# $\frac{dx}{dt}, \frac{dx}{d\frac{dx}{dt}}, \frac{dx}{d\frac{dx}{d\frac{dx}{dt}}}$ ad infinitum to $\sqrt{2x}$

Firstly, for a finite such tower, say $n$ deep, assuming there are no divisions by $0$ to get in the way

$$\frac{dx}{d\frac{dx}{d\frac{dx}{d ...}}} = \frac{dx}{dt}\Big(\frac d{dt}\big(\frac{dx}{d\frac{dx}{d\frac{dx}{d ...}}}\big)\Big)^{-1} = \frac{\frac{dx}{dt}}{\frac d{dt}\Big(\frac{\frac{dx}{dt}}{\frac d{dt}\Big(\frac{\frac{dx}{dt}}{\frac d{dt}(...)}\Big)}\Big)}$$

The bottom of this tower is

$$\frac{dx}{d \frac{dx}{dt}} = \frac{dx}{dt}\frac{dt}{d\frac{dx}{dt}} = \frac{x'}{x''}$$

and we can, at least in principle, work back up to the top to have the whole tower expressed as upto $n^{th}$ derivatives. I did write out what the next layers would look like but I couldn't see any nice pattern.

Ok so now my question is whether it would be in any sense meaningful to consider the infinite tower, in which case the following analysis happens:

Writing $F = \frac{dx}{d\frac{dx}{d\frac{dx}{d ...}}}$, we have

$$F = \frac{\frac{dx}{dt}}{\frac{dF}{dt}} \implies F = \sqrt{2x + c}$$

• Be careful with square roots. Why not $F = -\sqrt{2x+c}$ also? – GEdgar Apr 17 '18 at 13:56

If $x$ and $y$ are smooth real functions "in $t$", the derivative $\frac{dx}{dy}(t_0)$ makes sense only for those $t_0$ such that $y$ is a diffeomorphism in a neighborhood of $t_0$ that is contained in the domain of $x$. (I.e., when $y$ is locally a chart for a $1$-dimensional manifold around $t_0$ on which $x$ is defined.) In this case, we have the rule(**) $$\frac{dx}{dy} = \frac{dx/dt}{dy/dt}$$ Note that a real function $y$ defined on an interval is everywhere locally a diffeomorphism iff it is smooth and strictly monotone.

I can see only one way to define the infinitely iterated derivative $$\frac{dx}{d\frac{dx}{d\frac{dx}{\ddots}}}$$ which as you suggest is by a limit of a sequence $$(d_nx)_{n \geq 0}=t,\frac{dx}{dt},\frac{dx}{d\frac{dx}{dt}},\frac{dx}{d\frac{dx}{d\frac{dx}{dt}}},\ldots$$

There are multiple questions:

1. For which $x$ do all $d_nx$ make sense?
2. Does this converge? In what sense? There are many notions of convergence of functions.
3. If it converges to $F$, is $F = \frac{dx}{dF}$? Will this expression even make sense, i.e. is $F$ smooth? strictly monotone? ... continuous?

1. For all $d_nx$ to be well-defined on the same interval, we would need all of them to be smooth and strictly monotone. Smoothness is automatic. Examples:
• For $x = t$, we have $d_1 = 1$ and so $d_2$ is not defined.
• For $x = \exp t$, we have $d_1 = \exp t$, $d_2=1$, and the other $d_n$ don't make sense anymore.
• For $x = t^2$, we have $d_1 = 2t$, $d_2 = t$, $d_3 = 2t$, $d_4 = t, \ldots$ All of them are well-defined on $\mathbb R$.
• In general, there appears to be no nice formula for $d_n$ to test whether they are strictly monotone.
2. As in the examples above, even when all $d_n$ are defined, the sequence need not converge.
3. For continuous $f$ on a metric space, the limit $f(f(f( \cdots x)))$ if it exists is necessarily a fixed point of $f$. Is our $y \mapsto \frac{dx}{dy}$ continuous? For which metric? It is for example continuous for the locally uniform convergence of real analytic functions $y$, by Cauchy's integral formula.

(**) Note how writing $dx/dt$ for $x'$ is compatible with the philosophy to view a derivative as the derivative of a function with respect to another function, here the identity mapping $\operatorname{id} : \mathbb R \to \mathbb R$, denoted $t$. Compare with partial derivatives in differential geometry, or already in multivariable calculus, where a partial derivative is the derivative with respect to a projection map onto a component of the cartesian product.

That is, a 'variable' is really a function, usually the identity map.

The cancellation law is just the chain rule: we define $dx/dy(t_0)$ by $$(x \circ y^{-1})'(y(t_0))=x'(t_0)\cdot(y^{-1})'(y(t_0))$$ where $(y^{-1})'(y(t_0))=1/y'(t_0)$ again by the chain rule applied to $(y\circ y^{-1})'(y(t_0))$

• So the recursive definition is $d_n=\cfrac{\mathrm dx}{\mathrm dd_{n-1}}$, right? If so your definition $d_0=x$ doesn't quite fit the pattern; it should be $d_0=t$. – Rahul Apr 18 '18 at 4:43
• Thanks, edited. – punctured dusk Apr 18 '18 at 7:40