If you don't feel like reading the justification for my question, just skip down to the question.

Some recurrence relations can be solved by finding invariants, or quantities that remain unchanged. For example, consider the sequences defined by $$a_{n+1}=\frac{a_nb_n}{a_n+b_n+1}$$ $$b_{n+1}=a_n+b_n$$ $$a_0=b_0=-1$$ If one defines the function $\mu$ as $$\mu (x,y)=x+xy+y$$ then it is easily proven that $$\mu (x,y)=\mu\bigg(\frac{xy}{x+y+1}, x+y\bigg)$$ From this, it follows that $$\mu (a_{n+1}, b_{n+1})=\mu (a_n,b_n)$$ and $$\mu (a_n,b_n)=\mu(a_0,b_0)$$ $$\mu (a_n,b_n)=-1$$ $$a_n+a_nb_n+b_n=-1$$ $$a_n=\frac{-1-b_n}{1+b_n}$$ $$a_n=-1$$ and so $$b_{n+1}=-1+b_n$$ $$b_{n+1}=b_n-1$$ ...yielding explicit formulas for $a_n$ and $b_n$: $$a_n=-1,\space\space\space b_n=-1-n$$

This was perhaps a trivial (and manufactured) example, but it demonstrates how invariants are used to solve recursions.


I would like to know the following: what are some methods used to find invariants? That is, what methods might one use to find a particular nontrivial solution $\mu$ to the functional equation $$\mu(x,y)=\mu(f(x,y),g(x,y))$$ where $f,g$ are given?

Franklin Pezzuti Dyer
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  • Nice question. Experience and "*educated guesses*" provide reliable tools, but in general it might be a non-trivial problem. Related: https://math.stackexchange.com/questions/2459643/finding-an-integral-formula-for-intertwined-recursive-sequences – Jack D'Aurizio Nov 28 '17 at 01:00
  • The problem you want is even a little harder than you have written, since you want $f$ and $g$ to be functions of both $x$ and $y$, as per your example, not just single variable like you wrote. – Will Fisher Nov 28 '17 at 01:35
  • @WillFisher Ooh, good observation. I should probably clarify in my question. – Franklin Pezzuti Dyer Nov 28 '17 at 23:57

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