# 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)?

Aug 2, 2016

$b = 10 : y = 1.03987685 \mathmr{and} - 0.1085584058$

$b = \pi : y = 1.204799967 \mathmr{and} - 0.3187166453$

$b = e : y = 1.256760625 \mathmr{and} - 04767602146$

#### Explanation:

Making separate graphs for $x = {b}^{y} - b \mathmr{and} x = \frac{1}{y}$, 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 ).

${y}_{n} = \log \frac{b + \frac{1}{y} _ \left(n - 1\right)}{\log} b$, 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 $\left({b}^{y} - b - \frac{1}{y}\right)$ 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

${y}_{n} = \frac{1}{{b}^{{y}_{n - 1}} - b}$

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

Another iterative form has to be tried.

When $b \le 1$, there is no solution...