MATH319 Slides 134 Stable rational functions

{\cal S}

135 The stable rational functions form a differential ring

Lemma

(i) The set ${\cal S}$ of stable rational functions forms a commutative ring with $1$ in which one can differentiate with respect to $s$ , so ${\cal S}$ satisfies $(R),(C),(ID2),(Diff)$ .

(ii) However, ${\cal S}$ is not a field.

Proof. (i) (R) Multiplication and addition: Given $f_{1}(s)=g_{1}(s)/h_{1}(s)$ and $f_{2}(s)=g_{2}(s)/h_{2}(s)$ with ${\hbox{degree}}(g_{1}(s))\leq{\hbox{degree}}(h_{1}(s))$ and ${\hbox{degree}}(g_{2}(s))\leq{\hbox{degree}}(h_{2}(s))$ we have

f_{1}(s)f_{2}(s)=g_{1}(s)g_{2}(s)/h_{1}(s)h_{2}(s)

where ${\hbox{degree}}(g_{1}(s)g_{2}(s))\leq{\hbox{degree}}(h_{1}(s)h_{2}(s))$ . Also, the zeros of $h_{1}(s)h_{2}(s)$ are either zeros of $h_{1}(s)$ or zeros of $h_{2}(s)$ , hence are in LHP. By partial fractions, we can write $f\in{\cal S}$ as

f(s)=q+\sum_{j=1}^{N}a_{j}(s-\lambda_{j})^{-n_{j}}.