# Prove $n \rightarrow 2^+$ as $a \rightarrow 0$

While studying solids of revolution at college, I came across a problem related to physics that seems to have an answer difficult to prove mathematically, which I have not been able to obtain.

## Motivation

Consider the surface $$z=|x|^n+|y|^n$$ for $$0 \le z \le h$$ and $$n \ge 1$$. If this surface were a physical object standing in a table, it is very clear that for values of $$n \approx 1$$ it will be in an unstable equilibrium, while at great values, such as $$n=10$$ it will be at an stable equilibrium, that is, the object may or may not tumble/fall under any force $$F>0$$ depending on the value of $$n$$. The question is then at which value of $$n$$ does it transition from unstable to stable?

$$n=1.5$$ and $$n=10$$

## Problem statement

From the motivating problem I was able to obtain the equation below. Solving for $$n$$ as $$a \rightarrow 0$$ gives the answer, for any constant $$c>0$$

in the physical problem, $$c$$ is the z-coordinate of the center of mass of the object and $$a$$ is associated with at which point it will tumble. Taking $$a \rightarrow 0$$ provides a situation in which it may fall if moved by any infinitesimal amount, which is an unstable equilibrium

$$\left (c-a^n \right )na^{n-2}=1$$ However, there seems to be no feasible way to isolate $$n$$ in order to apply the limit $$a \rightarrow 0$$. Using Wolfram Alpha, it seems the value of $$n$$ approaches $$2$$ by using $$a \approx 0$$, but I see no way to prove it.

• I think it should not be $2$ in general, since let's just assume that $n=2$, then what will happen if we set $n=2$ and then take the limit as $a\rightarrow 0$. In this case, you will end up with something like $2c=1$. Thus, this implies that the $z$-coordinate of the center of mass is $\frac{1}{2}$ which I doubt is true since you are cutting off the $z$ value at an arbitrary height $h$. Aug 31, 2023 at 16:32
• I did not understand the exact meaning of $a$. For example, if $n=1$, then your equation has the solution $a = c/2$. What does that mean? Aug 31, 2023 at 17:57
• Please edit your question to provide a more meaningful and informative title. Thank you.
– D.W.
Aug 31, 2023 at 18:30
• Here is a solution that someone can formalise. Let $c\mapsto2c$ and rewrite as $a^n=c\pm\sqrt{c^2-a^2/n}$. Taking the positive root leads to $n\to0$ as $a\to0$. Taking the negative root, a series expansion gives $a^n=a^2/(2cn)+O(a^4)$ whence $n\to2$ as $a\to0$. There are important pieces of rigour needed between each of these steps, including justifying whether the limits come from above or below. Aug 31, 2023 at 19:18
• I have used ContourPlot from Mathematica, and it looks like if $c < 1/2$, then $n$ converges to $2$ from the left as $a\to 0^+$. Aug 31, 2023 at 20:46

To avoid problems with noninteger powers of negative numbers, we assume $$a>0$$. Let $$\varepsilon\in (0,1]$$ be any number. If $$n\ge 2+\varepsilon$$, $$a<\min\left\{e^{-1/2}, (c(2+\varepsilon))^{-1/\varepsilon}\right\}$$ then $$\left (c-a^n \right )na^{n-2} The previous to the last inequality holds because the function $$cna^{n-2}$$ has the derivative $$ca^{n-2}(1+n\ln a)$$, and so is decreasing when $$a$$ is a fixed numbers less then $$e^{-1/2}$$. On the other hand, if $$n\le 2-\varepsilon$$, $$a<\min\left\{1, \left(\frac c2\right)^{1/\varepsilon}\right\}$$ then $$\left (c-a^n \right )na^{n-2}\ge \frac{c}{2}a^{-\varepsilon}>1$$. It follows $$n\to 2$$ when $$a\to 0$$.