# While resolving a continued fraction by simplifying it into a quadratic equation, does the extraneous root have any significance?

Suppose we want to find the value of the the following expression, with infinite terms: $$y = {1\over 1+ {1\over 1+ {1\over 1+ {1\over \ddots}}}}$$ To solve this, we follow the following procedure, by replacing the continued fraction with the equivalent term itself, and then forming and solving the quadratic equation formed.: $$y = \frac{1}{1+y}$$ By solving the quadratic $y^2+y-1$, we get two roots, $y=\frac{-1+\sqrt5}{2}$ and $y=\frac{-1-\sqrt5}{2}$, out of which only the positive root is considered a solution of the equation.

Why do we get the extra negative root while solving this? Does this root have any significance relating to the original question?

• The quadratic equation should be $y^2\color{blue}{+}y-1=0$ so $y=\frac{-1\pm\sqrt5}{2}$ are the two solutions. If $x=1$, the innermost/bottom-most fraction is $\frac{1}{2}$ and the sequence (of fractions, simplifying from the bottom upwards) converges alternately to a positive value: $\frac{-1+\sqrt5}{2}$. – Marconius Sep 14 '15 at 21:45
• Corrected. Thanks. – Aditya Garg Sep 15 '15 at 18:08

From

$$y=\frac{1}{1+y} \tag{1}$$

the resulting quadratic equation should be

$$y^2+y−1=0 \tag{2}$$

with the two solutions being $y=\frac{−1\pm\sqrt5}{2}$.

It turns out that these values of $y$ are the fixed points of the function $f:\mathbb{R}\to\mathbb{R}$ where

$$f(x)=\frac{1}{1+x} \tag{3}$$

The continued fraction converges to whatever value the iterated function sequence

$$x,\ f(x),\ f(f(x)),\ f(f(f(x))), \ldots$$

converges to. The fixed points of $f$ are the only possible values to which the continued fraction can converge. Convergence of the continued fraction in the neighbourhood of either of the fixed points depends on whether the fixed point is attractive or not. Since $f$ is continuously differentiable in the neighbourhoods of both $\frac{−1-\sqrt5}{2}$ and $\frac{−1+\sqrt5}{2}$ (there is a discontinuity only at $-1$), either fixed point is attractive if and only if

$$|f'(x_0)|<1$$

From (3),

$$f'(x)=\frac{-1}{(1+x)^2} \tag{4}$$

Now at either fixed point, equation (1) is obeyed with $y=x_0$, so (4) reduces to

$$f'(x_0)=-\left(\frac{1}{1+x_0}\right)^2=-x_0^2 \tag{5}$$

So now calling $x_1=\frac{−1-\sqrt5}{2}$ and $x_2=\frac{−1+\sqrt5}{2}$ we must have

$$f'(x_1)=-\left(\frac{−1-\sqrt5}{2}\right)^2=\frac{-3-\sqrt5}{2} \implies |f'(x_1)|>1$$

so $x_1$ is an unstable fixed point, and the sequence will diverge away from this value.

Hence, the root $x_1=\frac{−1-\sqrt5}{2}$ arises as a non-attractive fixed point that can be safely disregarded as the continued fraction cannot possibly converge to this value.

For $x=\frac{-1-\sqrt5}{2}$ exactly, $f(x)=\frac{-1-\sqrt5}{2}$ but this is an unstable equilibrium (any perturbations cause a divergence).

Also,

$$f'(x_2)=-\left(\frac{−1+\sqrt5}{2}\right)^2=\frac{\sqrt5-3}{2} \implies |f'(x_2)|<1$$

so $x_2$ is an attractive fixed point. Because $f'(x_2)<0$, the sequence will converge in alternating fashion (oscillating about the fixed point) to the fixed point at $x_2=\frac{−1+\sqrt5}{2}$.

Since \begin{align} x\in(-\infty,-2) &\implies f(x)\in(-1,0) \\ x\in(-1,0) &\implies f(x)\in(0,\infty) \\ x\in(0,\infty) &\implies f(x)\in(0,\infty) \\ \end{align}

it is fairly straightforward to see that the sequence will converge to $\frac{-1+\sqrt5}{2}$ for $x\in(-\infty,-2)\cup(-1,0)\cup(0,\infty)$.

• @AdityaGarg - This comes from a first-order Taylor-Maclaurin expansion about the fixed point: $f(x)\approx f(x_0)+f'(x_0)(x-x_0)$ which can be rearranged to $f(x)-f(x_0)\approx f'(x_0)(x-x_0)$. When $|f'(x_0)|<1$, $f(x)$ is closer to $f(x_0)$ than $x$ is to $x_0$, so the sequence is converging. When $|f'(x_0)|>1$, $f(x)$ is further from $f(x_0)$ than $x$ is from $x_0$, so the sequence is diverging. – Marconius Sep 16 '15 at 10:11