3.5.1. Defining substitution
Definition 3.4.5 shows that if f ∈ K[x] is a polynomial, then almost anything (to be more precise: any element of aK-algebra) can be substituted into f.
In contrast, if f ∈ K[[x]] is a FPS, then there are far fewer things that can be substituted into f. Even elements ofK itself cannot always be substituted into f. For example, if we try to substitute 1 for x in the FPS 1+x+x2+x3+· · ·, then we get
1+1+12+13+· · · =1+1+1+1+· · · ,
which is undefined. Real analysis can help make sense of certain values of FPSs (for example, substituting 1
2 for x into the FPS 1+x+x2+x3+· · · yields the convergent series 1+1
2 + 1 22 + 1
23 +· · · = 2), but this is subtle and specific to certain numbers and certain FPSs.19
19For instance, it is not hard to see that there is no nonzero complex number that can be
Thus, polynomials have an advantage over FPSs.
However, not all is lost. Some things can be substituted into an FPS. For example:
• We can always substitute 0 for x in an FPS a0+a1x+a2x2+a3x3+· · ·. The result is
a0+a10+a202+a303+· · · =a0+0+0+0+· · · =a0.
• We can always substitute x for x in an FPS a0+a1x+a2x2+a3x3+· · ·. The result is the same FPS a0+a1x+a2x2+a3x3+· · · that we started with (obviously).
• We can always substitute 2x for x in an FPS a0+a1x+a2x2+a3x3+· · ·. The result is
a0+a1(2x) +a2(2x)2+a3(2x)3+· · · = a0+2a1x+4a2x2+8a3x3+· · · .
• We can always substitute x2+x for x in an FPS a0+a1x+a2x2+a3x3+
· · ·. This is less obvious, so let me explain why. If we try to substitute x2+x for xin an FPS a0+a1x+a2x2+a3x3+· · ·, then we obtain
a0+a1x+a2x2+a3x3+· · · hx+x2i
=a0+a1
x+x2
+a2
x+x22
+a3
x+x23
+· · ·
=a0+a1
x+x2
+a2
x2+2x3+x4
+a3
x3+3x4+3x5+x6 +· · ·
=a0+a1x+ (a1+a2)x2+ (2a2+a3)x3+ (a2+3a3+a4)x4+· · · . I claim that the right hand side here is well-defined. To prove this, I need to show that for eachn ∈ N, the coefficient of xn on this right hand side is afinite sum of ai’s. Indeed, fix n ∈ N. Recall that the right hand side is obtained by expanding the infinite sum
a0+a1
x+x2 +a2
x+x22
+a3
x+x23
+· · · .
Only the first n+1 addends of this infinite sum (i.e., only the addends ak x+x2k
with k ≤ n) can contribute to the coefficient of xn, since any of the remaining addends is a multiple of xn+1 (because it has the form ak x+x2k
=ak(x(1+x))k = akxk(1+x)k with k≥n+1) and thus has
substituted into the FPS ∑
n∈Nn!xn to obtain a convergent result. Thus, even though some complex numbers can be substituted into some FPSs, there is no complex number other than 0 that can be substituted intoeveryFPS.
a zero coefficient of xn. Hence, the coefficient of xn in this infinite sum equals the coefficient of xn in thefinitesum
a0+a1
x+x2 +a2
x+x22
+a3
x+x23
+· · ·+an
x+x2n
. But the latter coefficient is clearly a finite sum of ai’s. Thus, my claim is proved, and it follows that the result of substitutingx2+xforxin an FPS a0+a1x+a2x2+a3x3+· · · is well-defined.
The idea of the last example can be generalized; there was nothing special aboutx+x2 that we used other than the fact thatx+x2 is a multiple ofx (that is, an FPS whose constant term is 0). Thus, generalizing our reasoning from this example, we can convince ourselves that any FPS gthat is a multiple of x (that is, whose constant term is 0) can be substituted into any FPS. Let us introduce a notation for this, exactly like we did for substituting things into polynomials:
Definition 3.5.1. Let f and g be two FPSs in K[[x]]. Assume that x0
g = 0 (that is,g =g1x1+g2x2+g3x3+· · · for some g1,g2,g3, . . . ∈ K).
We then define an FPS f [g] ∈ K[[x]] as follows:
Write f in the form f = ∑
n∈Nfnxnwith f0, f1, f2, . . .∈ K. (That is, fn = [xn] f for each n∈ N.) Then, set
f [g] :=
∑
n∈N
fngn. (44)
(This sum is well-defined, as we will see in Proposition 3.5.2(b)below.) This FPS f [g] is also denoted by f ◦g, and is called the composition of f with g, or the result ofsubstituting gfor xin f.
Once again, it is not uncommon to see this FPS f [g] denoted by f (g), but I will eschew the latter notation (since it can be confused with a product).
In order to prove that Definition 3.5.1 makes sense, we need to ensure that the infinite sum ∑
n∈N fngn in (44) is well-defined. The proof of this fact is analogous to the reasoning I used in the last example; let me present it again in the general case:
Proposition 3.5.2. Let f and gbe two FPSs inK[[x]]. Assume that x0
g =0.
Write f in the form f = ∑
n∈Nfnxn with f0, f1, f2, . . .∈ K. Then:
(a)For each n∈ N, the first ncoefficients of the FPS gn are 0.
(b) The sum ∑
n∈N fngn is well-defined, i.e., the family (fngn)n∈N is summable.
(c)We have x0
n∑∈N fngn
= f0.
Proof of Proposition 3.5.2. (a)This is easily proved by induction on n. Here is a shorter alternative argument:
The FPS g has constant term x0
g = 0. Hence, Lemma 3.3.16 (applied to a= g) yields that there exists anh ∈ K[[x]] such that g=xh. Consider thish.
Now, let n ∈ N. From g = xh, we obtain gn = (xh)n = xnhn. However, Lemma 3.3.17 (applied to k = n and a = hn) yields that the first n coefficients of the FPSxnhn are 0. In other words, the firstn coefficients of the FPSgn are 0 (sincegn =xnhn). Thus, Proposition 3.5.2(a) is proved.
(b)This follows from part(a). Here are the details.
We must prove that the family (fngn)n∈N is summable. In other words, we must prove that the family figi
i∈Nis summable (since figi
i∈N = (fngn)n∈N).
In other words, we must prove that for eachn ∈N, all but finitely manyi ∈N satisfy[xn] figi
=0 (by the definition of “summable”). So let us prove this.
Fixn ∈N. We must prove that all but finitely manyi ∈ Nsatisfy[xn] figi
= 0.
Indeed, leti ∈Nsatisfyi>n. Then,n <i. Now, the firsticoefficients of the FPS gi are 0 (by Proposition 3.5.2 (a), applied to i instead of n). However, the coefficient[xn] gi
ofgiis one of these firsticoefficients (becausen <i). Thus, this coefficient [xn] gi
must be 0. Now, fi ∈ K; thus, (23) (applied to λ = fi
anda =gi) yields[xn] figi
= fi·[xn]gi
| {z }
=0
=0.
Forget that we fixed i. We thus have shown that all i ∈ N satisfying i > n satisfy[xn] figi
=0. Hence, all but finitely many i ∈Nsatisfy [xn] figi
=0 (because all but finitely many i ∈ N satisfy i > n). This is precisely what we wanted to prove. Thus, Proposition 3.5.2(b)is proved.
(c)Let nbe a positive integer. We shall first show that x0
(fngn) = 0.
Indeed, Proposition 3.5.2 (a) shows that the first n coefficients of the FPS gn are 0. However, the coefficient
x0
(gn) is one of these first n coefficients (since n is positive). Thus, this coefficient
x0
(gn) must be 0. Now, fn ∈ K;
thus, (23) (applied to fn, gn and 0 instead of λ, a and n) yields x0
(fngn) = fn·hx0i
(gn)
| {z }
=0
=0.
Forget that we fixedn. We thus have shown that hx0i
(fngn) = 0 for each positive integern. (45)
Now,
here, we have split off the addend for n=0 from the sum
This proves Proposition 3.5.2(c).
Example 3.5.3. The FPS x+x2 has constant term show that this indeed the case.
In Example 1 in Section 3.1, we had shown that f0+ f1x+ f2x2+ f3x3+· · · = x
Cancelling x from this equality (this is indeed allowed – make sure you un-derstand why!), we obtain
1
1−x−x2 = f1+ f2x+ f3x2+ f4x3+· · · .
However, it appears reasonable to expect that 1
1−x−x2 = 1 1−x
hx+x2i
, (46)
because substitutingx+x2forxin the expression 1
1−x results in 1 1−x−x2. This is plausible but not obvious – after all, we defined 1
1−x
x+x2 to be the result of substituting x+x2 for x into the expanded version of 1
1−x (which is 1+x+x2+x3+· · ·), not into the fractional expression 1
1−x. Nevertheless, (46) is true (and will soon be proved). If we take this fact for granted, then our claim easily follows:
f1+ f2x+f3x2+f4x3+· · · = 1
1−x−x2 = 1 1−x
h
x+x2i
=1+x+x2+x3+· · · hx+x2i (since 1
1−x =1+x+x2+x3+· · ·).
3.5.2. Laws of substitution
The plausible but nontrivial statement (46) that we have just used follows from part(c)of the following proposition:20
Proposition 3.5.4. Composition of FPSs satisfies the rules you would expect it to satisfy:
(a)If f1, f2,g∈ K[[x]]satisfy x0
g =0, then(f1+f2)◦g = f1◦g+ f2◦g.
(b) If f1, f2,g ∈ K[[x]] satisfy x0
g = 0, then (f1· f2)◦ g = (f1◦g)· (f2◦g).
(c) If f1, f2,g ∈ K[[x]] satisfy x0
g = 0, then f1 f2
◦g = f1◦g
f2◦g, as long as f2 is invertible. (In particular, f2◦g is automatically invertible under these assumptions.)
(d)If f,g∈ K[[x]]satisfy x0
g =0, then fk◦g = (f ◦g)k for eachk ∈N.
(e)If f,g,h ∈ K[[x]]satisfy x0
g=0 and x0
h =0, then x0
(g◦h) =0 and (f ◦g)◦h= f ◦(g◦h).
(f)We have a◦g =afor each a∈ K and g ∈K[[x]]. (g)We have x◦g =g◦x= gfor each g∈ K[[x]].
20We are treating the symbol “◦” similarly to the multiplication sign·in our PEMDAS conven-tion. Thus, an expression like “f1◦g+f2◦g” is understood to mean(f1◦g) + (f2◦g).
(h)If(fi)i∈I ∈K[[x]]I is a summable family of FPSs, and ifg∈ K[[x]]is an FPS satisfying
x0
g = 0, then the family (fi◦g)i∈I ∈ K[[x]]I is summable as well and we have
For our proof of Proposition 3.5.4, we will need the following lemma:
Lemma 3.5.5. Let f,g ∈ K[[x]]satisfy x0
g =0. Letk ∈ Nbe such that the first kcoefficients of f are 0. Then, the first k coefficients of f ◦gare 0.
Proof of Lemma 3.5.5. This is very similar to the proof of Proposition 3.5.2(a).
We have x0
g=0. Hence, Lemma 3.3.16 (applied toa =g) yields that there exists anh∈ K[[x]]such that g =xh. Consider thish.
Write the FPS f in the form f = (f0, f1, f2, . . .). Then, the first k coefficients of f are f0, f1, . . . ,fk−1. Hence, these coefficients f0, f1, . . . , fk−1 are 0 (since the firstk coefficients of f are 0). In other words,
fn =0 for eachn <k. (47)
But this ensures that the first k coefficients of f ◦g are 0 21. Thus, Lemma 3.5.5 follows.
Our proof of Proposition 3.5.4 will furthermore use the Kronecker delta nota-tion:
21Proof.We must show that[xm] (f ◦g) =0 for any nonnegative integerm<k. But we can do
Definition 3.5.6. If i and j are any objects, then δi,j means the element (1, if i= j;
0, if i6=j ofK.
For example,δ2,2=1 andδ3,8 =0.
Proof of Proposition 3.5.4. The proof is long and not particularly combinatorial.
I am merely writing it down because it is so rarely explained in the literature.
(a)This is an easy consequence of the definitions, and also appears in [Loehr11, Theorem 7.62] and [Brewer14, Proposition 2.2.2].
Here are the details: Let f1, f2,g ∈ K[[x]] satisfy equal-ities in (48) together, we find
f1+f2 =
∑
this directly: Ifmis a nonnegative integer such thatm<k, then
[xm] (f◦g) = [xm]
exactly as we wanted to show.
Adding these two equalities together, we obtain
(f) This is a near-trivial consequence of the definitions. To wit: Let a ∈ K.
Then,
(g)This is easy, too. Indeed, we have
n
∑
∈NIn other words,x◦g=g (sincex◦g is a synonym for x[g]).
(b)This appears in [Loehr11, Theorem 7.62] and [Brewer14, Proposition 2.2.2].
Here is the proof: Multiplying these two equalities together, we find
f1[g]· f2[g] =
∑
However, we can apply the same computations to x instead of g (since x is syn-onymous to f [g]). This proves Proposition 3.5.4(b).
Did you notice it? I have cheated. The above proof of Proposition 3.5.4(b) relied on some manipulations of infinite sums that need to be justified. Namely, we replaced
“∑
i∈N ∑
j∈N
” by “ ∑
(i,j)∈N2
”. This is an application of the “discrete Fubini rule”, and as we said above, this rule can only be used if we know that the family f1,if2,jgi+j
(i,j)∈N×N
is summable. In other words, we need to show the following statement:
Statement 1: For each m ∈ N, all but finitely many pairs (i,j) ∈ N×N satisfy[xm] f1,if2,jgi+j
=0.
We shall achieve this by proving the following statement:
Statement 2: For any three nonnegative integersm,i,jwith m < i+j, we have[xm] gi+j
=0.
[Proof of Statement 2: Letm,i,jbe three nonnegative integers withm<i+j. We must show that[xm] gi+j
=0.
We have x0
g= 0. Hence, Lemma 3.3.16 (applied toa =g) shows that there exists anh ∈K[[x]]such thatg= xh. Consider this h.
=0. This proves Statement 2.]
[Proof of Statement 1: Letm∈N. If(i,j)∈N×Nis a pair satisfyingm<i+j, then all but finitely many such pairs satisfym<i+j). This proves Statement 1.]
As explained above, Statement 1 shows that the family f1,if2,jgi+j
(i,j)∈N×Nis summable, and thus our interchange of summation signs made above is justified. This completes our proof of Proposition 3.5.4(b).
(c)This follows easily from parts (b) and (f). In detail: Let f1, f2,g ∈ K[[x]]
be such that x0
g = 0. Assume that f2 is invertible. Let us first show that f2◦gis invertible.
f2◦g is therefore well-defined.
It now remains to prove that f1 f2
◦g = f1◦g
f2◦g. To this purpose, we argue as follows: The expression f1
f2
is well-defined, since f2 is invertible. Proposition 3.5.4(b) (applied to f1 divide both sides of this equality by f2◦g (since f2◦g is invertible), and thus obtain f1◦g
(d)Let f,g ∈ K[[x]]satisfy x0
g = 0. We must prove that fk◦g = (f ◦g)k for eachk ∈N.
We prove this by induction onk:
Induction base: We have f0
|{z}=1
◦g =1◦g =1 (by Proposition 3.5.4 (f), applied toa =1). Comparing this with(f ◦g)0 =1, we find f0◦g = (f ◦g)0. In other words, fk◦g = (f ◦g)k holds for k=0.
Induction step: Let m∈ N. Assume that fk◦g= (f ◦g)k holds for k=m. We must prove that fk◦g = (f ◦g)k holds fork =m+1.
We have assumed that fk◦g = (f ◦g)k holds for k = m. In other words, we have fm◦g= (f ◦g)m. Now,
fm+1
| {z }
=f·fm
◦g= (f · fm)◦g = (f ◦g)·(fm◦g)
| {z }
=(f◦g)m
(by Proposition 3.5.4(b), applied to f1 = f and f2 = fm)
= (f ◦g)·(f ◦g)m = (f ◦g)m+1.
In other words, fk◦ g = (f ◦g)k holds for k = m+1. This completes the induction step. Thus, we have proven that fk◦g = (f ◦g)k for each k ∈ N.
Proposition 3.5.4(d)is now proven.
(h)This is just a generalization of Proposition 3.5.4(a)to (potentially) infinite sums. The proof follows the same method, but unfortunately requires some technical reasoning about summability. I will give the full proof for the sake of completeness, but be warned that it contains nothing of interest.
Let(fi)i∈I ∈ K[[x]]I be a summable family of FPSs. Let g ∈ K[[x]]be an FPS satisfying
x0
g =0.
First, we shall prove that the family(fi◦g)i∈I ∈ K[[x]]I is summable. Indeed, we recall that the family(fi)i∈I ∈ K[[x]]I is summable. In other words,
for eachn ∈ N, all but finitely manyi ∈ I satisfy [xn] fi =0
(by the definition of “summable”). In other words, for eachn∈ N, there exists a finite subset In of I such that
all i ∈ I\In satisfy [xn] fi =0. (54) Consider this subset In. Thus, all the sets I0,I1,I2, . . . are finite subsets of I.
Now, letn ∈Nbe arbitrary. The setI0∪I1∪ · · · ∪In is a union ofn+1 finite subsets of I (because all the sets I0,I1,I2, . . . are finite subsets of I), and thus itself is a finite subset of I. Moreover,
alli ∈ I\(I0∪I1∪ · · · ∪In) satisfy [xn] (fi◦g) = 0. (55)
[Proof of (55): Leti∈ I\(I0∪I1∪ · · · ∪In). We must show that[xn] (fi◦g) =0.
Let m ∈ {0, 1, . . . ,n}. Then, Im ⊆ I0∪I1∪ · · · ∪In, so that I0∪I1∪ · · · ∪In ⊇ Im
and thus I\(I0∪I1∪ · · · ∪In)
| {z }
⊇Im
⊆ I \Im. Hence, i ∈ I\(I0∪I1∪ · · · ∪In) ⊆ I \Im. Therefore, (54) (applied tominstead ofn) yields[xm]fi =0.
Forget that we fixedm. We thus have shown that[xm]fi =0 for eachm∈ {0, 1, . . . ,n}. In other words, the firstn+1 coefficients of fiare 0. Hence, Lemma 3.5.5 (applied to fi andn+1 instead of f andk) shows that the firstn+1 coefficients of fi◦gare 0. How-ever, [xn] (fi◦g) is one of these first n+1 coefficients (indeed, it is the last of them);
thus, this coefficient[xn] (fi◦g)must be 0. This proves (55).]
Now, recall that I0∪ I1∪ · · · ∪In is a finite subset of I. Hence, thanks to (55), we know that there exists a finite subset J of I such that all i ∈ I\ J satisfy [xn] (fi◦g) = 0 (namely, J = I0∪ I1∪ · · · ∪In). In other words, all but finitely manyi ∈ I satisfy[xn] (fi◦g) = 0.
Forget that we fixedn. We thus have shown that
for eachn ∈N, all but finitely manyi∈ I satisfy [xn] (fi◦g) =0.
In other words, the family (fi◦g)i∈I ∈ K[[x]]I is summable (by the definition of “summable”).
It now remains to prove that
i∑∈I
fi
◦g = ∑
i∈I
fi◦g.
For eachi∈ I, we write the FPS fi in the form fi = ∑
n∈N fi,nxn with fi,0, fi,1, fi,2, . . .∈ K. First, we shall show that
the family (fi,mgm)(i,m)∈I×N is summable. (56) [Proof of (56): Fix ann∈N. Let J denote the set I0∪I1∪ · · · ∪In. Hence, Jis a union of n+1 finite subsets of I (because all the sets I0,I1,I2, . . . are finite subsets of I), and thus itself is a finite subset of I. The set J× {0, 1, . . . ,n}must be finite (since it is the product of the two finite sets J and{0, 1, . . . ,n}).
Now, let(i,m)∈(I×N)\(J× {0, 1, . . . ,n}). We shall prove that[xn] (fi,mgm) =0.
We have(i,m)∈/ J× {0, 1, . . . ,n}(since(i,m)∈ (I×N)\(J× {0, 1, . . . ,n})).
We note that fi,m ∈ Kand thus [xn] (fi,mgm) = fi,m·[xn] (gm)(by (23)). However, we have
x0
g = 0. Thus, Proposition 3.5.2 (a) (applied to fi and fi,j and minstead of f and fj andn) yields that the firstmcoefficients of the FPSgm are 0. In other words, we
have h
xki
(gm) =0 for eachk∈ {0, 1, . . . ,m−1}. (57) Now, if we haven ∈ {0, 1, . . . ,m−1}, then we have [xn] (gm) = 0 (by (57), applied to k = n) and therefore [xn] (fi,mgm) = fi,m·[xn] (gm)
| {z }
=0
= 0. Hence, [xn] (fi,mgm) = 0 is proved in the case when n ∈ {0, 1, . . . ,m−1}. Thus, for the rest of this proof of [xn] (fi,mgm) = 0, we WLOG assume that n ∈ {/ 0, 1, . . . ,m−1}. Hence, n > m−1 (sincen ∈ N), so that n ≥ m(since n and mare integers). Therefore, m ≤ n, so that
m ∈ {0, 1, . . . ,n} and therefore Im ⊆ I0∪I1∪ · · · ∪In = J (since we defined J to be
Forget that we fixedn. We thus have shown that
for eachn∈N, all but finitely many (i,m)∈ I×Nsatisfy [xn] (fi,mgm) =0.
In other words, the family(fi,mgm)(i,m)∈I×N is summable. This proves (56).]
Now, we have shown that the family(fi,mgm)(i,m)∈I×N is summable. Renam-ing the index(i,m) as(i,n), we thus conclude that the family (fi,ngn)(i,n)∈I×N is summable. The same argument (but with g replaced by x) shows that the family(fi,nxn)(i,n)∈I×N is summable (since the FPS x satisfies summing the equalities fi = ∑
n∈N fi,nxn over all i∈ I, we obtain
(here, we have been able to interchange the summation signs, since the family (fi,nxn)(i,n)∈I×N is summable). Thus, Hence, Definition 3.5.1 (applied to f = ∑
i∈I
On the other hand, for eachi ∈ I, we have fi[g] = ∑
(again, we have been able to interchange the summation signs, since the family (fi,ngn)(i,n)∈I×N is summable). Thus,
Comparing this with (58), we find Proposition 3.5.4(h)is proven.
(e)This is [Loehr11, Theorem 7.63] and [Brewer14, Proposition 2.2.5].22 Again, let us give the proof:
Write the FPS g in the form g = ∑
22See also [19s, Proposition 7.6.14] for a similar property for polynomials.
In view of f ◦g= f [g] = ∑
n∈Nfngn, we can rewrite (59) as (f ◦g)◦h=
∑
n∈N
(fngn)◦h. (60) However, for eachn ∈N, we have fngn = fn·gn (by Theorem 3.2.6(d), applied toλ= fn and a=gn) and thus
(fngn)◦h =fn·gn
◦h= fn◦h
| {z }
=fn
(by Proposition 3.5.4(f), applied to fnandh
instead ofaandg)
· (gn◦h)
| {z }
=(g◦h)n (by Proposition 3.5.4(d),
applied togandh instead of fandg)
by Proposition 3.5.4(b),
applied to fn, gn and hinstead of f1, f2 and g
!
= fn·(g◦h)n = fn·(g◦h)n (61) (since Theorem 3.2.6 (d) (applied to λ = fn and a = (g◦h)n) yields fn · (g◦h)n = fn·(g◦h)n). Thus, (60) becomes
(f ◦g)◦h=
∑
n∈N
(fngn)◦h
| {z }
=fn·(g◦h)n (by (61))
=
∑
n∈N
fn·(g◦h)n
= f [g◦h] since f[g◦h] =
∑
n∈N
fn·(g◦h)n
!
= f ◦(g◦h).
This completes the proof of Proposition 3.5.4(e).
Example 3.5.7. Let us use Proposition 3.5.4(c)to justify the equality (46) that we used in Example 3.5.3. Indeed, we know that the FPS 1−x is invertible.
Thus, applying Proposition 3.5.4(c)to f1 =1 and f2 =1−xand g =x+x2, we obtain
1
1−x◦x+x2
= 1◦ x+x2 (1−x)◦(x+x2). Using the notation f[g] instead of f ◦g, we can rewrite this as
1 1−x
hx+x2i
= 1
x+x2 (1−x) [x+x2]. In view of 1
x+x2
= 1 and (1−x)x+x2
= 1− x+x2
= 1−x−x2, this rewrites as 1
1−x
x+x2
= 1
1−x−x2. Thus, (46) is proved.
Let us summarize: If f,g ∈ K[[x]]are two FPSs, then the composition f [g]is not always defined. However, it is defined at least in the following two cases:
• in the case when f is a polynomial (that is, f ∈ K[x]), and
• in the case when ghas constant term 0 (that is, x0
g =0).
This justifies some more of the things we did back in Section 3.1; in particular, Example 1 from that section is now fully justified. But we still have not defined (e.g.) the square root of an FPS, which we used in Example 2.
Before I explain square roots, let me quickly survey differentiation of FPSs.