Heron-Verfahren/Startglied 1/Funktion/Aufgabe/Lösung

Die Iteration im Heron-Verfahren zur Berechnung von ${}{\sqrt {b}}$ ist durch

{}x_{n+1}:={\frac {x_{n}+{\frac {b}{x_{n}}}}{2}}={\frac {x_{n}^{2}+b}{2x_{n}}}\,

gegeben. Deshalb ist

{}x_{1}(b)={\frac {1+b}{2}}\,,

{}{\begin{aligned}x_{2}(b)&={\frac {x_{1}(b)^{2}+b}{2x_{1}(b)}}\\&={\frac {\left({\frac {1+b}{2}}\right)^{2}+b}{2{\frac {1+b}{2}}}}\\&={\frac {{\frac {1+2b+b^{2}}{4}}+b}{1+b}}\\&={\frac {1+6b+b^{2}}{4(1+b)}},\end{aligned}}

{}{\begin{aligned}x_{3}(b)&={\frac {x_{2}(b)^{2}+b}{2x_{2}(b)}}\\&={\frac {\left({\frac {1+6b+b^{2}}{4(1+b)}}\right)^{2}+b}{2{\frac {1+6b+b^{2}}{4(1+b)}}}}\\&={\frac {1}{2}}{\left({\frac {1+6b+b^{2}}{4(1+b)}}+b{\frac {4(1+b)}{1+6b+b^{2}}}\right)}\\&={\frac {(1+6b+b^{2})(1+6b+b^{2})+16b(1+b)^{2}}{8(1+b)(1+6b+b^{2})}}\\&={\frac {b^{4}+28b^{3}+70b^{2}+28b+1}{8b^{3}+56b^{2}+56b+8}}.\end{aligned}}