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alexandræ · linear interpolation, sequences, and f(1/<i>x</i>) = 1/f(<i>x</i>) : the strange case of the square root

linear interpolation, sequences, and f(1/x) = 1/f(x) : the strange case of the square root

march 21st–22nd, 2023

    the usual approximation we use to compute the square root is the newton-raphson method. namely, √z is the limit of the sequence that goes an+1 = (an²−z)/2an, with a0 chosen to be sufficiently close to √z ; for example, if z's base-2 order of magnitude is 2n (meaning ⌊log2(z)⌉ = n), you can use 2⌊n/2⌉. yet, there are many other ways of computing the square root, and by messing around a bit, i found something quite weird.

    the idea was to use linear interpolation on the square root between n² and (n+1)² on the one hand, and between 1/(n+1)² and 1/n² on the other, for all n ≥ 1, so that gives us a function f such that f(n²(1−t)+(n+1)²t) = n(1−t)+(n+1)t and f((1−t)/(n+1)²+t/n²) = (1−t)/(n+1)+t/n for t ∈ [0,1]. it's not exactly top-notch approximation, but the real trick i hoped to exploit was the following : √1/x = 1/√x, meaning that √x = ½(√x+1/√1/x ). henceforth, i defined a sequence of functions (fn)n≥0 such that f0 = f, and fn+1 : x ↦ ½( f(x)+1/f(1/x)).

    surprisingly... it works wonders ! we have supx ∈ ℝ+ | f1(x)−√x | < 10-2, which is already quite good ! but then you have supx ∈ ℝ+ | f2(x)−√x | < 10-5, and even better, supx ∈ ℝ+ | f3(x)−√x | < 10-11, etc. all seems to indicate that fn(x)⟶√x for all x ∈ ℝ+. yet, √ seems to be the only such function among fairly similar ones.

    indeed, i tried to interpolate between successive cubes and inverses of cubes, and it seemed that the analogous functions and sequence would block above a certain threshold compared to ∛ or even x ↦ x2/3. i don't know if that's actually due to the distribution of squares being tighter than any other set of integers to the power of a fixed natural number, apart from 1 (or 0) obviously. i think we could investigate this phenomenon a bit further, who knows what that might yield.

    it's completely uninteresting to check for irrational powers, as all integer ≥ 2 to the power of an irrational is transcendantal (gelfond-schneider), so there's no integer in (x ↦ xα)-1(ℕ\{0,1}).

addendum 22/03/2023 :
in the end, it seems it doesn't even really depend on the distribution of the interpolation points. indeed, if we use, for n ≥ 1 and α ∈ ]0,1[, f0(n(1−t)+(n+1)t) = nα(1−t)+(n+1)αt and f0((1−t)/(n+1)+t/n) = (1−t)/(n+1)α+t/nα, fn(x) still seems to only converge towards xα for all x ∈ ℝ>0 if and only if α = 1/2. the plot thickens... x)
still, i prefer the following interpolation sequence :
  1. f0(nq(1−t)+(n+1)qt) = np(1−t)+(n+1)pt
  2. f0((1−t)/(n+1)q+t/nq) = (1−t)/(n+1)p+t/np
  3. fn+1(x) = ½(fn(x)+1/fn(1/x))
as it's much simpler to calculate by hand lol (don't let the huge symbol clusterfυck deceive you).
⟨density of αℕ fractional ...
a few finite group thingies (mainly...⟩
alexandræ
(ɔ) 2023 – 2024, intellectual property is a scam