(the motivation section turned out a little long, the mathematical question is at the end)
I need to work with electrical circuts at the moment, computing effective impedances etc. From electrodynamics, we have Kirchhoffs law and so on, which result in two rules: If you have two impedances $Z_1,Z_2$ in series, then the total impecance is given by $$s(Z_1,Z_2)=Z_1+Z_2.$$
If you have them in parallel you get $$p(Z_1,Z_2)=\frac{1}{\frac{1}{Z_1}+\frac{1}{Z_2}}$$
I was using the more general rule $\frac{1}{\frac{1}{Z_1}+\frac{1}{Z_2}+\frac{1}{Z_3}}$ until I dediced I should derive it from the above one, and indeed
$$p(Z_1,Z_2,Z_3):=p(Z_1,p(Z_2,Z_3))=\frac{1}{\frac{1}{Z_1}+\frac{1}{\frac{1}{\frac{1}{Z_2}+\frac{1}{Z_3}}}}=\frac{1}{\frac{1}{Z_1}+\frac{1}{Z_2}+\frac{1}{Z_3}}.$$
Now we have $s(Z_1,Z_1)=2Z_1$ and $p(Z_1,Z_1)=\tfrac{1}{2}Z_1$ and these operations are highly symmetric, in that for example $p(Z_1,p(Z_2,p(Z_3,Z_4))=p((Z_1,Z_2),p(Z_3,Z_4))$. And we can build complicated stuff, e.g. $p(p(Z_1,Z_2),s(Z_1,Z_2))=\frac{Z_1^2+Z_1Z_2}{3Z_1+2Z_2}$. Might be we can fit the coefficients there to any natural we like.
Now we have the passive elements resistor $R$, capacity $C$ and inductivity $L$ with
$$Z_R=R,\ \ Z_C(\omega)=\frac{1}{iC\omega},\ \ Z_L(\omega)=i\omega L.$$
and electrical circuits are used to compose these to various functions, so that the electrical resistence is selective w.r.t $\omega$. E.g. we have filters which effectively are functions which have only restricted support. Or see e.g. the two element LC-circuit for a simple construction. I was thinking if I can construct a frequency independend impedance without using constant elements $Z_R$, and the obvious idea would have been to let $\omega$ cancel out in
$$Z_C\cdot Z_L=\frac{1}{iC\omega}i\omega L=\frac{L}{C}.$$
For a moment I was thinking about which circuit combination would give my $Z_1Z_2$, but I quickly realized that both $s$ and $p$ map a unit Ohm back to Ohm (e.g. $\frac{1}{\frac{1}{Z_1}+\frac{1}{Z_2}}=(Z_1+Z_2)^{-1}Z_1Z_2$, which effectively still has units Ohm to the power of 1), and the unit of $Z_1Z_2$ would be Ohm to the power of 2. But it does for example make it possible to construct $\frac{1}{1+i\omega C}$.
So my question now is: What is the function space I can generate? I assume it's some extension of some restriction of the rational polynomials generated by symmetrical functions. Notice that we got "$+$" but not "$-$", and a weak version of "$/$" but not "$\cdot$". For $x,y\in \mathbb R_{>0}$, we have $p(x,y)=p(y,x)>x$ and $s(x,y)=s(y,x)<x$.
What is the function space in $\omega$ I can generate with the operations $$s(Z_1,Z_2)=Z_1+Z_2\ \text{and }\ p(Z_1,Z_2)=(Z_1+Z_2)^{-1}Z_1Z_2,$$ where $Z$'s are of the form $$a,\ b\ (i\omega),\ c\ (i\omega)^{-1}$$ where $a,b,c$ are e.g. in $\mathbb R_{>0}$ and $i^2=-1$?
This <link> is a more broad SE question on electrical circuits.
Edit (11.7.13):
I worked on it further and here some insights: In terms of $Z_j$'s, the expressions look always of the form $\tfrac{P}{Q}$, where both $P$ and $Q$ are polynmials of the form
$$\sum_i^N\prod_{j=\sigma(i)}^M Z_j,$$
e.g.
$$s(p(Z_1,Z_2),Z_3)=\frac{Z_1\cdot Z_2}{Z_1+Z_2}+Z_3=\frac{Z_1 \cdot Z_2+Z_1 \cdot Z_3+Z_2 \cdot Z_3}{Z_1 + Z_2}=\frac{\sum_{i=1}^3\prod_{j=i}^{i+1} Z_j}{\sum_{i=1}^2\prod_{j=i}^{i} Z_j},$$
An interesting subquestion pops up: Given $P$, can we determine the form of Q? (Up to overall multiplicative factors, which we know we can rescale) If yes, does to what extend does this still hold one we substiture the $Z$'s for $R$, $i\omega L$ or $\tfrac{1}{i\omega C}$?
The structure of every expression $s(p(*,*),*)$ clearly is of the form of a simple tree which we can enumerate. For any fixed number of inputs which all can be of the form $a,\ b\ (i\omega),\ c\ (i\omega)^{-1}$ (e.g. three $Z_1,Z_2,Z_3$ in the above case) and where the real values $a,b,c$ can be adjusted as we like, we can generate all possible expression.
This suggest a more explicit formulation of the question: For a given number of arguments, what is the function mapping the index of the tree to the polynomial $P/Q$?
I post this update now that I've discovered that that function exibits some nontrivial features: For exmaple, if two certain $Z$'s in the expression $s(*,p(*,*))$ have the same dependence in $i\omega$, then it turns out that (by rescaling of the numbers $a,b,c$), this is equivalent to the combination $p(*,s(*,*))$:
en.wikipedia.org/wiki/Equivalent_impedance_transforms