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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.

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    You can (and should!) use \$ and \$\$ to enclose TeX math formulas.2012-07-02
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    sorry I will do next time2012-07-02
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    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.2012-07-02
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    if you consider the differential equation2012-07-02
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    consider the differential equation $$z'(x) = c^2 + 2cz^2(x) + z^4(x) - (c+c^2)z(x) - 2cz^3(x) - z^5(x)$$ if you take $$c = 0$$ and you consider z as a real valued function you could write2012-07-02
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    consider the differential equation $$z'(x) = c^2 + 2cz^2(x) + z^4(x) - (c+c^2)z(x) - 2cz^3(x) - z^5(x)$$ if you take $$c = 0$$ and you consider z as a real valued function you could write $$z'(x) = z^4(x) - z^5(x)$$ that gave you the solution http://www.wolframalpha.com/input/?i=solve+y%27+%3D++y%5E4++-+y%5E5 I was wondering If I can do the same thing but in the general case knowing that if complex case the equation $$0 = c^2 + 2cz^2(x) + z^4(x) - (c+c^2)z(x) - 2cz^3(x) - z^5(x)$$ have always solutions2012-07-02
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    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.2012-07-02
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    I think the analogy of a recurrence relation $z_n = p(z_{n-1})$ to differential eqn. $z' = p(z)$ is misguided. Comparison of such a differential eqn. to a difference eqn. $z_n - z_{n-1} = p(z_{n-1})$ would seem better motivated. Perhaps you meant something of that kind?2012-07-08

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