On the relationship between differential equations and recurrences

Question

So recently I started looking into the relationship between differential equations and recurrence relations, and came across the following thread:

Link

I couldn't comment, neither could I find @Giorgio Mossa's email, so I had to open a thread. I have a few questions regarding his solution, as follows:

1. What is meant by $(a_n)_n$?
2. How am I supposed to use the isomorphism backward?
3. What does he mean by endowing the space of sequences with a product and derivation operator?
4. Could someone give me the first few steps on how to use the given isomorphism to convert the given recurrence into a differential equation?

Answer 1

1. $(a_n)$ or $\{a_n\}$ or $(a_n)_n$ or $(a_n)_{n\in\Bbb N}$ denote a sequence with running index $n$, usually over the natural integers. How much detail is given often depends on how obvious the default is.

2. If the forward transformation is to construct a power series from a sequence, then the reverse transformation is to read off the coefficient sequence from a power series.

3. These operations are well-defined on the side of (converging) power series, thus one can explore their effect on the coefficient sequences directly $$ \sum\frac{a_k}{k!}\sum\frac{b_l}{l!}x^l=\sum_{k+l=n}\frac{x^n}{n!}\sum\binom{n}{k}a_kb_{n-k} $$ so the product formula in the cited article is missing the binomial factor.

4. You multiply with $\frac{x^n}{n!}$ and sum up, using $$ a(x)=\sum\frac{a_n x^n}{n!},~~~a'(x)=\sum\frac{a_n nx^{n-1}}{n!}=\sum\frac{a_{n+1} x^{n}}{n!},~~~a''(x)=\sum\frac{a_{n+2} x^{n}}{n!} $$ This gives directly $$ a''(x)-5a'(x)+6a(x)=\sum_{n=0}^\infty\frac{x^n}{n!}2^n(4n-2) \\=\sum_{n=1}^\infty\frac{4(2x)^n}{(n-1)!}-\sum_{n=0}^\infty\frac{2(2x)^n}{n!} =(8x-2)e^{2x} $$

Answer 2

Here, $(a_n)_n=(a_0,a_1,\ldots)$ is an arbitrary sequence of real numbers.

Let $\mathbb{R}^\mathbb{N}$ equal the set of all sequences of real numbers $(a_0,a_1,\ldots)$.

Let $\mathbb{R}[[x]]$ denote the formal space of all power series $\sum_{n=0}^\infty \frac{a_n}{n!}^n$. Here, we don't ask anything about convergence.

Note that there is a one-to-one correspondence $$(a_0,a_1,\ldots)\leftrightarrow \sum_{n=0}^\infty \frac{a_n}{n!}x^n.$$ This is a linear bijection operator between these two spaces, which are vector spaces with the usual sum and scalar multiplication. Let $i:\mathbb{R}^\mathbb{N}\to \mathbb{R}[[x]]$ denote this linear bijection.

Note that $\mathbb{R}[[x]]$ already has a product structure: $$\Bigl(\sum_{n=0}^\infty \frac{a_n}{n!}x^n\Bigr)\Bigl(\sum_{m=0}^\infty \frac{b_m}{m!}x^m\Bigr)=\sum_{n,m=0}^\infty \frac{a_nb_m}{n!m!}x^{m+n}=\sum_{p=0}^\infty \frac{1}{p!}\Bigl[\sum_{q=0}^p \binom{p}{q}a_qb_{p-q}\Bigr]x^p.$$ Taking $i^{-1}$ of the last power series gives $\Bigl(\sum_{q=0}^n \binom{n}{q}a_qb_{n-q}\Bigr)_n$. So we define a product on $\mathbb{R}^\mathbb{N}$. Take two sequences $a,b$ and let $$(a_n)_n\otimes (b_n)_n = i^{-1}(i(a)\cdot i(b))=\Bigl(\sum_{q=0}^n \binom{n}{q}a_qb_{n-q}\Bigr)_n.$$ This is not the product defined in the link, which does not have the $\binom{n}{q}$ term. However, I don't see where that's used.

We also have the formal derivative: $$\frac{d}{dx}\sum_{n=0}^\infty \frac{a_n}{n!}x^n = \sum_{n=0}^\infty \frac{na_n}{n!}x^{n-1}=\sum_{n=1}^\infty \frac{na_n}{n!}x^{n-1}=\sum_{n=0}^\infty \frac{a_{n+1}}{n!}x^n.$$ Taking $i^{-1}$ of this is $(a_{n+1})$. Thus if we define $\Delta:\mathbb{R}^\mathbb{N}\to \mathbb{R}^\mathbb{N}$ by $\Delta((a_0,a_1,\ldots))=i^{-1}(\frac{d}{dx}i(a_0,a_1,\ldots)))=(a_1,a_2,\ldots)$, which is the shift operator.

Purely by definition, $i(a\otimes b)=i(a)\cdot i(b)$ and $i(\Delta(a))=\frac{d}{dx}i(a)$$.

Let's see an example of using this. Let's take a trivial differential equation. Say $y'=y$ and $y(0)=1$. We try a solution $y=\sum_{n=0}^\infty \frac{a_n}{n!}x^n$. Using $,y(0)=1$ implies $$1=y(0)=a_0+0+0+\ldots,$$ so $a_0=$. Note that $$y'=\sum_{n=0}^\infty \frac{na_n}{n!}x^{n-1}=\sum_{n=0}^\infty \frac{a_{n+1}}{n!}x^n.$$ But two power series with the same center are equal (on an open interval, and the solution is no good to us otherwise) iff all their coefficients are equal. So $a_n=a_{n+1}$, and $1=a_0=a_1=a_2=\ldots$ and $y=\sum_{n=0}^\infty \frac{1}{n!}x^n=e^x$. And we solved the DE.

Now let's try to solve a simple recurrence: $a_0=1$ and $a_{n+1}=\frac{1}{2}a_n$. Of course, this should be $a_n=2^{-n}$. This is $\Delta a=\frac{1}{2}a$, which we look at through the $i$ function to get $\frac{d}{dx}y=\frac{1}{2}y$, where $y$ is the function to which $i(a)$ converges. This is a linear homogeneous differential equation, which we rewrite as $y'-\frac{1}{2}y=0$. We have $$I(x)=e^{\int -1/2 xdx}=e^{-x/2}.$$ Then $$[Iy]'=I[y'-\frac{1}{2}y]=0,$$ $$Iy=C,$$ and $$y=C/I=Ce^{x/2}.$$ Checking $y(0)=1$ gives $C=1$. We know $e^{x/2}=\sum_{n=0}^\infty \frac{1/2^n}{n!}x^n$. Hitting this with $i^{-1}$ gives $(a_0,a_1,a_2,\ldots)=(1,1/2,1/2^2,\ldots)$.