Firstly, take the family of differential equations $\dot x = > \frac{dx}{dt}=x^\alpha$, for any $\alpha \in \mathbb R$
The solution to these equations is $$(\text{for } \alpha=1):x(t)=x_0e^t$$ $$(\text{for } \alpha\neq 1): x(t)=\left(x_0^{1-\alpha}+(1-\alpha)t\right)^{\frac{1}{1-\alpha}}$$
In the second equation we can rewrite $\alpha$ as the deviation from $1$: $\alpha = 1+\delta$:
$$(\text{for } \alpha\neq 1): x(t)=\left(x_0^{-\delta}+-\delta > t\right)^{\frac{1}{-\delta}}$$
Note first that for any positive $\delta$ (no matter how miniscule), $x(t)$ approaches infinity at a finite (!!) time $t$. I will call such behaviour "infinitely explosive".
To see this, note that $\frac{1}{x(t)}$ approaches $0$ at finite time $t$ as follows: $\frac{1}{x(t)}=\left(x_0^{-\delta}+-\delta t\right)^{\frac{1}{\delta}}=0 \implies t= \frac {1}{x_0^{\delta}\delta }$, which is a finite number for any $\delta \in \mathbb R$.
On the other hand, for negative $\delta$, there is no $t$ such that $1/x(t)=0$ (unless, $x_0=0$). In fact, for $\alpha = 0$, it grows linearly, for $\alpha = 0.5$, quadratically, etc.
So this gives a very weird result: if we take the differential equation, and slowly "slide" $\alpha$ towards $1$, $x(t)$ slowly converges to the exponential function. This is intuitive once we realize that $e^x=\sum_{n=1}^\infty(\frac{x^n}{n!})$. However, when we then slide $\alpha$ only slightly above $1$, x(t) turns into an "infinitely explosive function" (reaches infinity in finite time).
Now Compare this with the following family of equations:
$$\dot x = \text{hypexp}(x,n), \text{for any } n\in \mathbb N$$ Whose solutions are $$x(t)= {e^{e^{...^{x}}}}$$ with $n+1$ amount of $e$'s
Where
$\text{hypexp}(x,0)= x$
$\text{hypexp}(x,1)= x\cdot ln(x)$
$\text{hypexp}(x,2) = x \cdot ln(x)\cdot ln(ln(x))$
$\text{hypexp}(x,n) = x \cdot ln(x)\cdot ln(ln(x))\cdot ln(ln(ln(x)))\cdot ...$
Now given $\dot x = f(x)$, define a function $f$ to be "explosively larger" than $g$ if $f$ approaches reaches infinity at smaller $t$ than $g$ for the same $x_0$. ( I don't know if this concept already has another name)
Now here comes the weird conclusion: no matter how large $n$ and how miniscule $\delta$, $f(x)=x\cdot x^{\delta}$ is always explosively larger than $g(x)=x\cdot ln(x) \cdot ln(ln(x))\cdot ...$, whereas if $\delta$ reaches $0$, $g$ is immediately monumentally larger (in the normal sense) than $f$ for any $t>1$.
Discussion
- Does anyone else find this as weird and interesting as I do?
- Is there some kind of "infintely explosive supremum"? A kind of "smallest" infinitely explosive (or "largest" non-infinitely explosive) function such that all non-infinitely explosive functions have to be smaller than it. Note that it would have to be larger than even bizarrely quickly increasing functions, such as Ackermann's function.
- Is there theory that makes sense of all this?