snowplow problem

[Albert Chan]

(03-23-2021 05:50 PM)Albert Chan Wrote:  ln(k) = ln(-ln(1-y)/ln(1+y)) = y + y^3/4 + 19/144*y^5 + ...

Instead of getting a formula from k to x, it is better to do z = ln(k), to x.

→ y = z - z^3/4 + O(z^5) = z / (1 + z^2/4 + O(z^4)) ≈ z/(1+z^2/4)
→ x = 1/y - 1 = 1/z - 1 + z/4     // rough guess of x, from z = ln(k)

For k=1.5, guess x ≈ 1.568 (94 minutes)
For k=2.0, guess x ≈ 0.6160 (37 minutes)

This estimate matched pretty well. (Mathematica session)
Unmatched coefficients (z^5, z^7, z^9) have same sign, and similar size.

Code:

In[1]:= logk = Log[-Log[1 - y]/Log[1 + y]];

In[2]:= Series[logk, {y,0,10}]    (* z in terms of y *)

             3       5        7         9
            y    19 y    751 y    2857 y        11
Out[2]= y + -- + ----- + ------ + ------- + O[y]
            4     144     8640     44800

In[3]:= InverseSeries[%] /. y->z  (* y in terms of z *)

             3    5       7         9
            z    z    91 z    3379 z        11
Out[3]= z - -- + -- - ----- + ------- + O[z]
            4    18   8640    1814400

In[4]:= Series[z/(1+z^2/4), {z,0,10}] (* Pade[2,2] *)

             3    5    7    9
            z    z    z    z         11
Out[4]= z - -- + -- - -- + --- + O[z]
            4    16   64   256

Update: more terms, x in terms in z = ln(k)

x = 1/z - 1 + z/4 + z^3/144 - 7/4320*z^5 - 73/3628800*z^7 + 11/1451520*z^9 + ...