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Let's say for example we have $a_1=1/2$ and $a_{n+1}=(a_n+1)/(a_n+2)$,

How would one go about solving for the general formula for $a_n$ in terms of $n$?

From this I'd like to try to solve a different problem, wanting this as a mere example.

In just making up this question on the fly, it's remarkable that the limit of $a_n$ as $n \to \infty$ is $(1+\sqrt{5})/2-1$ and so this could be constructive in relation to the golden ratio.

Theo Bendit
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tyty
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    I think the very first step would be to compute some terms of the sequence to see if you can figure out a pattern (which you can attempt to prove by induction). This is a simple thing you can do, and a lot of the time, it will yield important insight. There are plenty of times where it won't, but its effort-to-success ratio is, over all, quite favourable! The appearance of the golden ratio could also come about from some connection with Fibonacci numbers, but who knows? Happy investigating! – Theo Bendit Dec 24 '21 at 23:57
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    I understand that and invoked a regression analysis and came up with nothing but an approximation. I'm hoping one could step up and show the steps in solving this to produce an exact solution. – tyty Dec 24 '21 at 23:59
  • Regression analysis is too involved for what I'm thinking. If $a_1 = 1/2$, then what is $a_2$? What about $a_3$? Then $a_4$, and $a_5$? You can compute all of these values exactly, and write them as exact rational numbers. If you can spot a pattern, then you can try to prove it using induction. – Theo Bendit Dec 25 '21 at 00:01
  • 1 5.00000000000000E-01 2 6.00000000000000E-01 3 6.15384615384615E-01 4 6.17647058823529E-01 5 6.17977528089888E-01 6 6.18025751072961E-01 7 6.18032786885246E-01 8 6.18033813400125E-01 9 6.18033963166707E-01 10 6.18033985017358E-01 – tyty Dec 25 '21 at 00:02
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    I would firmly recommend writing them as quotients of integers, rather than decimal expansions. If your calculator cannot handle this task, then I also firmly recommend performing the calculations by hand, as these decimal expansions are beyond unenlightening. – Theo Bendit Dec 25 '21 at 00:04
  • For example,$$a_2 = \dfrac{\frac{1}{2}+1}{\frac{1}{2} + 2} = \dfrac{1+1 \times 2}{1 + 2 \times 2} = \frac{3}{5}.$$ – Theo Bendit Dec 25 '21 at 00:07
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    I don't believe there is a general procedure for calculating $a(n)$ knowing the recurrence relation alone. The case where the recurrence is linear is very well known, but here it isn't. It is fairly trivial to prove that the limit of $a(n)$, if it exists, is a solution of $a=\frac{a+1}{a+2}$. i.e. the fixed point of the map $x\mapsto \frac{x+1}{x+2}$. The convergence may follow if this map is locally a contraction around the fixed point. See https://en.m.wikipedia.org/wiki/Banach_fixed-point_theorem . –  Dec 25 '21 at 00:09
  • OK. This ends up being simple, as a(n)=F(3+n)/F(4+n) where F is the Fibonacci function. I'll make a new sequence to get away from Fibonacci stuff so the solution will not be well characterized using rudimentary steps.

    The hope in this question would be to show how any sequence could be characterized, at least in general ways.

     – tyty Dec 25 '21 at 00:10
  • @tyty Feel free to try a different sequence, but I think you honestly got a good exposure to the relevant techniques to solving non-linear recurrence relations (and it's good that it turned out simple enough to solve). As Stinking Bishop points out, there really isn't any better, more general, cutting edge procedure for figuring out closed formulas for recurrence relations. Basically, if somebody spots a pattern in a particular sequence, good for them, but most of the time, that's too much to ask. – Theo Bendit Dec 25 '21 at 00:13
  • See https://en.wikipedia.org/wiki/Pell%27s_equation & https://en.wikipedia.org/wiki/Continued_fraction – PM 2Ring Dec 25 '21 at 00:14

1 Answers1

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Have a look at my answer to this question.

Applied to your problem $$a_n= \frac{2 \left(\left(-3-\sqrt{5}\right)^n-\left(\sqrt{5}-3\right)^n\right)}{\left(\sqrt{5 }-1\right) \left(\sqrt{5}-3\right)^n+\left(1+\sqrt{5}\right) \left(-3-\sqrt{5}\right)^n}$$ and the first terms will be $$\left\{\frac{1}{2},\frac{3}{5},\frac{8}{13},\frac{21}{34},\frac{55}{89},\frac{144}{ 233},\frac{377}{610},\frac{987}{1597},\frac{2584}{4181},\frac{6765}{10946},\frac {17711}{28657},\frac{46368}{75025},\cdots\right\}$$

Numerators and denominators correspond to known sequences (search in $OEIS$) and they are related to Fibonacci numbers.

Claude Leibovici
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  • Yes, I recognized this early on. I wanted to see people approach the problem. See one of my earlier posts. I just didn't put into terms beyond F(x) where F is the Fibonacci function, ie F(n)=1/sqrt(5)*(phi^n-(1-phi)^n) and in particular the correct index. – tyty Dec 25 '21 at 04:01