3
$\begingroup$

I study two similar nonlinear functional recurrence systems, given by

$$P_\pm:\qquad f_n\cdot(1\pm g f_{n-1}) = g\mp(1+2g)f_{n-1} \qquad (n>0)$$ and $$f_0=g$$

Here $f_n$ and $g$ are functions of one variable with

$$g(z)=\frac{z}{1-z}$$

I try to find the solution of $P_+$, but so far, only was able to solve $P_-$. I am grateful for any ideas.

What I tried so far: Although I am not sure if this goes into the right direction, here is how I did $P_-$: Substitute $q_n = (1-gf_n)/(1+g)$ and we obtain

$$q_n = 2-\frac{1}{q_{n-1}}\qquad(n>0)$$ with $$q_0 =1-g$$

Just by "looking at" the first functions $q_n$, it seems that

$q_n(z)=\frac{z(n+2)-1}{z(n+1)-1},$ which can indeed be verified to fulfill the recurrence.

However, for the system $P_+$, the analogous substitution $q_n=(1+gf_n)/(1+g)$ yields the slightly different recurrence $$q_n=\frac{1}{q_{n-1}}-2z$$ with $$q_0 = 1+g-2z.$$

I have no idea how to solve this, but one might exploit the knowledge about $P_-$. Finding a transformation between those very similar problems would allow to express the solutions of $P_+$ by those of $P_-$.

$\endgroup$
0
$\begingroup$

I finally found the answer, one can solve both problems

$$(P_-):\qquad q_n = 2-\frac{1}{q_{n-1}},\qquad q_0=1-g\\ (P_+):\qquad q_n=\frac{1}{q_{n-1}}-2z,\qquad q_0=1+g-2z$$

with a substitution also known from the algebraic Riccati equation (see, e.g. here),

$$q_n=\frac{a_{n+1}}{a_n},\qquad a_1=q_0, \qquad a_0=1.$$

This transforms the problems into linear equations

$$(P_-):\qquad a_n = 2a_{n-1}-a_{n-2},\\ (P_+):\qquad a_n = a_{n-2}-2za_{n-1}.$$

These can be solved by standard methods, e.g., introducing a generating function which in addition to $z$ depends on a new variable, that corresponds to the index $n$. The final solutions are

$$(P_+):\qquad q_n(z)=-\frac{x_+(z)^{n+2}(1-\Gamma(z))-x_-(z)^{n+2}(1+\Gamma(z))}{x_+(z)^{n+1}(1-\Gamma(z))-x_-(z)^{n+1}(1+\Gamma(z))}$$

where $$q_\pm(z) := z\pm\Gamma(z),\qquad \Gamma(z):=\sqrt{1+z^2},$$ and $$(P_-):\qquad q_n(z) = \frac{z(n+2)-1}{z(n+1)-1}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.