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F(n) = F(n-1) + F(n-2) 

Is this differentiable? How or why not?

I understand how to differentiate equations like x**3 + 2x**2 + 3x + 5. But this is a recurrence relation and I don't know how it is differentiated.

I'm trying to understand the rate of growth of this equation. Can you tell me what would be the rate of growth at n = 987654?

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    I have a feeling what you're really looking for is asymptotics.2012-12-28

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Equations can never be differentiable, but functions may or may not be.

Without any context, it looks like your equation is actually a recurrence equation, and $n$ is assumed to be an integer. Functions that are only defined for integer values are never differentiable.

On the other hand, your equation defines the so called Fibonacci numbers (or something similar, depending on the initial values), and a resonable interpretation of "rate of growth" would be $F(n+1)-F(n)$, which can be computed. If $F(0) = F(1) = 1$, then $$F(n) = \frac{1}{\sqrt 5}\left( \frac{1+\sqrt5}{2}\right)^n - \frac{1}{\sqrt 5}\left( \frac{1-\sqrt5}{2}\right)^n.$$

I'll leave it to you to compute $F(987655)-F(987654)$.

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    How did you arrive at those expressions for F(n+1) and F(n)?2012-12-28
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    @KshitizSharma: Just a lucky guess! (Actually, I solved the recurrence. There are a number of methods for solving such equations, see for example http://en.wikipedia.org/wiki/Recurrence_relation)2012-12-28
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    It is too harsh to say that things _initially_ defined only for integers are "never" differentiable... but only that it does not _initially_ make sense. Indeed, the closed-form expression makes perfect sense for index $n$ any real number, and can be differentiated. The real point is that there is no _initial_ sense, no _a-priori_ sense, of differentiation of a thing only defined _initially_ at integers... and/but the OP didn't really want differentiability, even though they thought they might. _Asymptotics_ is the right keyword, generally, even though here the closed-form is decisive.2012-12-28
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    @paulgarrett: I stand by my original comment. There are many ways to extend a function $F$ defined on $\mathbb{Z}$ to $\mathbb{R}$ and the extended functions' derivatives are usually not the same. What's not to say that $$F(n) = \left(\frac{1}{\sqrt 5}\left( \frac{1+\sqrt5}{2}\right)^n - \frac{1}{\sqrt 5}\left( \frac{1-\sqrt5}{2}\right)^n\right) + 124123551\sin(2\pi n)$$ is not the ''correct'' extension.2012-12-29
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    Of course extensions are typically not unique, although equally "of course" everyone chooses the expression without the "noise", because it is simpler. And "undefined" is not the same as "does not exist", depending on one's philosophical and practical outlooks. And "not defined" does not entail "not definable", etc. Since I am not a formalist, these things are distinguishable to me, but, I recognize, potentially meaningless distinctions to others.2012-12-29