# Relation between differential equations and sequence recursions

It's obvious that there is a strong relation between linear recursions of sequences and linear differential equations. The common methods for solving them are nearly identical. For example, the general solution to

$$a_{n+2} = 5 a_{n+1} - 6 a_n + 2^n (4n - 2)$$

is

$$a_n = 3^n A - 2^n (n^2 + 2n + B)$$

with arbitrary constants $A$ and $B$; if $a_0 = 1$ and $a_1 = 2$ were given,

$$a_n = 6 \cdot 3^n - 2^n (n^2 + 2n + 5)$$

would be the only solution. In terms of differential equations, the general solution to

$$\frac{d^2y}{dx^2} - 5 \frac{dy}{dx} + 6 y = e^{2x} (4x - 2)$$

is

$$y(x) = e^{3x} A - e^{2x} (2x^2 + 2x + B),$$

or, if $y(0) = 1$ and $y'(0) = 2$ are given,

$$y(x) = 2 \cdot e^{3x} - e^{2x} (2x^2 + 2x + 1).$$

There are some major parallels, but there are also differences. While I know why this basically works, i.e., that raising $n$ by one does the same thing to $n^3$ as differentiating with respect to $x$ does to $e^{3x}$, my question is: Is there an underlying connection between these two equations, something I missed? Or is it just that, that they have some similar properties?

• Yes, there is: finite difference equations are the discrete versions of differential equations. Remember a derivative is the limit of a variation rate $\Delta f/\Delta x$. In the discrete version, the variable is $x:=n$, $\Delta x=\Delta n=1$. May 5, 2016 at 19:40
• For another parallel: One can turn differential equations and recurrence into algebraic equations by means of a Laplace transform and a generating function respectively. (I think the latter is sometimes called a transform as well, but I don't recall the terminology.) May 5, 2016 at 20:18

There is indeed a deep connection between the two equations, that is the starting point for the theory of generating functions.

The connection is given by the following one-on-one correspondence between real-valued sequences and powerseries $$i \colon \mathbb R^{\mathbb N} \longrightarrow \mathbb R[[x]]$$ $$i((a_n)_n) = \sum_{n \in \mathbb N} \frac{a_n}{n!}x^n$$ which is an isomorphism between these $\mathbb R$-vector spaces.

By using this isomorphism backward you can endow the space of sequences with a product, defined as $(a_n)_n \cdot (b_n)_n=(\sum_{k=0}^n a_k b_{n-k})_n$, and a derivation operator, which coincides with the shifting operator: $\frac{d}{dx}((a_n)_n)=(a_{n+1})_n$ (it is an easy count to verify that $i\left(\frac{d}{dx}(a_n)_n\right)=\frac{d}{dx}i(a_n)_n$).

You can think of a recursive equation as a sequence of equations, parametrized by the index $n$, that you can fuse into a equation whose terms are expressions build up from sequences using sum, multiplication, scalar multiplication and the shifting/derivator operator.

For instance from recursive equation in your question you can get the following equation $$\frac{d^2}{dx^2}(a_n)_n=5\frac{d}{dx}(a_n)_n -6 (a_n)_n+4(2^nn)_n-2(2^n)_n$$ which through the isomorphism $i$, by letting $y=i(a_n)_n$, becomes $$\frac{d^2}{dx^2}y=5\frac{d}{dx}y-6y+4e^{2x}-2e^{2x}$$ that is the differential equation in you question.

Since these two equations correspond through the isomorphisms $i$ the solutions of the equations correspond one to each other through $i$ too: if $(a_n)_n$ is a solution to the sequence-equation then $i(a_n)_n$ is a solution to the differential equation.

For instance if you take the solution $a_n=6\cdot 3^n-2^n(n^2+2n+5)$ then $$i(a_n)_n = 6e^{3x}-e^{2x}(x^2+2x+5)\ .$$

There could be so much more to say about generating functions but I am afraid that would take us too far from the scope of the question.

I hope this helps.

Yes, you can use the difference operator on a sequence $$\Delta(a_n) = a_{n+1} - a_n$$ to rewrite the recurrence relations into difference equations, which are a discretized analog of differential equations, with similar method of solution. Just as an example, as $y'=y$ yields an exponential family, $y=Ae^x$, so $\Delta(a_n) = a_n$ yields an exponential family $a_n = A \cdot 2^n$…

For your recurrence relation, the analog would be $$\Delta^2(a_n) - 4\Delta(a_n) + 2a_n = 2^n(4n-2).$$

Actually, there is a calculus, peculiarly called "time scales", that contains both the discrete and continuous versions, as well as combinations and/or variations. To a large extent the connections between discrete and continuous are revealed in this calculus.

It was cleverly introduced by Hilger, and then it was developed by many others, although really not producing anything new. For example, the papers of Hilger are in my opinion wonderful works, very well written, and easy to read even if quite technical, but really they contain a reformulation of what already existed, with unified statements and proofs.

But yes:

Time scales do help revealing the similarities between discrete and continuous time. (Certainly, there are many other ways in which the similarities have been noticed, although perhaps not always in some organized manner.)

On the other hand, not really:

Taking as an example dynamical systems, there are many differences between discrete and continuous time, such as types of bifurcations that only occur for one of them, such as global topological properties that depend on something like the Jordan curve theorem, and such as ergodic properties that don't extend to suspensions, not to mention that to consider only a $1$-dimensional time is a considerable restriction (even for physical applications), and of course the theory of time scales does not address (neither it can address) any of these "objections".