(this follows by substituting −x forx on both sides of (88))
=
∑
Thus, Proposition 3.8.4 is proved.3.9. Integer compositions
3.9.1. Compositions
Next, let us count certain simple combinatorial objects known asinteger compo-sitions. There are easy combinatorial ways to do this (see, e.g., [19fco, §2.10.1]), but we shall employ generating functions, in order to see one more example of how these can be used.
Definition 3.9.1. (a)An (integer) compositionmeans a (finite) tuple of positive integers.
(b)The size of an integer compositionα = (α1,α2, . . . ,αm) is defined to be the integerα1+α2+· · ·+αm. It is written |α|.
(c) The length of an integer composition α = (α1,α2, . . . ,αm) is defined to be the integerm. It is written `(α).
(d)Let n∈ N. A composition of nmeans a composition whose size isn.
(e) Let n ∈ Nand k ∈ N. A composition of n into k parts is a composition whose size isn and whose length isk.
Example 3.9.2. The tuple (3, 8, 6) is a composition with size 3+8+6 = 17 and length 3. In other words, it is a composition of 17 into 3 parts.
The empty tuple()is a composition of 0 into 0 parts. It is the only compo-sition of 0, and also is the only compocompo-sition with length 0.
The followingquestionsarise quite naturally:
1. How many compositions of nexist for a givenn∈ N?
2. How many compositions of ninto kparts exist for given n,k ∈ N? Let us use generating functions to answer question 2.
Approach to question 2. Fixk ∈ N, but don’t fixn. Let
an,k = (# of compositions ofn intok parts). (89) We want to findan,k. We define the generating function
Ak :=
∑
n∈N
an,kxn = (a0,k,a1,k,a2,k, . . .) ∈ Q[[x]]. (90) Let us write P for the set {1, 2, 3, . . .}. Then, a composition of n into k parts is nothing but a k-tuple (α1,α2, . . . ,αk) ∈ Pk satisfying α1+α2+· · ·+αk = n.
Hence, (89) can be rewritten as
an,k =# of allk-tuples (α1,α2, . . . ,αk) ∈Pk satisfyingα1+α2+· · ·+αk =n
=
∑
(α1,α2,...,αk)∈Pk; α1+α2+···+αk=n
1. (91)
Thus, we can rewrite the equality Ak = ∑
by the same product rule that we used back in the proof of Theorem 3.8.3
=
∑
n∈P
xn
!k
(here, we have renamed all thek summation indices as n, and realized that all ksums are identical). Since
n
∑
∈Pthis can be rewritten further as Ak = now: Theorem 3.3.10 (applied to −kinstead ofn) yields
(1+x)−k =
∑
Substituting −x for x on both sides of this equality (i.e., applying the map
f 7→ f ◦(−x)), we obtain
(here, we have substitutedn−k for j in the sum). Hence, our above computa-tion of Ak can be completed as follows:
Ak =xk (1−x)−k
this did not change the sum, since all the newly introduced addends withn <k are 0). Comparing coefficients, we thus obtain
an,k =
n−1 n−k
for each n∈ N. (94)
Ifn >0, then we can rewrite the right hand side of this equality as
n−1 k−1
(using Theorem 2.0.6). However, ifn = 0, then this right hand side equals δk,0 instead (where we are using Definition 3.5.6). Thus, we can rewrite (94) as
an,k =
We have thus answered our Question 2. Let us summarize the two an-swers we have found ((94) and (95)) in the following theorem ([19fco, Theorem 2.10.1]):
Theorem 3.9.3. Let n,k∈ N. Then, the # of compositions of ninto k parts is
This theorem has other proofs as well. See [19fco, Proof of Theorem 2.10.1]
for a proof by bijection and [19fco, solution to Exercise 2.10.2] for a proof by induction.
As an easy consequence of Theorem 3.9.3, we can now answer Question 1 as well:
Theorem 3.9.4. Let n∈ N. Then, the # of compositions of nis (2n−1, ifn >0;
1, ifn =0.
Proof of Theorem 3.9.4 (sketched). If n = 0, then the # of compositions of n is 1 (since the empty tuple () is the only composition of 0). Thus, for the rest of this proof, we WLOG assume that n 6= 0. Hence, we must prove that the # of compositions ofnis 2n−1.
If (n1,n2, . . . ,nm) is a composition of n, then m ∈ {1, 2, . . . ,n} (why?). In other words, any composition of n is a composition ofn into k parts for some k∈ {1, 2, . . . ,n}. Hence,
(here, we have substitutedk forn−kin the sum). Comparing this with 2n−1 = (1+1)n−1 =
(by the binomial theorem)
=
3.9.2. Weak compositions
A variant of compositions are the weak compositions. These are like composi-tions, but their entries have to only be nonnegative rather than positive. For the sake of completeness, let us give their definition in full:30
Definition 3.9.5. (a) An (integer) weak composition means a (finite) tuple of nonnegative integers.
(b)Thesizeof a weak composition α = (α1,α2, . . . ,αm) is defined to be the integerα1+α2+· · ·+αm. It is written|α|.
(c) The length of a weak composition α = (α1,α2, . . . ,αm) is defined to be the integerm. It is written `(α).
(d) Let n ∈ N. Aweak composition of n means a weak composition whose size is n.
(e) Let n ∈ N and k ∈ N. A weak composition of n into k parts is a weak composition whose size isnand whose length is k.
Example 3.9.6. The tuple (3, 0, 1, 2) is a weak composition with size 3+0+ 1+2 = 6 and length 4. In other words, it is a weak composition of 6 into 4 parts. It is not a composition, since one of its entries is a 0.
Weak compositions are a rather natural analogue of compositions, but behave dissimilarly in one important way: Anyn ∈ N has infinitely many weak com-positions. Indeed, all the tuples (n), (n, 0), (n, 0, 0), (n, 0, 0, 0), . . . are weak compositions ofn(and of course, there are many more weak compositions ofn, unlessn=0). Thus, it makes no sense to look for an analogue of Theorem 3.9.4 for weak compositions. However, an analogue of Theorem 3.9.3 does exist:
Theorem 3.9.7. Let n,k ∈ N. Then, the # of weak compositions of n into k parts is
n+k−1 n
=
n+k−1 k−1
, if k>0;
δn,0, if k=0.
Proof of Theorem 3.9.7 (sketched). (See [19fco, Theorem 2.10.5] for details and al-ternative proofs.) Adding 1 to a nonnegative integer yields a positive integer.
Furthermore, if we add 1 to each entry of ak-tuple, then the sum of all entries of thek-tuple increases by k.
Thus, if(α1,α2, . . . ,αk) is a weak composition of ninto k parts, then
(α1+1,α2+1, . . . ,αk+1) is a composition ofn+k intokparts. Hence, we can define a map
{weak compositions of ninto kparts} → {compositions of n+kinto k parts}, (α1,α2, . . . ,αk) 7→(α1+1,α2+1, . . . ,αk+1).
30Beware that the word “weak composition” does not have a unique meaning in the literature.
Furthermore, it is easy to see that this map is a bijection (indeed, its inverse is rather easy to construct). Thus, by the bijection principle, we have
|{weak compositions of ninto kparts}|
=|{compositions of n+kinto k parts}|
= (# of compositions ofn+k into kparts)
=
n+k−1 n+k−k
by the first equality sign in Theorem 3.9.3, applied to n+kinstead ofn
=
n+k−1 n
.
Thus, we have shown that the # of weak compositions of n into k parts is n+k−1
n
. It remains to prove that this equals
n+k−1 k−1
, if k>0;
δn,0, if k=0 as well. This is similar to how we obtained (95): If k = 0, then it is clear by inspection; otherwise it follows from Theorem 2.0.6. Theorem 3.9.7 is proven.
3.9.3. Weak compositions with entries from {0, 1, . . . ,p−1}
Theorems 3.9.3 and 3.9.7 may stir up hopes that other tuple-counting problems also have simple answers. Let us see if this hope holds up.
An attempt. We fix three nonnegative integersn,kand p. We are looking for the
# ofk-tuples(α1,α2, . . . ,αk) ∈ {0, 1, . . . ,p−1}ksatisfyingα1+α2+· · ·+αk =n.
In other words, we are looking for the # of all weak compositions of n into k parts with the property that each entry is< p. Let us denote this # by wn,k,p.
Just as when counting compositions, we invoke a generating function. Forget that we fixedn, and define the FPS
Wk,p :=
∑
n∈N
wn,k,pxn ∈ Q[[x]]. For eachn∈ N, we have
wn,k,p =# of allk-tuples (α1,α2, . . . ,αk) ∈ {0, 1, . . . ,p−1}k satisfyingα1+α2+· · ·+αk =n)
=
∑
(α1,α2,...,αk)∈{0,1,...,p−1}k; α1+α2+···+αk=n
1.
Thus, we can rewrite the equalityWk,p = ∑ by the same product rule that we used back
in the proof of Theorem 3.8.3
(here, we have renamed all thek summation indices as n, and realized that all ksums are identical). Since
n∈{0,1,...,p
∑
−1}xn =x0+x1+· · ·+xp−1 = 1−xp 1−x
(the last equality sign here is easy to check31), this can be rewritten further as Wk,p =
The binomial theorem yields all j∈ N; this did not change the sum, since all the newly introduced addends are 0). Multiplying this by (93), we obtain
(1−xp)k(1−x)−k
here, we have renamed the summation index j asiin the second sum
Comparing coefficients in this equality, we find that eachn ∈Nsatisfies
is uniquely determined by its second entry j
here, we have extended the range of summation by dropping the “pj≤n” requirement; this does not change
the sum, since all newly introduced addends are 0
(here, we have removed all addends with j > k from the sum; this does not change the sum, since all these addends are 0).
Thus, we have proved the following fact:
Theorem 3.9.8. Let n,k,p ∈ N. Then, the # of k-tuples (α1,α2, . . . ,αk) ∈
In general, this expression is the simplest we can get. A combinatorial proof of Theorem 3.9.8 can be found in [19fco, Exercise 2.10.6].
However, the particular case when p = 2 is worth exploring, as it allows for a much simpler expression. Indeed, the k-tuples (α1,α2, . . . ,αk) ∈ {0, 1}k are just the “binary k-strings”, i.e., the k-tuples formed of 0s and 1s. Impos-ing the condition α1+α2+· · ·+αk = n on such a k-tuple is tantamount to requiring that it contain exactly n many 1s. Therefore, the # of all k-tuples (α1,α2, . . . ,αk)∈ {0, 1}k satisfyingα1+α2+· · ·+αk =n is
k n
, since we have to choose whichnof itskpositions will be occupied by 1s (and then all remain-ing k−n positions will be occupied by 0s). However, Theorem 3.9.8 (applied
to p = 2) yields that this # equals ∑k these two results, we obtain the following identity:
Proposition 3.9.9. Let n,k ∈ N. Then, k
We now return to general properties of FPSs.
Definition 3.10.1. Let n ∈ N. Let f,g ∈ K[[x]]be two FPSs. We write f x
n
≡ g if and only if
eachm ∈ {0, 1, . . . ,n} satisfies [xm] f = [xm]g.
Thus, we have defined a binary relation x
n coefficients of f agree with the first n+1 coefficients of g. Here are some examples:
One way to get an intuition for the relation x
n
≡ is to think of it as a kind of “approximate equality” up to degree n. (This makes the most sense if one thinks of x as an infinitesimal quantity, in which case a term λxk (with λ ∈ K) is the more “important” the lowerk is. From this viewpoint, f x
n
≡g means that the FPSs f andgagree in theirn+1 most “important” terms and differ at most in their “error terms”.) For this reason, the statement “f x
n
≡ g” is sometimes written as “f =g+o(xn)” (an algebraic imitation of Landau’s little-o notation from asymptotic analysis). Another intuition comes from elementary number theory: The relation x
n
≡ is similar to congruence of integers modulo a given integer. This is more than a similarity; the relationx
n
≡can in fact be restated as a divisibility in the same fashion as for congruences of integers (see Proposition 3.10.4 below). For this reason, the statement “f x
n
≡ g” is sometimes written as “f ≡ gmodxn+1”. We shall, however, eschew both of these alternative notations, and use the original notation “f x
n
≡ g” from Definition 3.10.1, as both intuitions (while useful) would distract from the simplicity of Definition 3.10.132.
Here are some basic properties of the relation x
n
≡(some of which will be used without explicit reference):
Theorem 3.10.3. Let n∈ N.
(a)The relation x
n
≡on K[[x]] is an equivalence relation. In other words:
• This relation is reflexive (i.e., we have f x
n
32Case in point: Definition 3.10.1 can be generalized to multivariate FPSs, but the two intu-itions are no longer available (or, worse, give the “wrong” concepts) when extended to this generality.
(c) If a,b ∈ K[[x]] are two FPSs satisfying a x families of FPSs such that
eachs ∈ Ssatisfies as
Proof of Theorem 3.10.3 (sketched). All of these properties are analogous to famil-iar properties of integer congruences, except for Theorem 3.10.3 (d), which is moot for integers (since there are not many integers that are invertible in Z).
The proofs are similarly simple (using (18), (19), (20) and (23)). Thus, we shall only give some hints for the proof of Theorem 3.10.3 (d) here; detailed proofs of all parts of Theorem 3.10.3 can be found in Section B.1.
(d)Let a,b ∈K[[x]]be two invertible FPSs satisfyinga x
The FPS a is invertible; thus, its constant term x0
a is invertible in K (by Proposition 3.3.7).
. We shall prove this by strong induction onm: We fix some k ∈ {0, 1, . . . ,n}, and we assume (as an induction hypothesis) that
[xm]a−1
= [xm]b−1
for each m∈ {0, 1, . . . ,k−1}. (104) We must now prove that
xk
(here, we have split off the addend fori=0 from the sum). Thus,
We can solve this equation for xk
a−1
(since x0
a is invertible), and thus obtain The same argument (applied tob instead ofa) yields
h
The right hand sides of the latter two equalities are equal (since each i ∈ {1, 2, . . . ,k} satisfies
xi
a = xi
b as a consequence of (103), and satisfies xk−i
a−1
=xk−i b−1
as a consequence of (104), and since we have x0
a= x0
bas a consequence of (103)). Hence, the left hand sides must also be equal.
In other words, xk
a−1
=xk b−1
, which is precisely what we wanted to prove. Thus, the induction step is complete, so thata−1 x
n
≡b−1 is proved. Thus, Theorem 3.10.3(d)is proved. (See Section B.1 for more details.)
Let us next characterize xn-equivalence of FPSs in terms of divisibility:
Proposition 3.10.4. Let n ∈N. Let f,g ∈ K[[x]]be two FPSs. Then, we have f x
n
≡ gif and only if the FPS f −g is a multiple ofxn+1.
Proof of Proposition 3.10.4. See Section B.1 for this proof (a simple consequence of Lemma 3.3.18).
Finally, here is a subtler property of xn-equivalence similar to the ones in Theorem 3.10.3(b):
Proof of Proposition 3.10.5 (sketched). Writeaandbasa= ∑ (a consequence of Lemma 3.3.18 using
x0
See Section B.1 for the details of this argument.