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I hadn't even noticed before entering the subject above the parallel with the title of an earlier question posted here titled Summation methods for divergent series. And it's just mutatis mutandis as far as the subject line goes, but not when you get to the question. This question is more primitive.

This is just a question posted fruitlessly on November 30th on stackexchange:

The solutions to $$ x^2-6x+10=0 \tag 1 $$ are $$ 3\pm i\tag2. $$ 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 $$ x=6 - \cfrac{5}{3-\cfrac{5}{6-\cfrac{5}{3-\cfrac{5}{6-\cdots}}}} \text{ (with 3 and 6 alternating).} $$

QUESTION: 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)?

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  • $\begingroup$ So, have you had a chance to look at any of the sources I cited in my answer? $\endgroup$
    – Gerry Myerson
    Commented Mar 5, 2015 at 10:47
  • $\begingroup$ @GerryMyerson : I see no posted answers here or on stackexchange, and your name appears in the comments there only because you linked to this present question. Where is your answer that you refer to? ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Commented Mar 5, 2015 at 16:23
  • $\begingroup$ Better get your computer checked. mathoverflow.net/questions/198880/… is the answer I posted here two days ago. $\endgroup$
    – Gerry Myerson
    Commented Mar 5, 2015 at 22:58
  • $\begingroup$ I'm seeing it now, viewing it from a different computer. I'll try the other one again later. $\endgroup$
    – Michael Hardy
    Commented Mar 5, 2015 at 23:28
  • $\begingroup$ So, have you had a chance to look at any of the sources I cited in my answer? $\endgroup$
    – Gerry Myerson
    Commented Oct 29, 2019 at 6:33

1 Answer 1

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I copy'n'paste the review by M. Znojil of the paper, Kachmar, V. S., Rusyn, B. P., Shmoĭlov, V. I., Algorithms for computing the values of continued fractions, Zh. Vychisl. Mat. Mat. Fiz. 38 (1998), no. 9, 1436--1451; translation in Comput. Math. Math. Phys. 38 (1998), no. 9, 1375–1390, MR1669146 (99m:65026):

In a pedagogically oriented communication the authors emphasize that divergent continued fractions may be analysed in a way resembling techniques of re-summation of divergent series. Having this type of application in mind, they review existing algorithms (with a few historical comments added), compare some of their properties (number of operations needed, etc.) and illustrate their characteristic features (precision, stability and the rate of convergence) on a few tabulated numerical examples.

There is a series of Russian publications that may do what you want to do:

Shmoĭlov, V. I.; Zayats, I. A.; Sloboda, M. Z. {\cyr Raskhodyashchiesya nepreryvnye drobi}. (Russian. Russian summary) [Diverging continued fractions] Merkator, Lʹviv, 2000. 820 pp. ISBN: 966-7563-03-0. MR1901494 (2003h:40005).

Shmoĭlov, V. I. {\cyr Nepreryvnye drobi. Tom} I. (Russian) [Continued fractions. Vol. I] {\cyr Periodicheskie nepreryvnye drobi}. [Periodic continued fractions] Merkator, Lʹviv, 2004. 645 pp. ISBN: 966-7563-06-3. MR2058216 (2006i:40007).

Shmoĭlov, V. I. {\cyr Nepreryvnye drobi. Tom} II. (Russian) [Continued fractions. Vol. II] {\cyr Raskhodyashchiesya nepreryvnye drobi}. [Diverging continued fractions] Merkator, Lʹviv, 2004. 558 pp. ISBN: 966-7563-07-3. MR2058217 (2006i:40006).

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    $\begingroup$ They describe algorithms and give some examples. But they don't even try do discuss convergence of their algorithms. The convergence is a big question because it depends on delicate diophantine properties of numbers. $\endgroup$
    – Alexey Ustinov
    Commented Oct 29, 2019 at 3:42

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