Perhaps a better way to describe this set is ‘all you can build in finitely many steps using addition, inverse, and multiplication, starting from only elements with finite support’.
Ah, now I see what you are after.
But if you use addition on one of them, things may go wrong.
This is exactly right, here’s an illustration:
Here is a construction of (…,1,1,1,…): We have that 1+x+x2+… is the inverse of 1−x. Moreover, 1x+1x2+1x3+…is the inverse of x−1. If we want this thing to be closed under inverses and addition, then this implies that
(1+x+x2+…)+(1x+1x2+1x3+…)=⋯+1x3+1x2+1x+1+x+x2+…
can be constructed.
But this is actually bad news if you want your multiplicative inverses to be unique. Since 1x+1x2+1x3+… is the inverse of x−1, we have that −1x−1x2−1x3… is the inverse of 1−x. So then you get
−1x−1x2−1x3−⋯=1+x+x2+…
so
0=⋯+1x3+1x2+1x+1+x+x2+…
On the one hand, this is a relief, because it explains the strange property that this thing stays the same when multiplied by x. On the other hand, it means that it is no longer the case that the coordinate representation (…,1,1,1,…) is well-defined—we can do operations which, by the rules, should produce equal outputs, but they produce different coordinates.
In fact, for any polynomial (such as 1−x), you can find one inverse which uses arbitrarily high positive powers of x and another inverse which uses arbitrarily low negative powers of x. The easiest way to see this is by looking at another example, let’s say x2+1x.
One way you can find the inverse of x2+1x is to get the 1 out of the x2 term and keep correcting: first you have (x2+1x)(1x2+?), then you have (x2+1x)(1x2−1x5+?), then you have (x2+1x)(1x2−1x5+1x8+?), and so on.
Another way you can find the inverse of x2+1x is to write its terms in opposite order. So you have 1x+x2 and you do the same correcting process, starting with (1x+x2)(x+?), then (1x+x2)(x−x4+?), and continuing in the same way.
Then subtract these two infinite series and you have a bidirectional sum of integer powers of x which is equal to 0.
My hunch is that any bidirectional sum of integer powers of x which we can actually construct is “artificially complicated” and it can be rewritten as a one-directional sum of integer powers of x. So, this would mean that your number system is what you get when you take the union of Laurent series going in the positive and negative directions, where bidirectional coordinate representations are far from unique. Would be delighted to hear a justification of this or a counterexample.
Here is a construction of (…,1,1,1,…): We have that 1+x+x2+… is the inverse of 1−x. Moreover, 1x+1x2+1x3+…is the inverse of x−1. [...]
Yeah, that’s conclusive. Well done! I guess you can’t divide by zero after all ;)
I think the main mistake I’ve made here is to assume that inverses are unique without questioning it, which of course doesn’t make sense at all if I don’t yet know that the structure is a field.
My hunch is that any bidirectional sum of integer powers of x which we can actually construct is “artificially complicated” and it can be rewritten as a one-directional sum of integer powers of x. So, this would mean that your number system is what you get when you take the union of Laurent series going in the positive and negative directions, where bidirectional coordinate representations are far from unique. Would be delighted to hear a justification of this or a counterexample.
So, I guess one possibility is that, if we let [x] be the equivalence class of all elements that are =x in this structure, the resulting set of classes is isomorphic to the Laurent numbers. But another possibility could be that it all collapses into a single class—right? At least I don’t yet see a reason why that can’t be the case (though I haven’t given it much thought). You’ve just proven that some elements equal zero, perhaps it’s possible to prove it for all elements.
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 :-).
Ah, now I see what you are after.
This is exactly right, here’s an illustration:
Here is a construction of (…,1,1,1,…): We have that 1+x+x2+… is the inverse of 1−x. Moreover, 1x+1x2+1x3+…is the inverse of x−1. If we want this thing to be closed under inverses and addition, then this implies that
(1+x+x2+…)+(1x+1x2+1x3+…)=⋯+1x3+1x2+1x+1+x+x2+…
can be constructed.
But this is actually bad news if you want your multiplicative inverses to be unique. Since 1x+1x2+1x3+… is the inverse of x−1, we have that −1x−1x2−1x3… is the inverse of 1−x. So then you get
−1x−1x2−1x3−⋯=1+x+x2+…
so
0=⋯+1x3+1x2+1x+1+x+x2+…
On the one hand, this is a relief, because it explains the strange property that this thing stays the same when multiplied by x. On the other hand, it means that it is no longer the case that the coordinate representation (…,1,1,1,…) is well-defined—we can do operations which, by the rules, should produce equal outputs, but they produce different coordinates.
In fact, for any polynomial (such as 1−x), you can find one inverse which uses arbitrarily high positive powers of x and another inverse which uses arbitrarily low negative powers of x. The easiest way to see this is by looking at another example, let’s say x2+1x.
One way you can find the inverse of x2+1x is to get the 1 out of the x2 term and keep correcting: first you have (x2+1x)(1x2+?), then you have (x2+1x)(1x2−1x5+?), then you have (x2+1x)(1x2−1x5+1x8+?), and so on.
Another way you can find the inverse of x2+1x is to write its terms in opposite order. So you have 1x+x2 and you do the same correcting process, starting with (1x+x2)(x+?), then (1x+x2)(x−x4+?), and continuing in the same way.
Then subtract these two infinite series and you have a bidirectional sum of integer powers of x which is equal to 0.
My hunch is that any bidirectional sum of integer powers of x which we can actually construct is “artificially complicated” and it can be rewritten as a one-directional sum of integer powers of x. So, this would mean that your number system is what you get when you take the union of Laurent series going in the positive and negative directions, where bidirectional coordinate representations are far from unique. Would be delighted to hear a justification of this or a counterexample.
Yeah, that’s conclusive. Well done! I guess you can’t divide by zero after all ;)
I think the main mistake I’ve made here is to assume that inverses are unique without questioning it, which of course doesn’t make sense at all if I don’t yet know that the structure is a field.
So, I guess one possibility is that, if we let [x] be the equivalence class of all elements that are =x in this structure, the resulting set of classes is isomorphic to the Laurent numbers. But another possibility could be that it all collapses into a single class—right? At least I don’t yet see a reason why that can’t be the case (though I haven’t given it much thought). You’ve just proven that some elements equal zero, perhaps it’s possible to prove it for all elements.
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 :-).