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    For a simple continued fraction:
     S = a_0 + \frac{1}{a_1+\frac{1}{a_2+\frac{  1}{a_3+\frac{1}{a_4 +\frac{1}{a_5+ ...}}}}} = [a_0; a_1, a_2, a_3, a_4, ... ]

    I was initially wondering what [0;1,1,1,1,1,1,1,...] (or [0; \overline{1}\ ]) would be [Turns out it's the golden ratio; whoopdy doo], then I thought that I might as well do it for [0; \overline{x}] while I'm at it.

    I did all of this crap:
    Spoiler:
    Show
     S = \frac{1}{x+\frac{1}{x+\frac{1}{x  +\frac{1}{x +\frac{1}{x+ ...}}}}}
    So I noticed S^{-1} = x+ \frac{1}{x+\frac{1}{x+\frac{1}{x  +\frac{1}{x +\frac{1}{x+ ...}}}}}= S + x
     1= S^2 + Sx \therefore S^2 + Sx - 1 = 0 \Rightarrow S = \frac{ -x \pm \sqrt{x^2 +4}}{2}
    Firstly, ^^ from this, What I don't understand is what this \pm meeaaans. Is the minus option just residual from the way the equation is set up or is it another valid solution for when x is negative or non-real or something!? [speaking of which, all non-zero complex numbers would still work right?]

    *after plotting the graph, it looks like the negative one is valid for negative numbers and positive one is valid for positive numbers; make sense, but what about complex numbers?

    Secondly, how would you convert a non-simple continued fraction into a simple one? is there an easy method at least for rational values of x or is it all too complicated?, like for if T(x) = \frac{x}{x+\frac{x}{x+\frac{x}{x  +\frac{x}{x +\frac{x}{x+ ...}}}}}, could T(x) be related to S(x).

    **Side note: the graph of T(x) with loads of terms looks rather interesting, with a tan(x) looking pattern in the interval (-4, 0] that squishes together with more terms. Then with infinitely many terms, a 'gap' is sort of produced (which I assumed was because it was undefined bc of infinities); when you plug in values from x = -4 to 0, you get non-real complex solutions!?!?

    ^So my question here is: Can a fraction with infinitely many real terms really produce a non-real number?! How does that even work!? It seems that with finitely many terms, infinities do pop up, but for whatever reason as soon as the number of terms becomes infinite, poof, finite, non-real solutions. Like...wtf?I'm assuming it's just an error in the process or something, since we're manipulating S as though it's finite, though if it genuinely maps to non-reals then that'd be pretty effing awesome xDLastly,
    I was taught a certain method for geometric series that basically ended up with
     S = \frac{1-r^n}{1-r} so that with the conditions that |r| < 1 and n -> infinity, then S = \frac{1}{1-r}.

    However, doing it this way seemingly avoids that difficulty and gets the same result; but r can be any number
    Spoiler:
    Show
    S = 1 + r + r^2 + r^3 + ...

S - 1 = r + r^2 + r^3 + ...

    \dfrac{S - 1}{r} = 1 + r + r^2 + r^3 + ...

\quad \quad \quad = S

    S = \frac{1}{1-r}
    I'm unsure as to which one to trust more, but the latter 'seems' just as good as the first one; if not, then I'd love an explanation as to why!
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    I'll try and address a few of your questions...

    (a) Note that the golden ratio is [1; 1,1,...] not [0; 1,1,...]

    (b) Consider x^2 - bx - 1 = 0. This implies that

    x = b + \frac{1}{x}

    Repeatedly substitute for x in the RHS to obtain the simple continued fraction.

    Now consider x^2 - bx - b = 0. This gives you


    x = b + \frac{b}{x} which gives you the continued fraction

    b + \frac{b}{b+\frac{b}{b..}} etc.. Now can you take it from here?

    (c) Don't understand your question about "non-real" numbers. Can you re-phrase?

    (d) Your method for deriving S depends upon the RHS of the expression converging. This is something that needs proof!
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    (Original post by Callum Scott)
    **Side note: the graph of T(x) with loads of terms looks rather interesting, with a tan(x) looking pattern in the interval (-4, 0] that squishes together with more terms. Then with infinitely many terms, a 'gap' is sort of produced (which I assumed was because it was undefined bc of infinities); when you plug in values from x = -4 to 0, you get non-real complex solutions!?!?
    I'm not totally clear what calculations you have/haven't done at this point, so I'm just going to talk in generalities, not about your specific function.

    If you have a sequence of values, it's quite common to be able to say "if this sequence of values converges to a single value, call it A, then A must satisfy a particular equation". (As you found with the simple continued fraction).

    But this doesn't mean that if you find A satisfying that equation, the sequence must converge to A. (it might not converge at all, or it might converge to a different root of the equation).

    As I said, I've not done the calculations, but I assume you've found that if your continued fraction converges to A, then A must be complex. So since there's no way your continued fraction can converge to anything complex (since every finite trucation is real), the only conclusion is that it doesn't converge at all.


    ^So my question here is: Can a fraction with infinitely many real terms really produce a non-real number?!
    No, it can't. However, in some contexts, mathematicians may do the equivalent of "let's ignore the fact it doesn't converge - we've got this lovely formula that we can evaluate even where it doesn't converge, so let's use the formula to define what the continued fraction equals (even though it still doesn't actually converge)".

    This is how you get things like the statement that "1 + 2 + 3 + ... = -1/12"; someone found some formulas to do with \sum_1^\infty n^{-s}, and used them when s = -1, ignoring the small technical issue that the sum no longer converges. The most surprising thing is that this often produces useful results!
 
 
 
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