# What kind of equations are differentiable?

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. –  Karolis Juodelė Dec 28 '12 at 22:03

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)? –  Kshitiz Sharma Dec 28 '12 at 22:11
@KshitizSharma: Just a lucky guess! (Actually, I solved the recurrence. There are a number of methods for solving such equations, see for example en.wikipedia.org/wiki/Recurrence_relation) –  mrf Dec 28 '12 at 22:15
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. –  paul garrett Dec 28 '12 at 23:59
@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. –  mrf Dec 29 '12 at 8:57
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. –  paul garrett Dec 29 '12 at 14:31