Recurrence relation. Why subtitute $A_{n} = cr^{n}$ for second order homogeneous recurrence?

Question

Recently I learned about recurrence relation and I understand the general solution for first order homogeneous recurrence relation is $A_{n} = cr^{n}$. But why do we substitute that for second order relation like $A_{n} = A_{n-1} + A_{n-2}$?

I know how to do it by substituting $A_{n} = cr^{n}$ into the equation and solve for the characteristic roots but I cant seem to see the reasons behind it...

Does this answer your question? Linear homogeneous recurrence relations with repeated roots; motivation behind looking for solutions of the form $nx^n$? — , Sep 11 '21 at 19:17
The real reason is much simpler: It works! By trying $r^n$, solutions to the recurrence are derived from the roots of a polynomial, with repeated roots corresponding to more complicated terms. This is similar to the way the Laplace transform turns integration and differentiation into multiplication and division. — marty cohen, Oct 22 '15 at 18:28

Thomas Andrews · Accepted Answer · 2015-10-22T18:30:46.913

Unfortunately, the main mathematical motivations to take this approach use math you perhaps haven't learned yet.

Idea 1: Linear Algebra

In a linear recurrence, we have a matrix which sends:

$$M:(A_{n},A_{n-1},\dots,A_{n-k+1})^T\to(A_{n+1},A_n,\dots,A_{n-k+2})^T$$

Then the values $r$ are eigenvalues of $M$. If you know the eigenvalues of a matrix $M$, it makes it much easier to represent matrix powers $M^n$.

Idea 2: Generating Functions

Alternatively, you can look at it in terms of generating functions. For example, if $A_{n+1}=A_n+A_{n-1}$, we get a power series called the generating function:

$$a(x)=\sum_{n=0}^{\infty} A_nx^n$$

Turns out the recurrence relation can be rewritten $(1-x-x^2)a(x)=A_0+(A_1-A_0)x$ -- the other terms cancel out. So $$a(x)=\frac{A_1+(A_1-A_0)x}{1-x-x^2}$$

Edit: Here' the "why."

We can see this as:

$$\begin{align} a(x)&=A_0&+&A_1x&+&A_2x^2&+&\dots&+&A_nx^n+\dots\\ xa(x)&=&\,&A_0x&+&A_1x^2&+&\dots&+&A_{n-1}x^n +\dots\\ x^2a(x)&=&\,&\,&\,&A_0x^2&+&\dots&+&A_{n-2}x^n+\dots \end{align}$$

Subtracting the second and third from the first gives us:

$$(1-x-x^2)a(x)=A_0+(A_1-A_0)x+(A_2-A_1-A_0)x^2+\dots+(A_n-A_{n-1}-A_{n-2})x^n$$

But note that $A_2-A_1-A_0=0$, and likewise for all the later terms $A_n-A_{n-1}-A_{n-2}$.

At first glance, knowing that $(1-x-x^2)a(x)=A_0+(A_1-A_0)x$ doesn't seem to give us much, but if we factor $1-x-x^2=(1-r_1x)(1-r_2x)$ we can use something called "partial fractions" to write:

$$\frac{a+bx}{1-x-x^2} = \frac{\alpha_1}{1-r_1x} + \frac{\alpha_2}{1-r_2x}$$

And we know the power series of these two terms: $$\frac{\alpha_i}{1-r_ix}=\sum (\alpha_ir_i^n)x^n$$

So we get, from adding those two power series, $A_n=\alpha_1r_1^n+\alpha_2r_2^n.$

In both cases, we get a problem when $r_1=r_2$ - then partial fractions doesn't work as above, and the matrix (usually) does not lead to a "diagonalizable" matrix. Then we get a different answer, but even then, the base case $r^n$ is an important case.

The case when $r_1=r_2$

In the simple case when there is only one root, we get a partial fraction decomposition like:

$$\frac{\alpha}{1-rx} + \frac{\beta}{(1-rx)^2}$$

(With a recurrence of higher degree, of course, the general terms are of the form $\frac{\alpha}{(1-rx)^k}$, but I'm sticking with the simplest recurrences.)

Then we use that:

$$\frac{1}{(1-rx)^2} = \sum_{n=0}^{\infty} (n+1)r^nx^n$$

So in some way $(n+1)r^n$ is the "more natural" choice, but since $A_n=r^n$ and $B_n=(n+1)r^n$ both satisfy our recursion, then $B_n-A_n=nr^n$ satisfies our recurrence.

The direct computational reason is related follows. If $p(x)$ is divisible by $(x-r)^n$ then $p'(x)$ is divisible by $(x-r)^{n-1}$.

Let's take a specific example, $p(x)=x^2-2x+1=(x-1)^2$. Now define $p_n(x)=x^np(x)=x^{n+2}-2x^{n+1}+x^n$, which is also divisible by $(x-1)^2$. Taking derivatives, we see that:

$$0=p_n'(r)=(n+2)r^{n+1}-2(n+1)r^n +nr^{n-1}$$

If $A_n=(n+1)r^n$ then this means that:

$$0=A_{n+1}-2A_n + A_{n-1}$$

In general, if your polynomial has a root $r$ that repeats $k$ times, then you can use $A_n=p(n)r^n$ where $p(n)$ is any polynomial of degree $k-1$.

Aside: From the linear algebra perspective, repeated roots means the matrix is not diagonalizable, so you are stuck using Jordan Normal Form, a more general approach. Exponentiation still is relatively easy to do for a matrix in Jordan Normal Form, but you get polynomials in $n$ for some of the terms.

For example, when $A_{n+1}=2rA_{n}-r^2A_{n-1}$ the matrix is:

$$M=\begin{pmatrix}0&1\\2r&-r^2\end{pmatrix}=S\begin{pmatrix}r&1\\0&r\end{pmatrix}S^{-1}$$

For some invertible matrix $S$. So:

$$M^n = S\begin{pmatrix}r&1\\0&r\end{pmatrix}^nS^{-1}=S\begin{pmatrix}r^n&nr^{n-1}\\0&r^n\end{pmatrix}S^{-1}$$

I am sorry but I am a little bit lost at $(1−x−x^{2})a(x)=A_{0}+(A_{1}−A_{0})x$ , how do you cancel the other terms? — GalaxyVintage, Oct 22 '15 at 07:22
Thank you so much. Also how do we handle it if the roots are the same? I was told that we can try $A_{n} = cnr^{n}$ without any reason to support it — GalaxyVintage, Oct 22 '15 at 17:58
you said $A_{n} = r^{n} $ and $ B_{n} = (n+1)r^{n}$, shouldn't it be $A_{n} = \alpha r^{n}$ and $ B_{n} = \beta (n+1)r^{n}$ — GalaxyVintage, Oct 22 '15 at 19:51
Well, I'm explaining why $nr^n$ is a solution, not explaining that it is the only solution. The constants are, intitially, completely uninteresting - you first need to find a "basis" for the set of all solutions, then worry about the constants. — Thomas Andrews, Oct 22 '15 at 19:58

Recurrence relation. Why subtitute $A_{n} = cr^{n}$ for second order homogeneous recurrence?

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