Is there a non-constant function $f$ such that $f'(x) = f(x - 1)$?

Question

In discrete calculus, where the difference operator $\Delta f = f(x + 1) - f(x)$ takes the place of $\frac{d}{dx}$, Fibonacci sequences are given by the functions satisfying:

$$ \Delta f(x) = f(x - 1) $$

Is there a non-constant function such that $\frac{d}{dx}f(x) = f(x - 1)$? If it exists, it would be the "continuous analogue of the Fibonacci sequence" in this sense, which seems cool.

Well, the smartalec answer would be that $f(x) = 0$ works, but I presume you meant non-zero $f$. — Thad Janisse, Oct 12 '15 at 03:34

score 11 · Accepted Answer · answered Oct 12 '15 at 03:36

11

Choose constant $C$ that satisfies $C=e^{-C}$. Note that $C=W(1)\approx 0.567$. We take

$$f(x)=e^{Cx}$$ $$f'(x)=Ce^{Cx}$$ $$f(x-1)=e^{C(x-1)}=e^{Cx}e^{-C}=Ce^{Cx}$$

answered Oct 12 '15 at 03:36

vadim123

82,796

Wow, wouldn't have thought of that. What is the function $W$ -- is that the Lambert W function? – Eli Rose Oct 12 '15 at 03:41
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Too bad this function doesn't agree with Fibonacci sequence for $x\in\mathbb{N}$. Would be great if that had happen. – Integral Oct 12 '15 at 03:42
@elirose This isn't very surprising given that Fibonacci's solutions are also exponential. – A.S. Oct 12 '15 at 03:46
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Yes, it's the Lambert W function. In fact this particular $C$ has a name, it's the Omega constant. – vadim123 Oct 12 '15 at 03:47

Is there a non-constant function $f$ such that $f'(x) = f(x - 1)$?

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