\[x_1=\frac{7+ \sqrt{2}}{2}\]
as a continued fraction:Let
\[a_1= int(\frac{7+ \sqrt{2}}{2})=4\]
, and \[\frac{7+ \sqrt{2}}{2}-4= \frac{\sqrt{2}-1}{2}\]
. Let \[x_2=\frac{2}{\sqrt{2}-1}\]
.Let
\[a_2= int(\frac{2}{\sqrt{2}-1})=4\]
, and\[ \frac{2}{\sqrt{2}-1}-4=\frac{6-4 \sqrt{2}}{\sqrt{2}-1}=\frac{6-4 \sqrt{2}}{\sqrt{2}-1} \frac{\sqrt{2}+1}{\sqrt{2}+1}= 2 \sqrt{2}-2\]
.\[x_3=\frac{1}{2 \sqrt{2}-2}\]
.
Let \[a_3= int(\frac{1}{2 \sqrt{2}-2}=1\]
.\[\frac{1}{2 \sqrt{2}-2}-1=\frac{3-2 \sqrt{2}}{2 \sqrt{2}-2}=\frac{3-2 \sqrt{2}}{2 \sqrt{2}-2}\]
.Let
\[x_4=\frac{2 \sqrt{2}-2}{3-2 \sqrt{2}}\frac{3+2 \sqrt{2}}{3+2 \sqrt{2}}=\frac{2}{\sqrt{2}-1}=x_2\]
.The cycle is repeating, so the continued fraction is
\[[ 4, \; \lt 4, \; 1 \gt ]\]
.