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I hadn't even noticed before entering the subject above the parellel 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)?

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)?

I hadn't even noticed before entering the subject above the parellel 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)?

I hadn't even noticed before entering the subject above the parellel 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)?

I hadn't even noticed before entering the subject above the parellel 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)?

I hadn't even noticed before entering the subject above the parellel 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)?