I'm supposed to find the value of the infinite continued fracton $[2;1,3,1,3,1,3,1,3...]$. How would I go about doing this?
Asked
Active
Viewed 42 times
0
-
1If the value is $x$, look to substitute $x-2$ in the expression for $x$. – saulspatz Feb 10 '19 at 19:46
1 Answers
1
Denote the continued fraction as $x$ and $x-2=y$, i.e.
$$y=\cfrac1{1+\cfrac1{3+\cdots}}$$
Moreover we get that
\begin{align*} y=\cfrac{1}{1+\cfrac{1}{3+y}}\implies y&=\frac1{\cfrac{4+y}{3+y}}\\ &=\frac{3+y}{4+y} \end{align*}
Which overall leads to the quadratic equation
$$y^2+3y-3=0$$
Solving this equation and choosing the positive solution further leads to
$$y=\frac{-3+\sqrt{21}}2\implies x=\frac{-3+\sqrt{21}}2+2$$
$$\therefore~x~=~2+\cfrac1{1+\cfrac1{3+\cdots}}~=~\frac{1+\sqrt{21}}2$$
The solution is confirmed by this calculator which produces the given continued fraction for the resulting value of $x$.
mrtaurho
- 16,103
-
1
-
@DrewWeisserman No problem, happy to help. Note that the idea of this whole procedure can be generalised in order to solve for the value of periodic continued fractions. – mrtaurho Feb 10 '19 at 20:17
-
1