A fluke or deep reason for this nice result: rigorously identify derivative from algebraic perturbation

Question

I came across a cool result and was wondering whether what I saw is a fluke or has a deeper reason. Long story short: can we find derivatives with algebraic perturbations?

Consider the algebraic equation $12 (x + \epsilon) - y^2 = 0$ that defines an implicit function $y=f(x)$. Now, the aim is to find derivatives. Focussing on the positive branch. We can do that analytically and find $\frac{dy}{dx} = \sqrt{3} x^{-\frac{1}{2}}$ as well as $\frac{d^2y}{dx^2} = -\frac{1}{2} \sqrt{3} x^{-\frac{3}{2}}$.

So far so good. Now, if we attempt to solve this for $y$ using a perturbation expansion, by letting $y^{\ast} = y^{(0)} + \epsilon y^{(1)} + \epsilon^2 y^{(2)} + \epsilon^3 y^{(3)} + \dots$, we obtain

$y\to 2 \sqrt{3} \sqrt{x}+\frac{\sqrt{3} \epsilon }{\sqrt{x}}-\frac{\sqrt{3} \epsilon ^2}{4 x^{3/2}}+\frac{\sqrt{3} \epsilon ^3}{8 x^{5/2}} $

From the first-order term, we can see the first derivative immediately. For the second-order term, they differ by a factor of two, specifically, it seems to me that the second order term is $\frac{1}{2} \frac{d^2y}{dx^2}$. Which would obviously vanish when applying $\frac{d}{d\epsilon}$ to the second-order term.

However, I think that is wrong. Could it be that it is a factor of $2!$ instead? So, to get the third derivative, I would multiply the term associated with $\epsilon^3$ by $3!$ and so on? It seems to work with fourth order, too.

Now, does this relationship hold more generally or did I just re-invent a special case of Taylor's theorem that breaks down when working with less friendly functions?

I would be very grateful if someone could help me make this more rigorous including a definition of the derivative in terms of a perturbation to the independent variable. I suppose a formal statement would be $\begin{equation} y \to y^{(0)} + \sum_{k=1}^n \frac{1}{k!} \frac{d^k y}{dx^k} \epsilon^{k} \end{equation}$.

Just to be clear, I know how to Taylor-expand. I am curious whether we can for an implicit function $F(x,y)=0$ employ a perturbation expansion $F(x + \epsilon,y)=0$ with the guess for $y$ above to find $\frac{d^k y}{dx^k} $.

Thank you so much!

PS: The Mathematica code to generate the output is $\begin{equation} \pmb{\text{AsymptoticSolve}[12(x+\epsilon )-y{}^{\wedge}2\text{==}0,\{y\},\{\epsilon ,0,3\}]} \end{equation}$

Answer 1

And if you would like to use perturbations in an implicit equation

$$F(x,y(x))=0,$$ you would have to use the chain rule $n $ times. For the first order

$$\frac{\partial F}{\partial x}(x,y(x)) +\frac{dy}{dx}\frac{\partial F}{\partial y}(x,y(x))=0$$ gives you the classical relation

$$\frac{dy}{dx}=- \frac{\frac{\partial F}{\partial x}(x,y(x))}{\frac{\partial F}{\partial y}(x,y(x))}.$$

And you can follow on step by step to get the following orders. But again in your particular case, the known Taylor series allows you to get every order easily.

Answer 2

You have

$$y=2 \sqrt{3} \sqrt{x}\left(1+ \frac{\epsilon}{\sqrt{x}}\right)^{\frac{1}{2}}.$$

Then just use Taylor expansion

$$(1+a)^{\beta}=\sum_{n=0}^\infty \binom{\beta}{n}a^n$$ which is valid for $0 \le \epsilon \lt \sqrt{x}$, where $$\binom{\beta}{n}= \frac{\beta(\beta-1)\dots(\beta-n+1)}{n!}$$

$\beta=\frac{1}{2}$ and $a=\frac{\epsilon}{\sqrt{x}}$.