Divergent continued fractions?

The solutions to $$ x^2-6x+10=0 \tag 1 $$ are $$ x = 3\pm i\tag2. $$ Rearranging $(1)$ just a bit, we get $$ x = 6 -\frac{10}x \tag3 $$ and then substituting the right side of $(3)$ for $x$ within the right side we get $$ x=6 - \cfrac{10}{6-\cfrac{10}x} $$ and iterating we have $$ x=6 - \cfrac{10}{6-\cfrac{10}{6-\cfrac{10}{6-\cfrac{10}{6-\cdots}}}} \tag 4 $$ (or in lowest terms $$ x=6 - \cfrac{5}{3-\cfrac{5}{6-\cfrac{5}{3-\cfrac{5}{6-\cdots}}}} $$ with $3$ and $6$ alternating).

Just as one speaks of "summation methods" by which $1+2+3+4+\cdots=\dfrac{-1}{12}$, etc., might there be some "division method" by which $(4)$ is equal to $(2)$?

PS: Might one prove that this continued fraction diverges in the usual sense by proving that if it converges then it must converge to the solution of $(1)$ (and obviously it does not)?

Solution 1:

Many regularization techniques are based on averaging. Let us define $a_0,b_0,m\in\Bbb C$ and

$$a_{n+1}=6-\frac{10}{a_n}$$

$$b_n=\frac{b_n+m\left(6-\frac{10}{b_n}\right)}{1+m}$$

If $a_n\to3\pm i$, then if $b_n$ converges, it also converges to $3\pm i$.

Choosing

$$m_\pm=-\frac{4\pm3i}5$$

gives us

$$b_{n,\pm}=\frac{(4\pm3i)\left(6-\frac{10}{b_n}\right)-5b_n}{-1\pm3i}$$

which should converge to one of the roots depending on the choice of sign.