We have that: $\epsilon$ is the smallest positive machine number that summed to $1$ resuts in $(\epsilon + 1)$: i.e. the smallest number greater than $1$:
if $p$ is precision and $\beta$ the base:
$$\begin{array}{lcl} 1 & = & 0.1\underbrace{00 \ldots 00}_{p-1 \text{ zeros}} \times \beta^1 \\ 1 + \epsilon & = & 0.1\underbrace{00 \ldots 0}_{p-2}1 \times \beta^1 \end{array}$$ we can consider $\epsilon$ as the distance between the above numbers: $$ \begin{array}{lcl} \epsilon & = & (1+\epsilon) - 1 \\ & = & 0.1\underbrace{00 \ldots 0}_{p-2}1 \times \beta^1 - 0.1\underbrace{00 \ldots 00}_{p-1} \times \beta^1 \\ & = & 0.\underbrace{000 \ldots 0}_{p-1}1 \times \beta^1 \\ & = & 1 \times \beta^{1} \times \beta^{-p} \\ & = & 1 \times \beta^{1-p} \\ & = & \beta^{1-p} \end{array} $$
so the only significant figure $1$, stays on the p-th position after the dot.
Calculating the value of that $p$, is equivalent to know the precision, i.e. the number of digits for the mantissa.
I know that the algorithm for finding Epsilon Machine is similar to this:
U = 1.0
while (1+U)>1 do
Umem = U
U = U/beta
end while
U = Umem
and I know that, from the given $\epsilon$, I can calculate the number $p$ of digits for the mantissa:
$$\begin{array}{lcl} \epsilon & = & \beta^{1-p} \\ \log\epsilon & = & \log\beta^{1-p} \\ \log\epsilon & = & (1-p)\log\beta \\ p & = & 1-\frac{\log \epsilon}{\log \beta} \end{array} $$
I don't know how elaborate this fact in Python, can you give me any hint? Thanks.
