Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the following recurrence relation

$$z_{n} = c^2 + 2cz_{n-1}^2 + z_{n-1}^4 - (c+c^2)z_{n-1} - 2cz_{n-1}^3 - z_{n-1}^5$$

where $z_{n}, c \in \mathbb{C}$.

I google a while but the formula for recurrence sequence, using the characteristic polynomial can be used for linear relation like Lucas Numbers.

I was thinking to moving on the continuos case so to solve the following non-linear differential equation over the complex field

$$z'(x) = c^2 + 2cz^2(x) + z^4(x) - (c+c^2)z(x) - 2cz^3(x) - z^5(x)$$

where $z$ is meromorphic function from complex to complex. Then come back to the discrete case.

If $z$ was a real-valued function this could be easly calculated by separating variables, but I do not know it I can apply the same procedure in the complex case.

share|cite|improve this question
You can (and should!) use \$ and \$\$ to enclose TeX math formulas. – tomasz Jul 2 '12 at 14:47
sorry I will do next time – user730712 Jul 2 '12 at 14:54
What sort of formula/asymptotics do you expect to see? Since your equation is non-linear, the sequence may depend strongly on the initial value. – Mark Bennet Jul 2 '12 at 15:27
if you consider the differential equation – user730712 Jul 2 '12 at 15:29
Even in the case $c=0$ your recurrence is of the form $z_{n+1} = z^4_n (1-z_n).$ From memory, even $a_{n+1}=a_n (1-a_n)$ has no explicit solution so you probably can't expect this one to have one either. I also recall that its behaviour in the case of $a_0 \in (0,1)$ is considered interesting and is well studied, so perhaps searching that up might lead to some relevant reading. – Ragib Zaman Jul 2 '12 at 15:52

One way to view the recurrence relation itself is as a fixed point iteration, z := p(z) where $p(z)$ is the complex polynomial:

$$ p(z) = c^2 - (c+c^2)z + 2cz^2 - 2cz^3 + z^4 - z^5 $$

The behavior of fixed point iterations depends essentially on the size of the derivative in a neighborhood of a fixed point, i.e. of a root of $z = p(z)$. If the derivatives are bounded in absolute value by 1 in such a neighborhood, then the iterates will converge to the fixed point once they enter that neighborhood ("basin of attraction"). But if the derivatives are bounded below (in absolute value) by 1, the root is a "repelling fixed point" and such iterations do not converge to that root.

A combination of repelling fixed points can easily lead to periodic cycles and hence to chaos, meaning that the dependence on initial conditions becomes too delicate for numerical accuracy in the solution.

The other approach suggested is to analyze the differential equation $z' = p(z)$, but as explained in my comment above, I would rather consider the equation $z' = p(z) - z$ as being more closely connected to the difference equation $z_n - z_{n-1} = p(z_{n-1}) - z_{n-1}$ as a forward Euler method.

Of course the differential equation is easily solved in either case by simple polynomial integration (which can take into account the initial condition as a constant of integration). But neither formulation guarantees any long range agreement between the solution of the recurrence relation and the integration of the polynomial.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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