A continued fraction with a delayed repeating part

Let's ignore the $3,6$ for a sec. The continued fraction that remains gives rise to the equation

$$x=1 + \frac{1}{4+\frac{1}{x}}$$

This simplified to $4x^2-4x-1=0$. Apply the quadratic equation to get $x=\frac{1+\sqrt{2}}{2}$. Then plug this into

$$y=3+\frac{1}{6+\frac{1}{x}}$$ to get $y=\frac{1}{4}(14-\sqrt{2})$ which intriguingly has a decimal expansion begining $3.14$


Assume a repeating section of fraction as $x$.

$$x=1+\dfrac{1}{4+\dfrac{1}{x}}$$

Now you can find the value of $x$.

Then the original fraction is:

$$3+\dfrac{1}{6+\dfrac{1}{x}}$$

Tags:

Number Theory

Continued Fractions

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