How to solve the following finite-differences equation: $$f(x) = f(x-1) + f(x-\sqrt{2}), \quad x\in [\sqrt{2}, +\infty) \,?$$ Let's say $f(x) = f_0(x)$ for $x \in [0, \sqrt{2})$ is a given function.

My attempt:

Let's write characteristic function $L(\lambda) = 1 - e^{-\lambda} - e^{-\lambda\sqrt{2}}$. We need to solve the equation $L(\lambda) = 0$. There should be a root $\lambda_0 > 0$ (because $L$ is monotonously increasing, $L(0) = -1$ and $\lim_{\lambda \to +\infty} L(\lambda) = 1$), but my first question there is: how many other roots it has? Is it true that it has infinite number of (complex) roots? why?..

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    $\begingroup$ What I only haven't tried... First of all I'm stuck at finding roots of the characteristic equation. How many roots it has? $\endgroup$ – Wormer Aug 18 '14 at 20:18
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    $\begingroup$ The characteristic equation does indeed have an infinite number of complex roots and a single real root (see here on WA). The single real root is greater than $0$, as expected. $\endgroup$ – abiessu Aug 18 '14 at 20:23
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    $\begingroup$ The real root is positive, because $z = -\lambda$. Okay, I know what to do with the real root: $f_0(x) = C_0 e^{\lambda_0 x}$ is one of the solutions ($C_0$ is a constant). But that's not a general representation of an arbitrary solution... The further, the worse: what are multiplicities of complex roots? $\endgroup$ – Wormer Aug 18 '14 at 20:29

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