June 2 Information

So I just finished watching an [ interesting video]. It basically shows that the function

${\displaystyle f(x)={\sqrt[{\phi }]{\phi ^{-1}}}x^{\phi }}$

has an unusual property. The derivative

${\displaystyle f'(x)=\phi {\sqrt[{\phi }]{\phi ^{-1}}}x^{\phi ^{-1}}}$

is equivalent to the compositional inverse of ${\displaystyle f(x)}$. In other words ${\displaystyle x=f'(f(x))}$.

For some reason that strikes me as quite amazing. Has this function been well studied? I can't think of any good search terms that might help either. Earl of Arundel ( talk) 21:53, 2 June 2021 (UTC) reply

(Note that ${\displaystyle \phi }$ stands for the golden ratio 1.61803···.) There is a mistake above in the exponent of ${\displaystyle x}$ in the rhs for the derivative; ${\displaystyle x^{\phi ^{-1}}}$ should be ${\displaystyle x^{\phi -1}.}$ I don't think this has function been studied to any degree; it strikes me as an isolated curiosity. By playing around a bit you can derive strange identities such as

${\displaystyle {\frac {d}{dx}}f(f(x))=xf'(x)}$

and

${\displaystyle f''(f(x))=(f'(x))^{{-}1},}$

but I spotted nothing that seemed to tie in with other stuff in the web of interesting mathematics. Differential equations often have a physical interpretation, but this function is necessarily between dimensionless domains.  -- Lambiam 22:47, 2 June 2021 (UTC) reply

"${\displaystyle x^{\phi ^{-1}}}$ should be ${\displaystyle x^{\phi -1}}$" In fact ${\displaystyle \phi ^{-1}=\phi -1}$. (Indeed it's this coincidence that makes the example work.) -- JBL ( talk) 23:44, 2 June 2021 (UTC) reply

Tricky ${\displaystyle \phi }$ strikes again! Earl of Arundel ( talk) 01:07, 3 June 2021 (UTC) reply

Also, the equation ${\displaystyle f(x)=f'(x)}$ in the unknown ${\displaystyle x}$ has two solutions (in the nonnegative reals):

${\displaystyle f(0)\,=f'(0)=0,}$

${\displaystyle f(\phi )=f'(\phi )=\phi .}$

 -- Lambiam 23:01, 2 June 2021 (UTC) reply

Ok so basically just an interesting curiosity. Now when you say "dimensionless", what exactly do you mean by that? Considering that the input and output of the function map to a two dimensional space that just seems like a very enigmatic statement. Earl of Arundel ( talk) 01:07, 3 June 2021 (UTC) reply

2D space? Not just ${\displaystyle \mathbb {R} }$?

In the sense of dimensional analysis, as in Dimensionless quantity. (I am not endorsing Lambiam's claim -- I haven't thought about it at all -- but that's the sense being invoked.) -- JBL ( talk) 02:35, 3 June 2021 (UTC) reply

In my reply I used "[[ dimensionless]]", which redirects to Dimensionless quantity. If the position of a physical object is given as a length quantity as a function of a time quantity, the position function maps T-quantities to L-quantities, or, for short, T to L. Its inverse, time as a function of position, maps L to T. Its derivative, the speed of the object, maps T to LT⁻¹. This generalizes to arbitrary dimensions. Let ${\displaystyle f}$ be a function mapping X to Y, where X and Y are dimensions (possibly 1, the dimemsion of dimensionless quantities). Then ${\displaystyle f^{{-}1}}$ maps Y to X and ${\displaystyle f'}$ maps X to YX⁻¹. So if now ${\displaystyle f'=f^{{-}1},}$ this function both maps Y to X and X to YX⁻¹, which gives us two equations: Y = X and X = YX⁻¹. Using the first of these to substitute X for Y in the second gives us X = 1, and so also Y = 1. Our function ${\displaystyle f}$ is a map between dimensionless quantities.  -- Lambiam 07:53, 3 June 2021 (UTC) reply

Well that truly IS mind-bending! (And brilliantly explained, I might add.) It might take a while for all of that to sink in anyhow. So that's vaguely analogous to how ${\displaystyle g(x)=g'(x)=e^{x}}$. I wonder if the two functions are somehow connected. Or is that a bit of a stretch? Earl of Arundel ( talk) 18:42, 3 June 2021 (UTC) reply

On second thought, maybe not. (Considering that ${\displaystyle g^{{-}1}(x)=ln(x)}$ .) Earl of Arundel ( talk) 18:51, 3 June 2021 (UTC) reply