Find the value of the periodic continued fraction with given terms

Question

Find the value of the periodic continued fraction with the terms $1, 3, 4, 3, 2, 3, 4, 3, 2, 3, 4, 3, 2, . . .$

We see that it starts to be periodic after $1$, i.e, $3,4,3,2$ then $3,4,3,2$ etc...

I know that $x= \frac{A_{k+1}}{B_{k+1}}$ = $\frac{A_{k-1}+x.A_K}{B_{k-1}+x.B_K}$ where $q_{k+1}=x$

I have $$x={3+\cfrac{1}{4+\cfrac{1}{3+\cfrac{1}{2+\cfrac{1}{x}}}}}$$ for the periodic part. If I use my formula to compute the right-hand side, I will end up with a quadratic in $x$. Solving for $x$ gives me the following quadratic formula: $5x^2-14x-7=0$, then I did $1+\frac1x$ to find the whole continued fraction, and I got the same quadratic equation. Is this correct?

Answer 1

if we take $u=x+1,$ we find that u > 2 has purely periodic c.f. $(2,3,4,3).$

Next $$ \left( \begin{array}{cc} 2 & 1 \\ 1 & 0 \\ \end{array} \right) \left( \begin{array}{cc} 3 & 1 \\ 1 & 0 \\ \end{array} \right) \left( \begin{array}{cc} 4 & 1 \\ 1 & 0 \\ \end{array} \right) \left( \begin{array}{cc} 3 & 1 \\ 1 & 0 \\ \end{array} \right) = \left( \begin{array}{cc} 7 & 2 \\ 3 & 1 \\ \end{array} \right) \left( \begin{array}{cc} 13 & 4 \\ 3 & 1 \\ \end{array} \right) = \left( \begin{array}{cc} 97 & 30 \\ 42 & 13 \\ \end{array} \right) $$

Then $$ u = \frac{97u+30}{42u + 13} $$ gives $$ 42u^2 + 13 u = 97 u + 30 $$ or $$ 7 u^2 - 14 u - 5 = 0. $$ Since $u > 0,$ $$ u = \frac{7 + 2 \sqrt{21}}{7} = 1 + \frac{ 2 \sqrt{21}}{7} $$ and $x=u-1$ is $$ x = \frac{ 2 \sqrt{21}}{7} $$

Answer 2

Yes, you're right. We have $5x^2-14x-7=0$. This is the answer.

\begin{eqnarray} x&=&3+\dfrac{1}{4+\dfrac{1}{3+\dfrac{1}{2+\dfrac{1}{x}}}}\\ &=&3+\dfrac{1}{4+\dfrac{1}{3+\dfrac{1}{\dfrac{2x+1}{x}}}}\\ &=&3+\dfrac{1}{4+\dfrac{1}{3+\dfrac{x}{2x+1}}}\\ &=&3+\dfrac{1}{4+\dfrac{1}{\dfrac{6x+3}{2x+1}+\dfrac{x}{2x+1}}}\\ &=&3+\dfrac{1}{4+\dfrac{1}{\dfrac{7x+3}{2x+1}}}\\ &=&3+\dfrac{1}{4+\dfrac{2x+1}{7x+3}}\\ &=&3+\dfrac{1}{\dfrac{28x+12}{7x+3}+\dfrac{2x+1}{7x+3}}\\ &=&3+\dfrac{1}{\dfrac{30x+13}{7x+3}}\\ &=&3+\dfrac{7x+3}{30x+13}\\ &=&\dfrac{90x+39}{30x+13}+\dfrac{7x+3}{30x+13}\\ &=&\dfrac{97x+42}{30x+13}\\ \end{eqnarray}

Now we have

\begin{eqnarray} x&=&\dfrac{97x+42}{30x+13}\iff\\ 30x^2+13x&=&97x+42\iff\\ 30x^2-84x-42&=&0\iff\\ 5x^2-14x-7&=&0 \end{eqnarray}

Solving this equation we have

$$x_{1,2}=\dfrac{14\pm\sqrt{56}}{10}=\dfrac{14\pm 2\sqrt{14}}{10}=\dfrac{7}{5}\pm \dfrac{1}{5}\sqrt{14}.$$