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Re: RSolve and complex solutions


On 27 Feb 2005, at 07:29, Skirmantas wrote:

> It may be a basic math question, but it would be helpful if somebody
> could briefly comment on it. If I have a system of recursion equations
> with real coefficients and solve it numerically step-by-step, all my
> solutions at all time steps are (obviously) real. If, however, I use
> RSolve to get the general formulas for these solutions, some of them
> become complex numbers. My understanding is the imaginary parts of
> these numbers are due to rounding errors and the actual solutions are
> only the real parts. Am I right?
>
>
>
Are all your equations linear? Otherwise the answer is "obviously no". 
A trivial example that satisifes all your stated conditions is

RSolve[a[n]^2 + 1 == 0, a[n], n]

{{a[n] -> -I, a[n] -> I}}

Andrzej Kozlowski


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