Heron-Verfahren/Wurzel 3/Wurzel 1 durch 3/Aufgabe/Lösung
Es ist
x
1
=
1
+
3
2
=
2
{\displaystyle {}x_{1}={\frac {1+3}{2}}=2\,}
und
x
2
=
x
1
+
3
x
1
2
=
2
+
3
2
2
=
7
4
.
{\displaystyle {}x_{2}={\frac {x_{1}+{\frac {3}{x_{1}}}}{2}}={\frac {2+{\frac {3}{2}}}{2}}={\frac {7}{4}}\,.}
Es ist
y
1
=
1
+
1
3
2
=
2
3
{\displaystyle {}y_{1}={\frac {1+{\frac {1}{3}}}{2}}={\frac {2}{3}}\,}
und
y
2
=
y
1
+
1
3
y
1
2
=
2
3
+
1
3
2
3
2
=
2
3
+
1
2
2
=
7
6
2
=
7
12
.
{\displaystyle {}y_{2}={\frac {y_{1}+{\frac {\frac {1}{3}}{y_{1}}}}{2}}={\frac {{\frac {2}{3}}+{\frac {\frac {1}{3}}{\frac {2}{3}}}}{2}}={\frac {{\frac {2}{3}}+{\frac {1}{2}}}{2}}={\frac {\frac {7}{6}}{2}}={\frac {7}{12}}\,.}
Es ist
x
0
⋅
y
0
=
1
⋅
1
=
1
,
{\displaystyle {}x_{0}\cdot y_{0}=1\cdot 1=1\,,}
x
1
⋅
y
1
=
2
⋅
2
3
=
4
3
,
{\displaystyle {}x_{1}\cdot y_{1}=2\cdot {\frac {2}{3}}={\frac {4}{3}}\,,}
und
x
2
⋅
y
2
=
7
4
⋅
7
12
=
49
48
.
{\displaystyle {}x_{2}\cdot y_{2}={\frac {7}{4}}\cdot {\frac {7}{12}}={\frac {49}{48}}\,.}
Die Heron-Folge
x
n
{\displaystyle {}x_{n}}
R
{\displaystyle {}\mathbb {R} }
3
{\displaystyle {}{\sqrt {3}}}
y
n
{\displaystyle {}y_{n}}
R
{\displaystyle {}\mathbb {R} }
1
3
{\displaystyle {}{\frac {1}{\sqrt {3}}}}
x
n
⋅
y
n
{\displaystyle {}x_{n}\cdot y_{n}}
1
{\displaystyle {}1}
Q
{\displaystyle {}\mathbb {Q} }
Q
{\displaystyle {}\mathbb {Q} }
