arcsinc( 1-y ), for small y

[Albert Chan]

To show weighted e method work, I use Mathematica:
To simplify, assumed m = 1, so e = 1/2 tan(x/2)

sinc[x_] := Sin[x] / x;
t = Series[1 - sinc[Sqrt[x]], {x, 0, 5}];
t = InverseSeries[t] /. x -> y;
exact5 = 1/2 Tan[Sqrt[t] / 2];

w = 4 - 27/56 y;
weighted = (w * Sqrt[z + z z] + Sqrt[z]) / (w+1);
weighted = weighted /. z -> 3/8d /. d -> y/(1-y);

ratio = Series[exact5 / weighted, {y, 0, 5}] // Simplify // Normal
ratio // N

>> 1 - 9459/25088000 y^3 + 2651751/309084160000 y^4
>> 1. - 3.77033e-4 y^3 + 8.57938e-6 y^4

weighted e estimate is simple, and very good. :-)

To check if above is valid, use ratio to improve Five Minute Challenge:

First, calculate exact e, with 30 significant digits:

FindRoot[sinc[x] == 5280/5281, {x, 0.1}, WorkingPrecision -> 50]
>> {x -> 0.033707758809879429348218498305111 ... }

e = N[5280/2 Tan[x / 2] /. %, 30] (* Reference: all good digits *)
>> 44.4984550191007992545541600167

(* Normally no need for scaling, but above assumed m=1 *)

e = N[5280 weighted /. y-> 1/5281, 30] (* 14 digits accurate *)
>> 44.4984550191009131676665673849

e = N[5280 weighted ratio /. y-> 1/5281, 30] (* 22 digits accurate *)
>> 44.4984550191007992545475544533