Sequences defined using a Linear Recurrence

January 6, 2014

(This article was first published on Freakonometrics » R-english, and kindly contributed to R-bloggers)

In the introduction to the time series course (MAT8181) this morning, we did spend some time on the expression of (deterministic) sequences defined using a linear recurence (we will need that later on, so I wanted to make sure that those results were familiar to everyone).

  • First order recurence

The most simple case is the first order recurence,{n-1} where\neq%201 (for convenience). Observe that we can remove the constant, using a simple translation\underbrace{[u_n-m]}_{v_n}%20=%20b%20\underbrace{[u_{n-1}-m]}_{v_{n-1}} if So, starting from this point, we will always remove the constant in the recurent equation. Thus,{v_n}%20=%20b{v_{n-1}}. From this equation, observe that{v_n}%20=%20b^n{v_{0}}, which is the general expression of{v_n}.

  • Second order recurence

Consider now a second order recurence,{v_n}%20=%20a{v_{n-1}}+b{v_{n-2}}. In order to find the general expression of{v_n}, define\boldsymbol{V}_n%20=(v_{n}},{v_{n-1}})^{\sffamily%20T}. Then\underbrace{\begin{bmatrix}v_n\\v_{n-1}%20\end{bmatrix}%20}_{\boldsymbol{V}_n%20}=%20\underbrace{\begin{bmatrix}a&%20b%20\\%201%20&%200\end{bmatrix}}_B\underbrace{\begin{bmatrix}v_{n-1}%20\\v_{n-2}%20\end{bmatrix}%20}_{\boldsymbol{V}_{n-1}%20} This time, we have a vectorial linear recurent equation. But what we’ve done previously still holds. For instance,{\boldsymbol{V}_n%20}=B{\boldsymbol{V}_{n-1}%20}=\cdots=B^n\boldsymbol{V}_{0} What could we say about^n ? If can be diagonalized, then\Delta%20P^{-1} and^n=P\Delta^n%20P^{-1}. Thus,\underbrace{\begin{bmatrix}v_n\\v_{n-1}%20\end{bmatrix}%20}_{\boldsymbol{V}_n%20}=%20B^n%20\underbrace{\begin{bmatrix}v_{0}%20\\v_{-1}%20\end{bmatrix}%20}_{\boldsymbol{V}_{0}%20}=%20P\underbrace{\begin{bmatrix}\lambda_1^n&%200%20\\%200%20&%20\lambda_2^n\end{bmatrix}}_{\Delta^n}%20P^{-1}\underbrace{\begin{bmatrix}v_{0}%20\\v_{-1}%20\end{bmatrix}%20}_{\boldsymbol{V}_{0}%20} so what we’ll get here is something like\alpha%20\lambda_1^n%20+\beta\lambda_2^n for some constant\alpha and\beta. Recall that\lambda_1 and\lambda_2 are the eigenvalues of matrix, and they are also the roots of the characteristic polynomial^2%20-%20ax%20-%20b. Since and are real-valued, there are two roots for the polynomial, possibly identical, possibly complex (but then conjugate). An interesting case is obtained when the roots are^{\pm%20i\theta}. In that case^n(\alpha\cos(n\theta)%20+%20\beta\sin(n\theta)) To visualize this general term, consider the following code. A first strategy is to define the sequence, given the two parameters, and two starting values. E.g.

> a=.5
> b=-.9
> u1=1; u0=1

Then, we iterate to generate the sequence,

> v=c(u1,u0)
> while(length(v)<100) v=c(a*v[1]+b*v[2],v)
> plot(0:99,rev(v))

It is also possible to use the generic expression we’ve just seen. Here, the roots of the characteristic polynomial are

> r=polyroot(c(-b, -a, 1))
> r
[1] 0.25+0.9151503i 0.25-0.9151503i
> plot(r,xlim=c(-1.1,1.1),ylim=c(-1.1,1.1),pch=19,col="red")
> u=seq(-1,1,by=.01)
> lines(u,sqrt(1-u^2),lty=2)
> lines(u,-sqrt(1-u^2),lty=2) Since,\alpha%20\lambda_1^n%20+\beta\lambda_2^n, then\begin{cases}%20\alpha%20+%20\beta%20=%20v_0%20\\%20\alpha%20r_1%20+%20\beta%20r_2%20=%20v_1%20\end{cases} it is possible to derive numerical expressions for the two parameters. If^n(A\cos(n\theta)%20+%20B\sin(n\theta)), then\lambda+\mu while\lambda-\mu). Thus,

> A=sum( solve(matrix(c(1,r[1],1,r[2]),2,2),c(u0,u1)))
> B=diff(solve(matrix(c(1,r[1],1,r[2]),2,2),c(u0,u1)))* complex(real=0,imaginary=1)

We can plot the sequence of points

> plot(0:99,rev(v))

and then we can also plot the sine wave, too

> t=seq(0,100,by=.1)
> lines(t,Vectorize(bv)(t-1),col="red",lty=2)
> lines(t,-Vectorize(bv)(t-1),col="red",lty=2)
> lines(t,Vectorize(fv)(t-1),col="blue")

We will see a lot of graph like this in the course, when looking at autocorrelation functions.

  • Higher order recurence

More generally, we can write\underbrace{\begin{bmatrix}v_n\\v_{n-1}\\v_{n-2}\\%20\vdots%20\\%20v_{n-p+1}%20\end{bmatrix}%20}_{\boldsymbol{V}_n%20}=%20\underbrace{\begin{bmatrix}b_{1}%20&%20b_{2}%20&b_3&%20\cdots%20&%20b_{p}%20\\%201%20&%200%20&%200&%20\cdots%20&0\\%200%20&%201%20&%200&%20\cdots%20&0\\%20\vdots%20&%20\vdots%20&%20\vdots%20&%20\ddots%20&%20\vdots%20\\%200%20&%200%20&%200&%20\cdots%20&%200\end{bmatrix}}_B\underbrace{\begin{bmatrix}v_{n-1}%20\\v_{n-2}\\v_{n-3}%20\\%20\vdots%20\\%20v_{n-p}%20\end{bmatrix}%20}_{\boldsymbol{V}_{n-1}%20} The matrix is a so called companion matrix. And similar results could be obtained for the expression of the general term of the sequence. If all that is not familar, I strongly recommand to read carefully a textbook on sequences and linear recurence.

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