Here, about 10 h ago, Cesaero R presented a super 28-sd y(2) as a zero of #2^y=2+1/y#. How do you find super strings y(b) as zeros of #b^y=b+1/y, b = 10, e and pi#? How do you prove uniqueness or otherwise of y(b)?

1 Answer
Aug 2, 2016

Answer:

#b = 10: y =1.03987685 and -0.1085584058#

#b = pi: y = 1.204799967 and -0.3187166453#

#b = e: y = 1.256760625 and -04767602146#

Explanation:

Making separate graphs for #x = b^y-b and x=1/y#, the we get

approximation(s) to zero(s) of the given equation as y at the

common point(s).

The second equation xy=1 represents two branches of a rectangular

hyperbola (RH), fanning to infinity, in Q1 and Q3, in between axes as

asymptotes.,

When b > 1, the first (exponential) graph cuts the RH in Q1, passes

through ((x, y) = (0, 1) and (-(b-1), 0), to meet the RH again in Q3. ,

x-axis is asymptotic to RH and, below x-axis, y = - b is

asymptotic to the exponential graph. So, there are two zeros with

opposite signs...

Here, the specified b values are all > 1.

With the limited facilities in my processor,, I could make only 10-sd

approximations, for the positive roots, in each case. I have used

the following iterative formula ( using logarithms ).

#y_n = log(b+1/y_(n-1))/log b#, n = 1, 2, 3, ...#, with a guess starter

#y_0#, obtained by root-bracketing method.

My approximations are:

#b = 10: y =1.03987685#

#b = pi: y = 1.204799967#

#b = e: y = 1.256760625#

The errors in #(b^y-b-1/y)# are all of order #10^(-8)#

The interested reader could hunt for 28-sd approximations, using

128-bit processor.

The negative zeros could emerge from yet another iterative

method.

The suggested iteration formula that gave the given answers for the

negative roots is

#y_n=1/(b^(y_(n-1))-b)#

The use of log anywhere here might not give the second zero..

Another iterative form has to be tried.

When #b <= 1#, there is no solution...