Let us now try to extend the concept of FPSs to allow negative powers ofx. As a motivating example, we recall the binary positional system:
Definition 3.14.1. Abinary representationof an integernmeans an essentially finite sequence(bi)i∈N = (b0,b1,b2, . . .) ∈ {0, 1}N such that
n=
∑
i∈N
bi2i.
(Recall: “Essentially finite” means “all but finitely many i ∈ N satisfy bi = 0”.)
The following theorem is well-known (and has been proved in Subsection 3.11.1):
Theorem 3.14.2. Each n∈ Nhas a unique binary representation.
Note that we are encoding the digits (actually, bits) of a binary representation as essentially finite sequences instead of finite tuples. This way, we don’t have to worry about leading zeros breaking the uniqueness in Theorem 3.14.2.
Let us now define a variation of binary representation:
Definition 3.14.3. A balanced ternary representation of an integer n means an essentially finite sequence (bi)i∈N = (b0,b1,b2, . . .)∈ {0, 1,−1}N such that
n=
∑
i∈N
bi3i.
Examples:
• The integer 19 has a balanced ternary representation(1, 0,−1, 1, 0, 0, 0, . . .), because
19=1−9+27=30−32+33 =1·30+ (−1)·32+1·33.
• The integer 42 has a balanced ternary representation(0,−1,−1,−1, 1, 0, 0, 0, . . .), because
42 =81−27−9−3=34−33−32−31.
Note that (unlike with binary representations) even negative integers can have balanced ternary representations.
In the Soviet Union of the 1960s/70s, balanced ternary representations have been used as a foundation for computers (see the Setun computer). The idea was shelved in the 1970s, but a noticeable amount of algorithms have been in-vented for working with balanced ternary representations. (See [Knuth2, §4.1]
for some discussion of these.) The following theorem (which goes back to Fi-bonacci) is crucial for making balanced ternary representations useful:
Theorem 3.14.4. Each integernhas a unique balanced ternary representation.
There are various ways to prove this (see, e.g., [20f, solution to Exercise 3.7.8]
for an elementary one). Let us here try to prove it using FPSs (imitating our proof of Theorem 3.14.2 in Subsection 3.11.1). Since thebi can be−1s, we must allow for negative powers ofx.
Let us first argue informally and see later whether we can make sense of what we have done.
Here is the idea42: We shall compute the product
1+x+x−1 1+x3+x−3 1+x9+x−9
· · · =
∏
i≥0
1+x3i+x−3i in two ways:
• On the one hand, this product equals
∏
i≥0
1+x3i +x−3i
| {z }
= ∑
b∈{0,1,−1}
xb·3i
=
∏
i≥0
∑
b∈{0,1,−1}
xb·3i
=
∑
(b0,b1,b2,...)∈{0,1,−1}N essentially finite
xb030xb131xb232· · ·
(here, we have just expanded the product)
=
∑
(b0,b1,b2,...)∈{0,1,−1}N essentially finite
xb030+b131+b232+···
=
∑
n∈Z
(# of balanced ternary representations of n)·xn, (130) since a balanced ternary representation of n is precisely an essentially finite sequence(b0,b1,b2, . . .) ∈ {0, 1,−1}Nsatisfyingb030+b131+b232+
· · · =n.
• On the other hand, we have 1+x+x−1 = 1−x3
x(1−x). Substituting x3i for x in this equality, we obtain
1+x3i +x−3i = 1−x3i+1 x3i
1−x3i for eachi ≥0.
42which, again, goes back to Euler [Euler48, §331]
Hence,
∏
i≥01+x3i+x−3i
=
∏
i≥0
1−x3i+1 x3i
1−x3i
= 1−x3
x(1−x) · 1−x9
x3(1−x3) · 1−x27
x9(1−x9) · 1−x81
x27(1−x27) · · · ·
= 1
xx3x9x27· · ·
| {z }
=x−∞ (whatever this means)
· 1
1−x
| {z }
=1+x+x2+x3+···
(by a somewhat daring use of the telescope principle)
=x−∞
1+x+x2+x3+· · ·
=· · ·+x−2+x−1+x0+x1+x2+· · ·
with some artistic license,
since xi 1+x+x2+x3+· · · =xi+xi+1+xi+2+· · · for eachi ∈Z
=
∑
n∈Z
xn. (131)
Comparing (130) with (131), we find
n
∑
∈Z(# of balanced ternary representations of n)·xn =
∑
n∈Z
xn. Comparing coefficients, we thus conclude that
(# of balanced ternary representations of n) =1
for each n ∈ Z. This “proves” Theorem 3.14.4; we just need to make our computations rigorous – i.e., define the ring in which we have been computing, explain whatx is, and justify the well-definedness of our infinite products and sums.
Let us first play around a bit further. We have (1−x)· · ·+x−2+x−1+x0+x1+x2+· · ·
=· · ·+x−2+x−1+x0+x1+x2+· · ·−x
· · ·+x−2+x−1+x0+x1+x2+· · ·
| {z }
=···+x−1+x0+x1+x2+x3+···
=···+x−2+x−1+x0+x1+x2+···
=· · ·+x−2+x−1+x0+x1+x2+· · ·−· · ·+x−2+x−1+x0+x1+x2+· · ·
=0.
Thus, dividing by 1−x, we obtain
· · ·+x−2+x−1+x0+x1+x2+· · · =0.
Comparing coefficients, we conclude that
1=0 for eachn ∈Z.
Oops! Looks like we have overtaxed our artistic license.
So we need to be careful with negative powers of x. Not everything that looks like a valid computation actually is one. So let us try to be rigorous and delimit what can and what cannot be done with negative powers ofx.
Let us try to define “FPSs with negative powers of x” formally. First, let’s define the largest possible space of such FPSs:
Definition 3.14.5. Let K[[x±]] be theK-moduleKZ of all families (an)n∈Z = (. . . ,a−2,a−1,a0,a1,a2, . . .) of elements of K. Its addition and its scaling are defined entrywise:
(an)n∈Z+ (bn)n∈Z = (an+bn)n∈Z;
λ(an)n∈Z = (λan)n∈Z for each λ∈ K.
An element of K[[x±]] will be called adoubly infinite power series. This name is justified by the fact that we will later use the notation ∑
n∈Z
anxn for a family (an)n∈Z ∈ K[[x±]].
Now, let us try to define a multiplication on thisK-module K[[x±]], in order to turn it into aK-algebra (likeK[[x]]). This multiplication should satisfy
(an)n∈Z·(bn)n∈Z = (cn)n∈Z, where cn =
∑
i∈Z
aibn−i
(since this is what we would get if we expanded
n∑∈Zanxn ∑
n∈Zbnxn
and combined like powers of x). Unfortunately, the sum ∑
i∈Zaibn−i is now infinite (unlike forK[[x]]), and is not always defined. Hence,K[[x±]]is not aK-algebra (even though it is a K-module). In general, we cannot multiply its elements.
This explains why our above computations have led us astray.
If we cannot multiply two arbitrary elements of K[[x±]], can we perhaps restrict ourselves to a smaller K-submodule of K[[x±]]whose elements can be multiplied? Of course, K[[x]] is such a submodule, but there are some others.
Here is one:
Definition 3.14.6. LetK[x±]be theK-submodule ofK[[x±]]consisting of all essentially finitefamilies(an)n∈Z. This is indeed aK-submodule (check it!).
It should be thought of as an analogue of the ring of polynomials K[x], but now allowing for negative powers of x.
The elements of K[x±] are called Laurent polynomials in the indeterminate xover K.
We define a multiplication onK[x±] by setting
(an)n∈Z·(bn)n∈Z = (cn)n∈Z, where cn =
∑
i∈Z
aibn−i. Note that the sum ∑
i∈Zaibn−i is now well-defined, because it is essentially finite.
We define an element x ∈ K[x±]by x= (δi,1)i∈Z.
Theorem 3.14.7. The K-module K[x±], equipped with the multiplication we just defined, is a commutativeK-algebra. Its unity is(δi,0)i∈Z. The element x is invertible in thisK-algebra.
This K-algebra K[x±] is called the Laurent polynomial ring in one indetermi-natex over K. It is often denoted byK
x±1
or K
x,x−1
as well.
Proposition 3.14.8. Any doubly infinite power series a = (ai)i∈Z ∈ K[[x±]]
satisfies
a =
∑
i∈Z
aixi.
Here, the powers xi are taken in the Laurent polynomial ring K[x±], but the infinite sum ∑
i∈Z
aixi is taken in the K-module K[[x±]]. (The notions of summable families and infinite sums are defined in K[[x±]] in the same way as they are defined inK[[x]].)
Examples of Laurent polynomials are
• any polynomial in K[x];
• x−15;
• x2+3+7x−3.
There are other, equivalent ways to define the Laurent polynomial ringK[x±]:
• as the group algebra of the cyclic groupZover K;
• as the localization of the polynomial ringK[x] at the powers ofx.
(These are done in some textbooks on abstract algebra – e.g., see [Ford21, Exercise 3.6.31] for a quick overview.)
Now let us see if we can make our above proof of Theorem 3.14.4 rigorous using Laurent polynomials. Unfortunately, K[x±] is “too small” to contain the infinite product ∏
i≥0
1+x3i+x−3i
. However, we can try using its partial products, which are finite. For eachk ∈N, we have
∏
k i=0
1+x3i+x−3i
=1+x+x−1 1+x3+x−3
· · ·1+x3k+x−3k
= 1−x3
x(1−x) · 1−x9
x3(1−x3) · · · 1−x3k+1 x3k
1−x3k
this is somewhat unrigorous, since 1−x3i are not invertible inK[x±], but this will soon be made rigorous
!
= 1
xx3x9· · ·x3k
| {z }
=x−(30+31+···+3k)
· 1−x3k+1 1−x
| {z }
=1+x+x2+···+x3k+1−1
=x−(30+31+···+3k)·1+x+x2+· · ·+x3k+1−1
=x−(30+31+···+3k)·1+x+x2+· · ·+x2(30+31+···+3k) since 3k+1−1=2·30+31+· · ·+3k
(check this!)
=x−(30+31+···+3k) +x−(30+31+···+3k)+1+x−(30+31+···+3k)+2+· · ·+x30+31+···+3k
=
∑
n∈Z;
|n|≤30+31+···+3k
xn.
On the other hand,
∏
k i=0
1+x3i+x−3i
| {z }
= ∑
b∈{0,1,−1}
xb·3i
=
∏
ki=0
∑
b∈{0,1,−1}
xb·3i
=
∑
(b0,b1,...,bk)∈{0,1,−1}k+1
xb030xb131· · ·xbk3k
by expanding the product using Proposition 3.11.22
=
∑
(b0,b1,...,bk)∈{0,1,−1}k+1
xb030+b131+···+bk3k
=
∑
n∈Z
(# ofk-bounded balanced ternary representations of n)·xn,
where a balanced ternary representation(b0,b1,b2, . . .)is said to bek-bounded if bk+1=bk+2 =bk+3=· · · =0. Comparing the two results, we find
n
∑
∈Z(# of k-bounded balanced ternary representations ofn)·xn
=
∑
n∈Z;
|n|≤30+31+···+3k
xn.
Comparing coefficients, we thus see that each n ∈ Z satisfying |n| ≤ 30+ 31+· · ·+3k has a unique k-bounded balanced ternary representation. Let-ting k → ∞ now quickly yields Theorem 3.14.4 (because any balanced ternary representation isk-bounded for a sufficiently largek).
Thus we have proved Theorem 3.14.4 up to the fact that we have divided by the polynomials 1−x, 1−x3, 1−x9, . . ., which are not invertible in the Laurent polynomial ringK[x±]. In order to fill this gap, we need a new K-algebra: The Laurent polynomial ring K[x±] is too small, whereas the original K-module K[[x±]] is not a ring. We need some kind of middle ground: some K-module lying between K[x±] and K[[x±]] that is a ring but allows division by 1−x, 1−x3, 1−x9, . . . (and, more generally, by 1−xi for each positive integeri).
This middle ground is calledK((x))and is defined as follows:
Definition 3.14.9. We let K((x)) be the subset of K[[x±]] consisting of all families(ai)i∈Z ∈ K[[x±]] such that the sequence(a−1,a−2,a−3, . . .) is essen-tially finite – i.e., such that all sufficiently low i∈ Zsatisfy ai =0.
The elements of K((x)) are called Laurent series in one indeterminate x over K.
For example (forK =Z):
• the “series” x−3+x−2+x−1+x0+x1+· · · belongs to K((x));
• the “series” 1+x−1+x−2+x−3+· · · does not belong to K((x));
• the “series” ∑
n∈Z
xn =· · ·+x−2+x−1+x0+x1+x2+· · · does not belong toK((x)).
Theorem 3.14.10. The subsetK((x)) is aK-submodule of K[[x±]]. But it has a multiplication (unlike K[[x±]]). This multiplication is given by the same rule as the multiplication of K[x±]: namely,
(an)n∈Z·(bn)n∈Z = (cn)n∈Z, where cn =
∑
i∈Z
aibn−i. The sum ∑
i∈Zaibn−ihere is well-defined, because it is essentially finite (indeed, for all sufficiently lowi ∈Z, we haveai =0 and thusaibn−i =0; on the other hand, for all sufficiently highi ∈Z, we have bn−i =0 and thus aibn−i =0).
Now, the ring K((x)) contains both the FPS ring K[[x]] and the Laurent polynomial ringK[x±]as subrings (and actually asK-subalgebras). This makes it one of the most convenient places for formal manipulation of FPSs. (However, it has a disadvantage compared to the FPS ring K[[x]]: Namely, you cannot easily substitute something forx in a Laurent series f ∈ K((x)).)
Now, our above computation of ∏k
i=0
1+x3i+x−3i
makes perfect sense in the Laurent series ring K((x)) (indeed, for each positive integer i, the power series 1−xi is invertible in K[[x]] and thus also invertible in K((x))). Hence, at last, we have a rigorous proof of Theorem 3.14.4.
Actually, you could also make sense of our original argument for prov-ing Theorem 3.14.4, with the infinite product ∏
i≥0
1+x3i+x−3i
, as long as you made sure to interpret it correctly: First, compute the finite products
∏k i=0
1+x3i +x−3i
in the ringK((x)). Then, take their limit lim
k→∞
∏k i=0
1+x3i+x−3i inK[[x±]](this is not a ring, but the notion of a limit inK[[x±]] is defined just as it was inK[[x]]).
More can be said about the K-algebra K((x)) when K is a field: Indeed, in this case, it is a field itself! This fact (whose proof is Exercise A.2.14.2) is not very useful in combinatorics, but quite so in abstract algebra.
One more remark. As we have explained, K[[x±]] is not a ring. However, some semblance of multiplication can be restored inK[[x±]]. Namely, a “dou-bly infinite power series” ∑
n∈Zanxn = (an)n∈Z ∈ K[[x±]] can be multiplied by a Laurent polynomial ∑
n∈Zbnxn = (bn)n∈Z ∈ K[x±], because in this case the sums
i∈∑Z
aibn−i will be essentially finite. Thus, while we cannot always multiply two elements of K[[x±]], we can always multiply an element of K[[x±]] with an element ofK[x±]. This makes K[[x±]] into a K[x±]-module. The equality
(1−x)· · ·+x−2+x−1+x0+x1+x2+· · ·=0 reveals that this module has torsion.