Goog calc link for Rydberg Poloynomial
Notice the Rydberg Polynomial only adds up to 0.999455406
$1={{m_ee^4\over 8{\epsilon_0}^2h^3cR_H}}-{\pi r_pR_H\over \alpha^2}$
This is due to the orignal Rydberg Constant equation, which is incomplete, has adjusted ALL of the constants so that it equals 1:
This ignores the proton radius and how it is connected/correlated to the other fundamental constants.
Note that it is tuned (ignoring precision of value used for proton radius for the moment) to:
...within 1.862e-10 of 1.0, thus, mainstream's most prized constant is a tweaked hack of theoretical and emperical work and measurements.
Once the correction is made to the equation to account for the proton radius, then the theory agrees with measurements more closely and some anomalies are simulatenously resolved.
Still to be peer reviewed, however, by inspection, dropping the electron mass to proton mass ratio from the derivation like the mainstream does with their reduced mass assumption (to allow for an analytical solution to the wave equaitons and the creation of the Solid-State theory, which is of major practical importance and for all practical purposes works but for all of the finest of measurements) was a bad idea. Good for technology, not so good for theory.
The stabilty and uniqueness of the solution is being revisited, and plan to present a working Excel spreadsheet to demonstrate this numerical methods solution. So, the error can be kind of forgiven since it requires iteration on a computer, computers which were not available (or readily available) when the original work was done back in the 30s, 40s, 50s. No excuse in the 60s and 70s when some of the theory was "finalized". It hardly seems like science, more like a captured dogma.
*Fred gets credit for asking me if I could solve for what mass is, a Feynam problem, which happened to coincide with my search for a derivation of the proton to electron mass ratio.
Moar later.
(forgive the repeats from earlier posts, however, this approach is more concise and less scattered)