Wikipedia:Reference desk/Archives/Mathematics/2021 June 2

= June 2 =

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

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

has an unusual property. The derivative

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

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

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)


 * (Note that $$\phi$$ stands for the golden ratio 1.61803···.) There is a mistake above in the exponent of $$x$$ in the rhs for the derivative; $$x^{\phi^{-1}}$$ should be $$x^{\phi-1}.$$ 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
 * $$\frac{d}{dx}f(f(x))=xf'(x)$$
 * and
 * $$f''(f(x))=(f'(x))^{{-}1},$$
 * 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)
 * "$x^{\phi^{-1}}$ should be $x^{\phi-1}$"|undefined In fact $$\phi^{-1} = \phi-1$$. (Indeed it's this coincidence that makes the example work.) --JBL (talk) 23:44, 2 June 2021 (UTC)
 * Tricky $$\phi$$ strikes again! Earl of Arundel (talk) 01:07, 3 June 2021 (UTC)
 * Also, the equation $$f(x)=f'(x)$$ in the unknown $$x$$ has two solutions (in the nonnegative reals):
 * $$f(0)\,=f'(0)=0,$$
 * $$f(\phi)=f'(\phi)=\phi.$$
 * --Lambiam 23:01, 2 June 2021 (UTC)
 * 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)
 * 2D space? Not just $\R$?
 * 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)


 * 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 $$f$$ be a function mapping X to Y, where X and Y are dimensions (possibly 1, the dimemsion of dimensionless quantities). Then $$f^{{-}1}$$ maps Y to X and $$f'$$ maps X to YX−1. So if now $$f'=f^{{-}1},$$ 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 $$f$$ is a map between dimensionless quantities.  --Lambiam 07:53, 3 June 2021 (UTC)
 * 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 $$g(x) = g'(x) = e^x$$. 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)
 * On second thought, maybe not. (Considering that $$g^{{-}1}(x) = ln(x)$$ .) Earl of Arundel (talk) 18:51, 3 June 2021 (UTC)