I want to find the analytic form of the recurrence relation $$a_{n}=\frac{a_{n-1}}{a_{n-1}-a_{n-2}},\,a_{1}=2,\,a_{2}=1 $$ but when looking at the results they seem chaotic. Is it possible that it simply doesn't have any analytic form?
2026-05-14 15:59:36.1778774376
Recurrence relation solution $a_{n}=\frac{a_{n-1}}{a_{n-1}-a_{n-2}}$
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3
This answer just gives more detail on my comments. Define $f:\mathbb{R}^2\rightarrow\mathbb{R}^2\cup \{\phi\}$ by: $$ f(a,b) = \left\{ \begin{array}{ll} \left(b, \frac{b}{b-a}\right) &\mbox{ if $a \neq b$} \\ \phi & \mbox{ otherwise} \end{array} \right.$$ where "$\phi$" formally represents "undefined." Clearly there are no fixed points $f(a,b)=(a,b)$. A point $(a,b)$, initiates a period $k$ orbit if: $$ (a,b) = \underbrace{f \circ f \circ \cdots \circ}_{\mbox{$k$ times}} f (a,b) $$ It suffices to look for periodic orbits with $a\neq b$.
period-2:
Search for $(a,b) = f(f(a,b))$, equivalently: $$ (a,b) \overset{f}{\rightarrow} (b,c) \overset{f}{\rightarrow} (a,b) $$ Then: $(a,b)=f(f(a,b))=f(b, b/(b-a))= (b/(b-a),\frac{b/(b-a)}{b/(b-a)-b})$. Thus: \begin{align} a &= b/(b-a)\\ b &=\frac{b/(b-a)}{b/(b-a)-b} \end{align} These two equations reduce to $-a^2 = (1-a)^2$. There are no real solutions to this equation. However, $a=(1/2)(1-i)$ and $a=(1/2)(1+i)$ are complex solutions. Using $a_1=(1/2)(1-i)$ and $a_2=(1/2)(1+i)$ produces a complex-valued 2-periodic orbit.
period 3:
Search for $(a,b) = f(f(f(a,b)))$, equivalently: $$ (a,b) \rightarrow (b,c) \rightarrow (c,d) \rightarrow (a,b) $$ Then the variables $a,b,c,d$ must satisfy: $$ c=\frac{b}{b-a} , d = \frac{c}{c-b}, a=d, b = \frac{d}{d-c} $$ It does not look like there are any real solutions.
period 4:
It does not look like there are any real solutions.
period 5:
Search for $(a,b) = f(f(f(f(f(a,b)))))$, equivalently: $$ (a,b) \rightarrow (b,c) \rightarrow (c,d) \rightarrow (d, x) \rightarrow (x,f) \rightarrow (a,b) $$ Then $a,b,c,d,x,f$ satisfy (where I use "$x$" instead of "$e$" to avoid the loaded meaning for $e = exp(1)$ when plugging into wolfram): $$ c=\frac{b}{b-a} , d = \frac{c}{c-b} , x = \frac{d}{d-c} , f = \frac{x}{x-d} , a=f, b = \frac{f}{f-x} $$
Numerically solving on wolfram finds 5 real-valued solutions (all cycles of each other), one is: \begin{align} a &\approx 0.39025856959944318578 \\ b &\approx 3.6493359517257782682 \\ c &\approx 1.1197451069249619854 \\ d &\approx -0.44265858616085099250 \\ x &\approx 0.28331895791066755308 \end{align}
Before I suggested using $a_1=0.390259$ and $a_2=3.64934$. The Steven Harding simulation produced what looks like an unstable period-5 orbit (with deviations due to finite decimal truncation error). Using $a_1=a$ and $a_2=b$ with the more accurate 20-digit-precision numbers above will likely produce a trajectory that stays close to the period-5 orbit for a longer time before deviating.
If we had infinite decimal precision then we could produce an orbit that forever stays period 5. My guess, based on numerical observations of sensitivity to initial conditions, and intuition gained from the "Period 3 Implies Chaos" result of Li and Yorke, is that this is a chaotic system. [The Li and Yorke result holds for period 3, not period 5, and to maps of the type $x_{k+1} = g(x_k)$ for $x_k \in \mathbb{R}$, rather than $x_k \in \mathbb{R}^2$, but I would expect the system here to also be chaotic.]