I'm looking at the examples on the Wiki Page Generalized Continued Fractions . The introduction to general root-finding states a formula:
But then, looking at the second example for cube root of 2,
What am I missing? If x = 5 and y = 3, I get $ z = 5^3 + 3$ which is clearly not correct, and then the statement $2z-y = 253$ makes no sense either.
If I keep y = 3, this implies a "new" z = 128 and the calculation is actually acting on $128^{\frac{1}{3}}$ , leading to m = 5 . Is that what's implied here?
By using the suggested parameters: $x=5,y=3$, for $n=3$ we have $z=128=2^7=2^6.2$. Then, to express $2^{1/3}$ as a continued fraction, it is sufficient to take $m=1$, as \begin{equation} 2^{1/3}=\frac{128^{1/3}}{2^2}=\frac14\left( 5^3+3 \right)^{1/3} \end{equation} The proposed continued fraction with these parameters reads \begin{equation} \left( 5^3+3 \right)^{1/3}=5+{\cfrac {3}{3\cdot5^2+{\cfrac {2\cdot3}{2\cdot5+{\cfrac {4\cdot3}{3\cdot3\cdot5^2+{\cfrac {5\cdot3}{2\cdot5+{\cfrac {7\cdot3}{5\cdot3\cdot5^2+{\cfrac {8\cdot3}{2\cdot5+\ddots }}}}}}}}}}}} \end{equation} Then several simplifications are in order: divide the numerator and denominators of the first fraction by a factor $6$, the second one by $2$, the third one by $3/2$,the fourth one by $2$, then $3/2$, $2$,... It gives \begin{equation} 2^{1/3}=\frac{\left( 5^3+3 \right)^{1/3}}{4}=\frac54+{\cfrac {1/2}{2\cdot5^2+{\cfrac {2}{5+{\cfrac {4}{2\cdot3\cdot5^2+{\cfrac {5}{5+{\cfrac {7}{5\cdot2\cdot5^2+{\cfrac {8}{5+\ddots }}}}}}}}}}}} \end{equation} which is the proposed result.
The second form comes from the decomposition given in the same page \begin{equation} \sqrt[n]{z^{m}}=x^{m}+{\cfrac {2x^{m}\cdot my}{n(2x^{n}+y)-my-{\cfrac {(1^{2}n^{2}-m^{2})y^{2}}{3n(2x^{n}+y)-{\cfrac {(2^{2}n^{2}-m^{2})y^{2}}{5n(2x^{n}+y)-{\cfrac {(3^{2}n^{2}-m^{2})y^{2}}{7n(2x^{n}+y)-{\cfrac {(4^{2}n^{2}-m^{2})y^{2}}{9n(2x^{n}+y)-\ddots }}}}}}}}}} \end{equation} Remarking that $2x^n+y=2z-y$, here $2x^n+y=253$ which immediately gives \begin{equation} 2^{1/3}=\cfrac54+\cfrac14{\cfrac {2\cdot5\cdot 3}{3\cdot253-3-{\cfrac {(3^{2}-1)3^{2}}{3\cdot3\cdot253-{\cfrac {(2^{2}3^{2}-1)3^{2}}{5\cdot3\cdot253-{\cfrac {(3^{2}3^{2}-1)3^{2}}{7\cdot3\cdot253-{\cfrac {(4^{2}3^{2}-1)3^{2}}{9\cdot3\cdot253-\ddots }}}}}}}}}} \end{equation} Here, by dividing the numerators and denominators of all the fractions by $3$, it comes \begin{equation} 2^{1/3}=\cfrac54+{\cfrac {1\cdot5/2}{253-1-{\cfrac {3^{2}-1}{3\cdot253-{\cfrac {2^{2}3^{2}-1}{5\cdot253-{\cfrac {3^{2}3^{2}-1}{7\cdot253-{\cfrac {4^{2}3^{2}-1}{9\cdot253-\ddots }}}}}}}}}} \end{equation} noting that $n^2\cdot3^2-1=(3n-1)(3n+1)$, we obtain the given expression.