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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)?

Michael Hardy
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    I know my approach may be very simple, but I graphed $y=x$ and $y=6-\frac{10}{x}$ and they have no real point of intersection. That's expected because you already indicated there are comples solutions. Your iteration also reminds me of Newton Rhapson process where in this case I would assume the iteration diverges. (An iteration does not always have to converge to a number!) And from that standpoint, (2) does not have to be equal to (4) By the way: +1 – imranfat Nov 30 '14 at 17:18
  • I would make the following distinction: "summation methods" work because an infinite divergent series can be shown to equal any number you wish (I don't remember who proved this). A similar "division method" likely exists, but it seems your question is really: does $x=6 - \cfrac{5}{3-\cfrac{5}{6-\cfrac{5}{3-\cfrac{5}{6-\cdots}}}} = 3\pm i$. – Ari Nov 30 '14 at 17:26
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    @Ari : I suspect you are confused if you say "infinite divergent series can be shown to equal any number you wish". There is a result that says _conditionally convergent_ series can be made to converge to any number you want or $+\infty$ or $-\infty$, by rearranging the terms. But that is certainly not the reason why "summation methods" work. The divergent series $1+2+3+\cdots$ cannot be made to converge in the usual sense to $-1/12$ by rearranging the terms. Rather, it can only converge in a different sense. ${}\qquad{}$ – Michael Hardy Nov 30 '14 at 22:21
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    Alternately, we might also ascribe finite values to [infinite tetration](http://en.wikipedia.org/wiki/Tetration#Extension_to_infinite_heights) for values which lie outside of its convergence interval, $(1/e)^{\large e}\le x\le e^{\large1/e}$. For instance, using the recurrence relation $i=e^{^{{\Large i}\tfrac\pi2}}$, we have $i\equiv~^{^{^{^{\Large\infty}}}}\bigg(e^{^\tfrac\pi2}\bigg)$. – Lucian Dec 01 '14 at 06:30
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    @imranfat Yeah I would say it reminds me of the square root algorithm that comes from the newton-raphson. So if you plug in $-1$ so you can see if there's a sequence converging to $i$, it just hops around, unless you make a complex guess to start off with. So it doesn't converge in a traditional sense since the convergents will all be real right? – snulty Dec 05 '14 at 19:06
  • Now posted to MO, http://mathoverflow.net/questions/198880/division-methods-for-divergent-continued-fractions/ – Gerry Myerson Mar 02 '15 at 23:17
  • Looking at this from a different machine a couple of days ago, I saw an answer posted here. From this machine, I don't see it. Can others see an answer posted here? – Michael Hardy Mar 08 '15 at 00:37

1 Answers1

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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.

Michael Hardy
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Simply Beautiful Art
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