\pm\sqrt{ah}+\mathcal{O}(x^{3/2}) \end{equation}

To get from the first to the second, I have done the following:

  1. Multiply both sides by $a$: $$ah=x^2+a\cdot\mathcal{O}(x^3)=x^2+\mathcal{O}(x^3)$$
  2. Take square root of both sides: $$\pm\sqrt{ah}=\sqrt{x^2+\mathcal{O}(x^3)}$$
  3. Expand $\sqrt{x^2+\mathcal{O}(x^3)}$ $$\pm\sqrt{ah}=\sqrt{x^2+\mathcal{O}(x^3)}=x+\mathcal{O}(x^{3/2})$$
  4. Rearrange $$x=\pm\sqrt{ah}-\mathcal{O}(x^{3/2})=\pm\sqrt{ah}+\mathcal{O}(x^{3/2})$$

Step 3 is the problem. If this expression is true, then the result follows, but I'm not sure how to derive this.

If I attempt to do so with Taylor expansion about $x=0$, for instance, I get the following: $$ \sqrt{x^2+\mathcal{O}(x^3)}=\sqrt{x^2}\sqrt{1+\mathcal{O}(x)}=|x|\left(1+\frac{1}{2}\mathcal{O}(x)-\frac{1}{8}\mathcal{O}(x^2)+\cdots\right)=x+\mathcal{O}(x^2) $$ Where I removed the $|\cdot|$ in the last step since the positive and negative cases are taken care of by the $\pm$ in $\pm\sqrt{ah}$. Clearly, a Taylor expansion in $x$ won't work, since by definition, it will only give positive integers for the exponents.

So I next tried the substitution $Z=\sqrt{x}$, leaving $$\sqrt{Z^4+\mathcal{O}(Z^6)}=Z^2\sqrt{1+\mathcal{O}(Z^2)}$$ If I then expand the $\sqrt{\;\cdot\;}$ in powers of $Z$ about $Z=0$, (I think) I will get $$\sqrt{1+\mathcal{O}(Z^2)}=\sqrt{1+\mathcal{O}(Z^2)}\bigg\rvert_{Z=0}+\frac{\mathcal{O}(Z)}{\sqrt{1+\mathcal{O}(Z^2)}}\bigg\rvert_{Z=0}Z+\mathcal{O}(Z^2)=\sqrt{1+\mathcal{O}(1)}+\frac{\mathcal{O}(1)}{\sqrt{1+\mathcal{O}(1)}}Z+\mathcal{O}(Z^2)$$ Where I have taken $$\mathcal{O}(Z^n)\bigg\rvert_{Z=0}=\mathcal{O}(1)\hspace{1cm}\text{(Is this correct?)}$$ At this point, I see that if I multiply back through by $Z^2$, I will get $$Z^2\sqrt{1+\mathcal{O}(Z^2)}=\sqrt{1+\mathcal{O}(1)}Z^2+\frac{\mathcal{O}(1)}{\sqrt{1+\mathcal{O}(1)}}Z^3+\mathcal{O}(Z^4)$$

However, I'm not sure how to proceed from here. I can see that the second term on the RHS is $\mathcal{O}(Z^3)=\mathcal{O}(x^{3/2})$, and that the first term is at least proportional to $Z^2=x$, so I know that I'm close, but I'm not sure what to do about the $\mathcal{O}(1)$ term in $\sqrt{1+\mathcal{O}(1)}$. Is my derivation valid up to this point? If so, how do I finish out the proof?

It seems that using a binomial expansion may also have worked, though with an exponent of $\frac{1}{2}$, it would be infinite, and I run into other problems with that approach.

A similar problem shows up later in the paper, in the following form:

$$\sqrt{Z+\mathcal{O}(\vec{X}^2)}=\sqrt{Z}+\mathcal{O}(|\vec{X}|),$$ For $Z$ a component of the vector $\vec{X}=\langle X,Y,Z\rangle$, $\vec{X}^2=|\vec{X}|^2$, and $|\vec{X}|\rightarrow 0$. The multivariate nature of the problem seems to confound the issue further, so I wonder whether the same sort of expansion can be used, and if so, what would need to be changed.

In general, these are expansions of the type $$\sqrt{f(x)+g(x)}=\sqrt{f(x)}+\mathcal{O}\left(\sqrt{g(x)}\right)$$ for small $x$, so it seems it may be a more general statement that looks like a binomial-type equality, but I am struggling to prove even these particular cases.

Lastly, I am just curious on the reasoning behind choosing to expand in powers of $\sqrt{x}$ rather than $x$. Is this dependent on the physical problem at hand? Is it valid to equate an expansion in $\sqrt{x}$ to an expansion in $x$, if both are done on the same function and to the same order?

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Perhaps it would be helpful to be more concrete, and start from $h(x) = \frac{x^2}{a} + g(x)$ where $g(x)\in \mathcal O(x^3)$. Inverting then gives

$$x = \pm \sqrt{a(h - g(x))} = \pm \sqrt{a h} \sqrt{1 - \frac{g(x)}{h}} \approx \pm \sqrt{ah} \big(1 - \frac{g(x)}{2h}\big)$$ $$ = \pm \sqrt{ah} \mp \sqrt{a} \frac{g(x)}{\sqrt{h}} $$

Examining the asymptotics of this last term we note that in the limit as $x \rightarrow 0$, $|g(x)| \leq A x^3$ and $|h|\leq B x^2$, which implies that $$\left|\frac{g(x)}{\sqrt{h}} \right|\leq \frac{A}{\sqrt{B}} x^2 \in \mathcal O(x^2)$$

Since $\mathcal O(x^2)\supset \mathcal O(x^{3/2})$ as $x\rightarrow 0$, the claim in the paper is true. I'm not sure why they used the $x^{3/2}$ bound rather than the tighter $x^2$ bound; it may be useful later in the paper, or it may have been a mistake. Either way, it is true.


More generally, assuming $f(x)$ and $g(x)$ go to zero as $x\rightarrow 0$ and $g(x) \in \mathcal o(f(x))$,

$$\sqrt{f(x) + g(x)} = \sqrt{f(x)} \sqrt{1 - \frac{g(x)}{f(x)}} \approx \sqrt{f(x)} - \frac{g(x)}{2\sqrt{f(x)}}$$

and so

$$\sqrt{f(x)+g(x)} = \sqrt{f(x)} + \mathcal O\big(\frac{g(x)}{\sqrt{f(x)}}\big)$$

However,

$$\mathcal O \big(\frac{g}{\sqrt{f}}\big) = \mathcal O\left(\sqrt{\frac{g}{f}} \cdot \sqrt{g}\right) \supset \mathcal O\left(\sqrt{g}\right)$$

because we assumed that $g\in o(f)$. Consequently, your assertion that

$$\sqrt{f(x) + g(x)} = \sqrt{f(x)} + \mathcal O(\sqrt{g(x)})$$

is true, but weaker than what we could actually show.

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  • $\begingroup$ So is it just a matter of knowing that $\mathcal{O}(x^{3/2})\subset\mathcal{O}(x^2)$? I do think it's useful later in the paper based on the form of other equations, but I guess I'm just a bit disappointed that there's no explicit proof for the $\mathcal{O}(x^{3/2})$ term. Thanks for your answer. It's the most helpful input I've gotten so far, so I will accept it as an answer. $\endgroup$ – Luke May 30 at 19:43
  • $\begingroup$ @Luke I mean, it is explicit proof for the $\mathcal O(x^{3/2})$ term. I take it your disappointment is that you don't know why the authors chose to make a weaker statement than they could have; this is more or less analogous to a claim like "all polynomials are differentiable everywhere." All polynomials are of course $C^\infty$ everywhere, not just differentiable, but if differentiability is the only condition that is needed or relevant in the paper, then it's not unreasonable to weaken the claim for readability. $\endgroup$ – J. Murray May 30 at 19:54
  • $\begingroup$ I see your point. I was loose with my wording. You have provided an explicit proof, and I appreciate it. What I meant was that I have not seen this result strictly via a particular expansion, where an $O(x^(3/2))$ term falls out naturally, as opposed to saying that it is a weaker result than a certain Taylor expansion and so must be true. My attempt with the substitution $Z=\sqrt{x}$ is the closest I've gotten. When I saw an expansion with a noninteger power, I wondered if there was more machinery going on behind it, e.g. generalized binomial, Puiseux, etc. $\endgroup$ – Luke May 30 at 20:05
  • $\begingroup$ @Luke I think it's important to bear in mind that this asymptotic notation does not imply a particular expansion or another. Having an $\mathcal O(x^{3/2})$ term does not mean that you are neglecting a term proportional to $x^{3/2}$; it means that you are neglecting a term bounded by a term proportional to $x^{3/2}$, which could equally well be of the form $ax^{3/2}$ or $bx^{\pi}$. In other words, these asymptotic bounds are somewhat upstream of any expansion you may wish to perform. It is simply quick and convenient to note that $\sqrt{1+\mathcal O(x^n)} = 1 + \mathcal O(x^{n/2})$. $\endgroup$ – J. Murray May 30 at 20:20
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You are just missing the fact that your solution with the Tayler expansion is already better than what the thesis claims! For $x<1$, we have $x^2 < x^\frac32$, so $\mathcal{O}(x^2) \supset \mathcal{O}(x^\frac32)$ when $x$ is near 0.

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  • $\begingroup$ I see. I am still interested in a solution that shows how something of the form $\sqrt{f(x)+\mathcal{O}(g(x)}=\sqrt{f(x)}+\mathcal{O}(\sqrt{g(x)})$ can be derived, as well as some of the other questions in my post, but your answer does help my understanding. $\endgroup$ – Luke May 20 at 19:52

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