Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If we have a linear homogeneous recurrence relation, such as $t_{k+1}=4t_k-4t_{k-1}$, and attempt to find solutions of the form $t_n=x^n$ for some $x \in \mathbb{R} \setminus \{0\}$, we obtain the characteristic polynomial $(x-2)^2=0$. The characteristic polynomial has repeated roots, so the standard technique is to look for solutions of the form $t_n=nx^n$ too.

It's routine to check that $t_n=n x^n$ is indeed another solution to the recurrence. But, if I didn't already know that $t_n=n x^n$ is a solution, why would I think to look at solutions of this form?

Question: What motivation is there to consider solutions such as $t_n=n x^n$ in the case of repeated roots?

I find it easy to motivate looking for solutions of the form $t_n=x^n$, since e.g. for the Fibonacci numbers $F(n)$, we can prove exponential upper and lower bounds on $F(n)$, and it seems reasonable to suspect that similar inequalities hold more generally.

share|improve this question
    
It'd be clearer to not use $x$ in both the characteristic polynomial and in $x^n$, e.g. use $c^n$. –  Bill Dubuque Apr 10 '12 at 0:46

2 Answers 2

up vote 10 down vote accepted

Linear Algebra:

To set the stage, let $V$ be the vector space of sequences $a_0, a_1, ... $ (say of elements of an algebraically closed field) indexed by non-negative integers. On this vector space define the shift operator $S : V \to V$, which sends a sequence $a_i$ to the sequence $(Sa)_i = a_{i+1}$. To say that $a$ satisfies a linear recurrence with characteristic polynomial $p$ is then precisely to say that $$p(S) a = 0.$$

Since we're working over an algebraically closed field, we can factor $$p(S) = \prod_{i=1}^n (S - \lambda_i I)^{m_i}$$

and then write our recurrence as $$\prod_{i=1}^n (S - \lambda_i I)^{m_i} a = 0.$$

In particular, any sequence satisfying $(S - \lambda_i I)^{m_i} a = 0$ (a generalized eigenvector of $S$ with eigenvector $\lambda_i$) satisfies the recurrence, and standard results in linear algebra imply that generalized eigenvectors span the entire space of solutions.

So it suffices to solve the generalized eigenvector equation. First, I'm not sure if you know this, so it's worth saying: if $m_i = 1$, then the space of solutions to $$(S - \lambda_i I) a = 0$$

is spanned by $a_n = \lambda_i^n$ and this is the clearest reason I can think of (eigenvector decomposition) to look for exponential solutions.

Before we tackle the general case, another special case: $\lambda_i = 1$. In this case $S - I$ is just the forward difference operator, and the sequences satisfying $$(S - I)^n a = 0$$

are precisely the polynomial sequences of degree less than $n$. This should be a familiar fact about the forward difference operator, and it turns out to imply the general result. Indeed, suppose $a$ satisfies $$(S - \lambda_i I)^n a = 0$$

and write $a_n = \lambda_i^n b_n$ for some sequence $b_n$.

Lemma: $((S - \lambda_i I)a)_n = \lambda_i^{n+1} ((S - I)b)_n$.

Proof. Straightforward computation.

By induction, it follows that $$(S - \lambda_i I)^n a = 0 \Leftrightarrow (S - I)^n b = 0$$

and the conclusion follows. (I haven't discussed the case $\lambda_i = 0$ but the solutions are easy to describe here.)

A Limiting Argument:

We'll work through a simple representative example since the goal here is motivation. Consider a sequence $a$ satisfying $(S - \lambda_1 I)(S - \lambda_2 I) a = 0$ where $\lambda_1 \neq \lambda_2$, and let's see what happens if we let $\lambda_1 \to \lambda_2$. To take this limit sensibly we'll fix initial conditions $a_0 = 0, a_1 = 1$. This will give $$a_n = \frac{\lambda_1^n - \lambda_2^n}{\lambda_1 - \lambda_2}$$

Now taking the limit $\lambda_2 \to \lambda_1$ and applying, for example, l'Hopital's rule, we obtain $$a_n = n \lambda_1^{n-1}.$$

Differential Equations:

The shift operator $S$ defined above has another incarnation as differentiation $D$: indeed, we just need to associate to a sequence $a_0, a_1, ...$ the exponential generating function $$A(t) = \sum_{n=0}^{\infty} a_n \frac{t^n}{n!}.$$

Solutions to $p(S) a = 0$ then become solutions to differential equations $p(D) A = 0$. But now that we can work in the formalism of differential equations we can appeal to physical intuition: in physical terms, the relevant phenomenon is resonance. More concretely, let's work with a forced harmonic oscillator $$m \frac{d^2 x}{dt^2} + kx = \sin \omega t.$$

(Think of periodically pushing a swing.) Note that $x$ satisfies $(m D^2 + k)(D^2 + \omega^2) x = 0$. This polynomial has repeated roots when $\omega^2 = \frac{m}{k}$, that is, when the frequency of the forcing function matches the "natural frequency" of the system (if physicists have better terms for these I've forgotten them). In that case one can observe experimentally in the relevant physical systems (e.g. the Tacoma Narrows bridge) that the amplitude of the oscillations increase linearly with time, and one can observe mathematically that the solution involves terms like $t \sin \omega t$ (and one simple way to see this is to use the limiting argument above). Computing the Taylor series of such solutions then gives the familiar extra factor of $n$.

share|improve this answer
    
I must say, this is a pretty awesome answer! –  Douglas S. Stones Apr 10 '12 at 22:39
2  
Fantastic answer! Only the Tacome Narrows bridge example seems imprecise since recent research suggests that not ressonance, but a non-linear phenomenum was the cause of the bridge collapse: "Large-Amplitude Periodic Oscillations in Suspension Bridges: Some New Connections with Nonlinear Analysis" Lazer, McKenna, SIAM Review, vol.32, no.4. (1990) or, for a simpler model, "The Collapse of the Tacoma Narrows Suspension Bridge: A Modern Viewpoint", Lewis, ODE book –  Lucas Seco Oct 3 '13 at 3:40

One way to look at this would be using generating functions.

The generating function comes out to

$$F(x) = \frac{Q(x)}{P(x)}$$

where the polynomial $P(x)$ is got from the characteristic polynomial, by replacing $x$ by $\frac{1}{x}$ and multiplying by an appropriate power of $x$.

Now using the partial fraction decomposition

$$ \frac{Q(x)}{(x-a_1)^{n_1} \dots (x-a_k)^{n_k}} = \sum_{i=1}^{k} \sum_{j=1}^{n_i} \frac{A_{ij}}{(x-a_1)^j}$$

tells us that we should get a polynomial in $n$.

share|improve this answer
1  
This answer uses ordinary generating functions whereas my answer above uses exponential generating functions; the connection between the two is given (essentially) by the Laplace transform. –  Qiaochu Yuan Apr 10 '12 at 1:43
    
The polynomial in $n$ comes from $\binom{-j}{n} = (-1)^n \binom{j + n - 1}{j - 1}$, a polynomial of degree $j - 1$ in $n$. –  vonbrand Apr 21 at 17:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.