[fatbrowndog]

his formula: α⁻¹ = (4π³ + π² + π) - (α/24)

α⁻¹ = 137.0359996 Experimental (CODATA): 137.0359991 Δ: < 0.005 ppm

Critical Problem: This is CIRCULAR The formula is: α⁻¹ = S - (α/24)

But α appears on both sides! This is not a closed-form solution. To solve it, you need: α⁻¹ = S - (α/24) α⁻¹ + α/24 = S α⁻¹(1 + 1/(24·α⁻¹)) = S This requires knowing α already to solve for α. It's circular.

[albert_roca]

It's not circular. It rearranges into a standard quadratic equation: x^2 − 24Sx + 24 = 0. α is derived as the root of this equation.

[fatbrowndog]

α⁻¹ = S - 1/(24α) α⁻¹·24α = 24αS - 1 24α²S - 24α - 1 = 0

α = (24 ± √(576 + 96S))/(48S) α = (24 + √(576 + 96·137.036...))/(48·137.036...) α = (24 + √13,723.66...)/(6577.74...) α = (24 + 117.12...)/(6577.74...) α = 141.12.../6577.74... α ≈ 0.021454...

α⁻¹ ≈ 46.61 ??? That's wrong!

[fatbrowndog]

α⁻¹ = S - 1/(24α)

α⁻¹·24α = 24αS - 1

24α²S - 24α - 1 = 0

α = (24 ± √(576 + 96S))/(48S)

α = (24 + √(576 + 96·137.036...))/(48·137.036...)

α = (24 + √13,723.66...)/(6577.74...)

α = (24 + 117.12...)/(6577.74...)

α = 141.12.../6577.74...

α ≈ 0.021454...

α⁻¹ ≈ 46.61 ???

That's wrong!

[albert_roca]

You transcribed the formula incorrectly.

The term is -(alpha / 24). You calculated -1 / (24 · alpha).

The correct derivation is:

  1 / alpha = S - (alpha / 24)
  1 = S · alpha - (alpha^2) / 24
  alpha^2 - 24 · S · alpha + 24 = 0

Solving this with S = 4 · π^3 + π^2 + π yields the correct value.

[fatbrowndog]

Doesn't fix or predict Fan et al. 2024 latest dataset.

Try harder.

[albert_roca]

This is moving the goalposts, but ok. The model matches the international standard of CODATA 2022 to 0.005 ppm. If and when this value is updated, the prediction can be re-evaluated. Until then, I stick to the standard.

[fatbrowndog]

hi, the formula should allow both.

you have an integral and order^ power.