Expansion of reciprocal of quadratic

Question

Can I expand $\frac{1}{1-.7B-.3B^2}$ into an infinite series? Where B is the backwards operator in time series.

I was thinking $\frac{1}{1-(-.3B)}\frac{1}{1-B}$. Express this as a product of a geometric series?

Help greatly appreciated

Answer 1

Yes, you could express this as a product of two geometric series.

In particular, we have \begin{align*} \dfrac{1}{1 - (-0.3B)} \cdot \dfrac{1}{1-B} &= \sum_{i=0}^\infty (-0.3 B)^i \sum_{j=0}^\infty B^j \\ &= \sum_{i=0}^\infty \sum_{j=0}^\infty (-0.3)^i B^{i+j} \\ &= \sum_{k=0}^\infty \sum_{i=0}^k (-0.3)^i B^k \\ &= \sum_{k=0}^\infty \dfrac{1 - (-0.3)^{k+1}}{1- (-0.3)} B^k \\ &= \sum_{k=0}^\infty \frac{1 - (-0.3)^{k+1}}{1.3} B^k \end{align*}

However, it's more straightforward to use partial fractions: if

$$\frac{1}{(1-B)(1+0.3B)} = \frac{s}{1-B} + \frac{t}{1+0.3B}$$

then, clearing denominators,

$$1 = s(1+0.3B) + t(1-B) = (s+t) + (0.3s - t) B.$$

Solving the system

$$\left\{ \begin{array}{ccc} s + t &=& 1, \\ 0.3 s - t &=& 0 \end{array}\right.$$

gives us $s = 10/13, t = 3/13,$

so we have

$$\frac{1}{1-.7B-.3B^2} = \frac{10/13}{1-B} + \frac{3/13}{1-(0.3)B}$$

Then, instead of multiplying two geometric series, we can just add them.

Answer 2

Your idea is correct: this is best expressed as a product of two geometric series. We don't have to worry about commutation in this case, so (convergence aside) we can disregard the fact that B is an operator and proceed directly with the expansion.

$$\frac{1}{1+.3B}\frac{1}{1-B} = (1-.3B+.09B^2-\ldots)(1+B+B^2+\ldots)$$

Of course, we want this as a single power series. The coefficient of $B^n$ in the product will be

$$\sum_{i=0}^n (-.3)^n = \frac{1-(-.3)^{n+1}}{1+.3} = \frac{10}{13}\left(1-\left(-\frac{3}{10}\right)^{n+1}\right).$$

So, your series is

$$\frac{1}{1-.7B-.3B^2} = \frac{10}{13}\sum_{k=0}^\infty \left(1-\left(-\frac{3}{10}\right)^{k+1}\right)B^k.$$