Exercise 7.7.2 (Value of $\langle 9, \overline{9,18} \rangle$)

Question

Find the irrational number having continued fraction expansion $\langle 9, \overline{9,18} \rangle$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $	heta = \langle \overline{9,18} angle$. Then $	heta = \langle 9,18, 	heta angle$, so
$$	heta = 9 + \frac{1}{18 + \frac{1}{	heta}}.$$
Therefore
\begin{align*}
	heta &= \frac{h_1 	heta + h_0}{k_1 	heta + k_0}\
&=\frac {163 \,	heta + 9}{18\,  	heta + 1},
\end{align*}
thus
\begin{align}
18 \, 	heta^2 - 162\,  	heta - 9 = 0,\qquad 	heta >0.
\end{align}
so
$$	heta = \frac{9 + \sqrt{83}}{2} .$$
Then 
\begin{align*}
\xi &= \langle 9, \overline{9,18} angle \
&= 9 + \frac{1}{	heta}\
&= 9 + \frac{2}{9 + \sqrt{83}}\
&= \sqrt{83}.
\end{align*}
So
$$\langle 9, \overline{9,18} angle = \sqrt{83}.$$

\end{proof}
With Sagemath:
\begin{verbatim}
sage: cfe = continued_fraction([(9,), (9,18)])
sage: cfe.value()
sqrt83
sage: verif = continued_fraction(sqrt(83)); verif
[9; 9, 18, 9, 18, 9, 18, 9, 18, 9, 18, 9, 18, 9, 18, 9, 18, 9, 18, 9, ...]
\end{verbatim}

Exercise 7.7.2 (Value of $\langle 9, \overline{9,18} \rangle$)

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Exercise 7.7.2 (Value of $\langle 9, \overline{9,18} \rangle$)

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