If you allow series that are infinite in both directions, then you have a new problem which is that multiplication may no longer be possible: the sums involved need not converge. And there’s also the issue already noted, that some things that don’t look like they equal zero may in some sense have to be zero. (Meaning “absolute” zero = (...,0,0,0,...) rather than the thing you originally called zero which should maybe be called something like ε instead.)
What’s the best we could hope for? Something like this. Write R for RZ, i.e., all formal potentially-double-ended Laurent series. There’s an addition operation defined on the whole thing, and a multiplicative operation defined on some subset of pairs of its elements, namely those for which the relevant sums converge (or maybe are “summable” in some weaker sense). There are two problems: (1) some products aren’t defined, and (2) at least with some ways of defining them, there are some zero-divisors—e.g., (x-1) times the sum of all powers of x, as discussed above. (I remark that if your original purpose is to be able to divide by zero, perhaps you shouldn’t be too troubled by the presence of zero-divisors; contrapositively, that if they trouble you, perhaps you shouldn’t have wanted to divide by zero in the first place.)
We might hope to deal with issue 1 by restricting to some subset A of R, chosen so that all the sums that occur when multiplying elements of A are “well enough behaved”; if issue 2 persists after doing that, maybe we might hope to deal with that by taking a quotient of A—i.e., treating some of its elements as being equal to one another.
Some versions of this strategy definitely succeed, and correspond to things just_browsing already mentioned above. For instance, let A consist of everything in R with only finitely many negative powers of x, the Laurent series already mentioned; this is a field. Or let it consist of everything that’s the series expansion of a rational function of x; this is also a field. This latter is, I think, the nearest you can get to “finite or periodic”. The periodic elements are the ones whose denominator has degree at most 1. Degree ⇐ 2 brings in arithmetico-periodic elements—things that go, say, 1,1,2,2,3,3,4,4, etc. I’m pretty sure that degree <=d in the denominator is the same as coefficients being ultimately (periodic + polynomial of degree < d). And this is what you get if you say you want to include both 1 and x, and to be closed under addition, subtraction, multiplication, and division.
Maybe that’s already all you need. If not, perhaps the next question is: is there any version of this that gives you a field and that allows, at least, some series that are infinite in both directions? Well, by considering inverses of (1-x)^k we can get sequences that grow “rightward” as fast as any polynomial. So if we want the sums inside our products to converge, we’re going to need our sequences to shrink faster-than-polynomially as we move “leftward”. So here’s an attempt. Let A consist of formal double-ended Laurent series ∑n∈Zanxn such that for n<0 we have |an|=O(t−n) for some t<1, and for n>0 we have |an|=O(nk) for some k. Clearly the sum or difference of two of these has the same properties. What about products? Well, if we multiply together a,b to get c then cn=∑p+q=napbq. The terms with p<0<q are bounded in absolute value by some constant times t−pqk where t gets its value from a and k gets its value from b; so the sum of these terms is bounded by some constant times ∑q>0tq−nqk which in turn is a constant times t−n. Similarly for the terms with q<0<p; the terms with p,q both of the same sign are bounded by a constant times t−n when they’re negative and by a constant times n(ka+kb) when they’re positive. So, unless I screwed up, products always “work” in the sense that the sums involved converge and produce a series that’s in A. Do we have any zero-divisors? Eh, I don’t think so, but it’s not instantly obvious.
Here’s a revised version that I think does make it obvious that we don’t have zero-divisors. Instead of requiring that for n<0 we have |an|=O(tn) for some t<1, require that to hold for allt<1. Once again our products always exist and still lie in A. But now it’s also true that for small enough t, the formal series themselves converge to well-behaved functions of t. In particular, there can’t be zero-divisors.
I’m not sure any of this really helps much in your quest to divide by zero, though :-).
If you allow series that are infinite in both directions, then you have a new problem which is that multiplication may no longer be possible: the sums involved need not converge. And there’s also the issue already noted, that some things that don’t look like they equal zero may in some sense have to be zero. (Meaning “absolute” zero = (...,0,0,0,...) rather than the thing you originally called zero which should maybe be called something like ε instead.)
What’s the best we could hope for? Something like this. Write R for RZ, i.e., all formal potentially-double-ended Laurent series. There’s an addition operation defined on the whole thing, and a multiplicative operation defined on some subset of pairs of its elements, namely those for which the relevant sums converge (or maybe are “summable” in some weaker sense). There are two problems: (1) some products aren’t defined, and (2) at least with some ways of defining them, there are some zero-divisors—e.g., (x-1) times the sum of all powers of x, as discussed above. (I remark that if your original purpose is to be able to divide by zero, perhaps you shouldn’t be too troubled by the presence of zero-divisors; contrapositively, that if they trouble you, perhaps you shouldn’t have wanted to divide by zero in the first place.)
We might hope to deal with issue 1 by restricting to some subset A of R, chosen so that all the sums that occur when multiplying elements of A are “well enough behaved”; if issue 2 persists after doing that, maybe we might hope to deal with that by taking a quotient of A—i.e., treating some of its elements as being equal to one another.
Some versions of this strategy definitely succeed, and correspond to things just_browsing already mentioned above. For instance, let A consist of everything in R with only finitely many negative powers of x, the Laurent series already mentioned; this is a field. Or let it consist of everything that’s the series expansion of a rational function of x; this is also a field. This latter is, I think, the nearest you can get to “finite or periodic”. The periodic elements are the ones whose denominator has degree at most 1. Degree ⇐ 2 brings in arithmetico-periodic elements—things that go, say, 1,1,2,2,3,3,4,4, etc. I’m pretty sure that degree <=d in the denominator is the same as coefficients being ultimately (periodic + polynomial of degree < d). And this is what you get if you say you want to include both 1 and x, and to be closed under addition, subtraction, multiplication, and division.
Maybe that’s already all you need. If not, perhaps the next question is: is there any version of this that gives you a field and that allows, at least, some series that are infinite in both directions? Well, by considering inverses of (1-x)^k we can get sequences that grow “rightward” as fast as any polynomial. So if we want the sums inside our products to converge, we’re going to need our sequences to shrink faster-than-polynomially as we move “leftward”. So here’s an attempt. Let A consist of formal double-ended Laurent series ∑n∈Zanxn such that for n<0 we have |an|=O(t−n) for some t<1, and for n>0 we have |an|=O(nk) for some k. Clearly the sum or difference of two of these has the same properties. What about products? Well, if we multiply together a,b to get c then cn=∑p+q=napbq. The terms with p<0<q are bounded in absolute value by some constant times t−pqk where t gets its value from a and k gets its value from b; so the sum of these terms is bounded by some constant times ∑q>0tq−nqk which in turn is a constant times t−n. Similarly for the terms with q<0<p; the terms with p,q both of the same sign are bounded by a constant times t−n when they’re negative and by a constant times n(ka+kb) when they’re positive. So, unless I screwed up, products always “work” in the sense that the sums involved converge and produce a series that’s in A. Do we have any zero-divisors? Eh, I don’t think so, but it’s not instantly obvious.
Here’s a revised version that I think does make it obvious that we don’t have zero-divisors. Instead of requiring that for n<0 we have |an|=O(tn) for some t<1, require that to hold for all t<1. Once again our products always exist and still lie in A. But now it’s also true that for small enough t, the formal series themselves converge to well-behaved functions of t. In particular, there can’t be zero-divisors.
I’m not sure any of this really helps much in your quest to divide by zero, though :-).