Closed form of a Continued Fraction

Question

I am trying to find a closed form of a continued fraction: $$[1,2,3,5,3,5,3,5...]$$ I understand I have to form a quadratic of some sort and factorise and to give me an irrational number but I am not sure how to construct this quadratic. Any help would be appreciated. Thanks

Answer 1

You are right that you have to find a quadratic. In general, let the recursive part be your variable; here, we get the result by choosing

$$x = \frac{1}{3+\frac{1}{5+\frac{1}{3+\dots}}}$$

Then, to solve for $x$, we make the equation

$$x=\frac{1}{3+\frac{1}{5+x}}$$

which yields the quadratic $3x^2+15x-5=0$, whose solutions turn out to be

$$x=-\frac{5}{2}\pm\frac{\sqrt{285}}{6}$$

The negative solution is discarded (since we want our fraction to be positive). This, however, is not our original continued fraction. The closed form for the fraction we want to find follows from $$1+\frac{1}{2+x}$$ the calculation of which is trivial.

Answer 2

The fraction in question is $1/(1+1/(2+x))$ where $$x=1/(3+1/(5+x))$$ Denesting the fractions gives us $$x=(5+x)(1-3x)\implies \\ 3x^2+15x-5=0\implies \\ x=\frac{-15+\sqrt{285}}6$$ where the $-$ solution is rejected since $x>0$. Then it is a matter of working up the final steps to arrive at $\frac{49-\sqrt{285}}6$.

Answer 3

Just looking at the last part, that leaves some work for you (and it's not clear what your "$\ldots$" mean (i.e. what contiues).

Define $$ y=3+\frac{1}{5+\frac{1}{y}} $$

We can rewrite that as: \begin{eqnarray} y\left(5+\frac{1}{y}\right) =& 3\left(5+\frac{1}{y}\right)+1\\ 5y+1 =& 16+\frac{3}{y}\\ 5y^2+y =& 16y+3\\ 5y^2-15y-3 =& 0 \end{eqnarray}

That's a quadratic in standard form.