Convention 3.7.1. Throughout Subsection 3.7, we assume thatKis not just a commutative ring, but actually a commutativeQ-algebra.
This entails that elements ofKcan be divided by the positive integers 1, 2, 3, . . ..
We can use this to define three specific (and particularly important) FPSs over K:
Definition 3.7.2. Define three FPS exp, log and exp inK[[x]] by exp :=
∑
n∈N
1 n!xn, log :=
∑
n≥1
(−1)n−1 n xn, exp :=exp−1=
∑
n≥1
1 n!xn.
(The last equality sign here follows from exp = ∑
n∈N
1
n!xn = 1
0!
|{z}
=1
x0
|{z}
=1
+ ∑
n≥1
1
n!xn =1+ ∑
n≥1
1 n!xn.)
Note that the FPS exp is the usual exponential series from analysis, but now manifesting itself as a FPS. Likewise, log is the Mercator series for log(1+x), where log stands for the natural logarithm function. The natural logarithm function itself cannot be interpreted as an FPS, since log 0 is undefined.
I will prove that
exp◦log=log◦exp=x. (63)
This is an algebraic analogue of the well-known fact from analysis which states that the exponential and logarithm functions are mutually inverse.
There is a short way of proving (63), which I will not take: Namely, one can show that any equality between holomorphic functions on an open disk around the origin leads to an equality between their Taylor series (viewed as FPSs). Thus, if you have proved in complex analysis that log◦exp = id on an open disk around 0 and exp◦log =id on an open disk around 1, then you automatically get (63) (indeed, exp and log are the Taylor series of the functions exp and log around 1, with the reservation that the point 1 has been moved to the origin by a shift). This approach uses nontrivial results from complex analysis, so I will not follow it and instead start from scratch.
The main tool in the proof of (63) will be the following useful proposition ([Grinbe17, Lemma 0.4]):
Proposition 3.7.3. Let g∈ K[[x]]with x0
g=0. Then:
(a)We have
(exp◦g)0 = (exp◦g)0= (exp◦g)·g0. (b)We have
log◦g0
= (1+g)−1·g0.
Proof of Proposition 3.7.3. (a)Let us first show that exp0 = exp0 = exp. Indeed,
n!xn. Hence, the definition of a derivative yields exp0 =
∑
Comparing this with (64), we find
exp0 =exp . (66)
Now, we can apply the chain rule (Theorem 3.6.2 (g)) to f = exp (since x0 g0. Thus, we have proved Proposition 3.7.3 (a).
(b)We have log= ∑ applied toxinstead ofg)
(by Theorem 3.6.2(b))
(here, we have substitutednforn−1 in the sum). On the other hand,
Now, we can apply the chain rule (Theorem 3.6.2 (g)) to f = log (since x0
Thus, we have proved Proposition 3.7.3(b).
We will need a very simple lemma, which says (in particular) that if two FPSs have constant terms 0, then so does their composition:
Lemma 3.7.4. Let f,g ∈ K[[x]] be two FPSs with Proposition 3.5.2 (c) yields
x0
Now, we can prove the equalities we promised ([Grinbe17, Theorem 0.1]):
Theorem 3.7.5. We have
exp◦log=x and log◦exp=x.
Proof of Theorem 3.7.5. Let us first prove that log◦exp=x.
Indeed, the idea of this proof is to show that log◦exp and x are two FPSs with the same constant term (namely, 0) and with the same derivative. Once this is proved, Theorem 3.6.2(h)will easily yield that they are equal.
Let us fill in the details. We have x0 (applied to g=exp) yields
log◦exp0
Now it remains to prove that exp◦log = x. There are (at least) two ways to do this:
• 1st way: A homework exercise (Exercise A.2.5.1) says that any FPS f with x0
f = 0 and with x1
f invertible has a unique compositional inverse (i.e., there is a unique FPS g with invertible), and thus see that log has a unique compositional inverse g.
This compositional inverse g must be exp, since log◦exp = x (indeed, comparing
g◦log
| {z }
=x
◦exp=x◦exp=exp with
g◦log
◦exp=g◦log◦exp
| {z }
=x
(by Proposition 3.5.4(e))
=g◦x=g
yieldsg=exp). But this entails that exp◦log=x as well.
• 2nd way:Here is a more direct argument. We shall first show that exp◦log= 1+x.
To wit: The FPS 1+x is invertible (by Proposition 3.3.9). Thus, applying the quotient rule (Theorem 3.6.2 (e)) to f = exp◦log and g = 1+x, we obtain
exp◦log 1+x
!0
=
exp◦log0
·(1+x)−exp◦log
·(1+x)0
(1+x)2 .
In view of exp◦log0
=exp◦log
· log0
|{z}
=(1+x)−1 (by (67))
by Proposition 3.7.3(a), applied to g=log
=exp◦log·(1+x)−1 and(1+x)0 =1, we can rewrite this as
exp◦log 1+x
!0
=
exp◦log
·(1+x)−1·(1+x)−exp◦log
·1 (1+x)2
=
exp◦log
−exp◦log
(1+x)2 =0=00. Thus, Theorem 3.6.2(h) (applied to f = exp◦log
1+x and g = 0) yields that exp◦log
1+x −0 is constant. In other words, exp◦log
1+x is constant. In other
words, exp◦log
1+x =afor some a∈ K. Consider thisa. From exp◦log 1+x =a, we obtain exp◦log =a(1+x) = a(1+x). Thus,
hx0i
exp◦log
=hx0i
(a(1+x)) =a.
However, it is easy to see that x0
exp◦log
=1 23. Comparing these two equalities, we finda =1. Thus, exp◦log = a
|{z}
=1
(1+x) =1+x.
Now, exp=exp−1=exp+−1. Hence, exp◦log
= (exp+−1)◦log=exp◦log
| {z }
=1+x
+ −1◦log
| {z }
=−1
(by Proposition 3.5.4(f),
applied to−1 and log instead ofaandg)
by Proposition 3.5.4(a),
applied to f1 =exp and f2 =−1 andg =log
!
=1+x+−1= (1+−1)
| {z }
=0
+x =x.
Either way, we have shown that exp◦log = x. Thus, the proof of Theorem 3.7.5 is complete.
In Definition 3.7.2, we have found algebraic versions of the exponential and logarithm functions as FPSs. Next, we shall define analogues of these functions as operators acting on FPSs (i.e., analogues not of the functions exp and log themselves, but rather of composition with these functions):
Definition 3.7.6. (a)We let K[[x]]0denote the set of all FPSs f ∈ K[[x]]with x0
f =0.
(b)We letK[[x]]1 denote the set of all FPSs f ∈ K[[x]] with x0
f =1.
(c)We define two maps
Exp :K[[x]]0 →K[[x]]1, g 7→exp◦g
23Proof.Recall that x0
log=0. Hence, Lemma 3.7.4 (applied to f =exp and g=log) yields hx0i
exp◦log
=hx0i
exp= 1
0! since exp=
∑
n∈N
1 n!xn
!
= 1 1 =1.
and
Log : K[[x]]1 →K[[x]]0, f 7→ log◦(f −1).
(These two maps are well-defined according to parts (c) and (d) of Lemma 3.7.7 below.)
The maps Exp and Log are algebraic analogues of the maps in complex anal-ysis that take any holomorphic function f to its exponential and logarithm, respectively (at least within certain regions in which these things are well-defined). As one would hope, and as we will soon see, they are mutually inverse. Let us first check that their definition is justified:
Lemma 3.7.7. (a)For any f,g∈ K[[x]]0, we have f ◦g∈ K[[x]]0. (b)For any f ∈ K[[x]]1 and g∈ K[[x]]0, we have f ◦g∈ K[[x]]1. (c)For any g ∈ K[[x]]0, we have exp◦g∈ K[[x]]1.
(d)For any f ∈ K[[x]]1, we have f−1∈ K[[x]]0and log◦(f −1)∈ K[[x]]0. Proof of Lemma 3.7.7. (a) Let f,g ∈ K[[x]]0. In view of the definition of K[[x]]0, this entails that
x0
f =0 and x0
g=0. Hence, Lemma 3.7.4 yields x0
(f ◦g) = x0
f = 0. In other words, f ◦g ∈ K[[x]]0 (by the definition of K[[x]]0). This proves Lemma 3.7.7(a).
(b)This is analogous to the proof of Lemma 3.7.7(a).
(c) Let g ∈ K[[x]]0. From exp = ∑
n∈N
1
n!xn, we obtain x0
exp = 1 0! = 1, so that exp ∈ K[[x]]1. Hence, Lemma 3.7.7 (b) (applied to f = exp) yields exp◦g∈ K[[x]]1. This proves Lemma 3.7.7 (c).
(d) Let f ∈ K[[x]]1. Thus, x0
f = 1. Now, (19) yields x0
(f −1) = hx0i
f
| {z }
=1
−hx0i 1
| {z }
=1
=1−1= 0, so that f −1 ∈ K[[x]]0. Furthermore, x0
log =0
(since log = ∑
n≥1
(−1)n−1
n xn) and thus log ∈ K[[x]]0. Hence, Lemma 3.7.7 (a) (applied to log and f −1 instead of f and g) yields log◦(f −1) ∈ K[[x]]0. Thus, Lemma 3.7.7(d)is proven.
Lemma 3.7.8. The maps Exp and Log are mutually inverse bijections between K[[x]]0 and K[[x]]1.
Proof of Lemma 3.7.8. First, we make a simple auxiliary observation: Each g ∈ K[[x]]0satisfies24
exp◦g =exp◦g+1. (70)
24As before, the “◦” operation behaves like multiplication in the sense of PEMDAS conventions.
Thus, the expression “exp◦g+1” means(exp◦g) +1.
[Proof of (70): Let g ∈ K[[x]]0. Recall that exp = exp−1, so that exp = exp+1=exp+1. Hence,
exp◦g= (exp+1)◦g=exp◦g+1◦g
(by Proposition 3.5.4(a), applied to f1 =exp and f2 =1). However, Proposition 3.5.4 (f) (applied to a = 1) yields 1◦ g = 1 = 1. Hence, exp◦g = exp◦g+ 1◦g
| {z }
=1
=exp◦g+1. This proves (70).]
Now, let us show that Exp◦Log=id. Indeed, we fix some f ∈ K[[x]]1. Then, f −1 ∈ K[[x]]0 (by Lemma 3.7.7 (d)). Hence, Proposition 3.5.4 (e) (applied to exp, log and f −1 instead of f, g and h) yields
exp◦log
◦(f −1) = exp◦log◦(f −1). Thus,
exp◦log◦(f −1)= exp◦log
| {z }
=x (by Theorem 3.7.5)
◦(f −1) = x◦(f −1)
= f −1 (71)
(by Proposition 3.5.4(g), applied tog = f −1). However, (Exp◦Log) (f) =Exp(Logf)
=exp◦(Log f) (by the definition of Exp)
=exp◦ (Log f)
| {z }
=log◦(f−1) (by the definition of Log )
+1
(by (70), applied to g=Logf)
=exp◦log◦(f −1)
| {z }
=f−1 (by (71))
+1
= (f −1) +1= f =id(f).
Forget that we fixed f. We thus have shown that(Exp◦Log) (f) =id(f) for each f ∈ K[[x]]1. In other words, Exp◦Log=id.
Using a similar argument, we can show that Log◦Exp = id. Indeed, let us fix some g ∈ K[[x]]0. Hence, Proposition 3.5.4 (e) (applied to log, exp and g instead of f, gand h) yields
log◦exp
◦g =log◦(exp◦g). Thus, log◦(exp◦g) = log◦exp
| {z }
=x (by Theorem 3.7.5)
◦g= x◦g
= g (72)
(by Proposition 3.5.4 (g)). But the definition of Exp yields Expg = exp◦g = exp◦g+1 (by (70)). Hence, Expg−1=exp◦g. Now,
(Log◦Exp) (g) =Log(Expg)
=log◦(Expg−1) (by the definition of Log)
=log◦(exp◦g) (since Expg−1=exp◦g)
= g (by (72))
=id(g).
Forget that we fixed g. We thus have shown that (Log◦Exp) (g) = id(g) for each g ∈ K[[x]]0. In other words, Log◦Exp = id. Combining this with Exp◦Log = id, we see that the maps Exp and Log are mutually inverse bijec-tions betweenK[[x]]0 and K[[x]]1. This proves Lemma 3.7.8.
We will now prove another lemma, which says that the Exp and Log maps deserve their names:
Lemma 3.7.9. (a)For any f,g∈ K[[x]]0, we have Exp(f +g) = Expf ·Expg.
(b)For any f,g∈ K[[x]]1, we have
Log(f g) =Logf +Logg.
Proof of Lemma 3.7.9 (sketched). (a)Like many of our arguments involving FPSs, this will be a short computation followed by lengthy technical arguments justi-fying the interchanges of summation signs. (In this aspect, our algebraic replica of the analysis of infinite sums doesn’t differ that much from the original.) We begin with the computation; the justifying arguments will be sketched after-wards.
Let f,g ∈ K[[x]]0. Thus, x0
f = 0 and x0
g = 0. Hence, f +g ∈ K[[x]]0 (since (18) yields
x0
(f +g) =hx0i f
| {z }
=0
+hx0i g
| {z }
=0
=0).
By the definition of Exp, we have
Expf =exp◦f =exp[f] =
∑
n∈N
1 n!fn
(by Definition 3.5.1, since exp= ∑
n∈N
1
n!xn). Similarly, Expg=
∑
n∈N
1 n!gn
and
Exp(f +g) =
∑
n∈N
1
n!(f +g)n. Now, the latter equality becomes
Exp(f +g) =
∑
This is sufficient to prove Lemma 3.7.9(a)if we can justify the above manipulations of infinite sums. Actually, there is just one manipulation that we need to justify, and that is our replacement of “ ∑
n∈N
”. This is an application of the
“discrete Fubini rule” (specifically, of a version thereof in which the summation is over all pairs (n,k) ∈ N×N satisfying k ≤ n). In order to justify this manipulation, we need to show that the family
1
k!(n−k)!fkgn−k
(n,k)∈N×Nsatisfyingk≤n
is summable. In other words, we need to show the following statement:
Statement 1: For each m ∈ N, all but finitely many pairs (n,k) ∈ N×N
We shall achieve this by proving the following statement:
Statement 2: For any three nonnegative integersm,k,`with m< k+`, we have[xm] fkg`
=0.
[Proof of Statement 2: Let m,k,` be three nonnegative integers with m < k+`. We must show that[xm] fkg`
=0.
We have x0
f =0. Hence, Lemma 3.3.16 (applied toa= f) shows that there exists anh ∈ K[[x]] such that f = xh. Consider thish and denote it by u. Thus, u ∈ K[[x]]
and f = xu.
We have x0
g= 0. Hence, Lemma 3.3.16 (applied toa =g) shows that there exists anh ∈ K[[x]]such that g = xh. Consider this h and denote it by v. Thus,v ∈ K[[x]]
andg=xv.
Now, from f =xuandg= xv, we obtain fkg` = (xu)k(xv)` =xkukx`v` = xk+`ukv`. However, Lemma 3.3.17 (applied tok+` and ukv` instead ofk and a) shows that the firstk+`coefficients of the FPSxk+`ukv`are 0. In other words, the firstk+`coefficients of the FPS fkg` are 0 (since fkg` = xk+`ukv`). But[xm] fkg`
is one of these firstk+` coefficients (since m < k+`). Thus, we conclude that [xm] fkg`
= 0. This proves Statement 2.]
Note that Statement 2 entails that the family fkg`
(k,`)∈N×N is summable (because whenm ∈ N is given, all but finitely many pairs (k,`) ∈ N×N satisfym < k+`).
However, we need to prove Statement 1, so let us do this:
[Proof of Statement 1: Let m ∈ N. If (n,k) ∈ N×N is a pair satisfying k ≤ n and m<n, then
[xm]
1
k!(n−k)!fkgn−k
= 1
k!(n−k)! [xm]fkgn−k
| {z }
=0 (by Statement 2 (applied to`=n−k), sincem<n=k+(n−k))
(by (23))
=0.
Thus, all but finitely many pairs(n,k)∈N×Nsatisfyingk≤ nsatisfy [xm]
1
k!(n−k)!fkgn−k
=0 (because all but finitely many such pairs satisfy m< n).
This proves Statement 1.]
As explained above, Statement 1 shows that the family 1
k!(n−k)!fkgn−k
(n,k)∈N×Nsatisfyingk≤n
is summable, and thus our interchange of summation signs made above is justified. This completes our proof of Lemma 3.7.9 (a).
(b) This easily follows from part (a), since we know that Log is inverse to Exp. Here are the details:
Let f,g ∈ K[[x]]1. Set u = Logf and v = Logg; then, u,v ∈ K[[x]]0 (since Log is a map fromK[[x]]1toK[[x]]0). Hence, Lemma 3.7.9(a)(applied touand vinstead of f and g) yields Exp(u+v) = Expu·Expv.
However, Lemma 3.7.8 says that the maps Exp and Log are mutually inverse
We can neatly pack the last few lemmas into a single theorem through the use of group isomorphisms. To this purpose, we need to observe thatK[[x]]0is a group under addition andK[[x]]1is a group under multiplication:
Proposition 3.7.10. (a) The subset K[[x]]0 of K[[x]] is closed under addition and subtraction and contains 0, and thus forms a group(K[[x]]0,+, 0).
(b)The subsetK[[x]]1ofK[[x]]is closed under multiplication and division and contains 1, and thus forms a group (K[[x]]1,·, 1).
Proof of Proposition 3.7.10. (a) It is clear that the set K[[x]]0 contains the FPS 0 (since
Thus,K[[x]]0is closed under addition and subtraction. This proves Proposition 3.7.10(a).
(b) Any a ∈ K[[x]]1 is invertible in K[[x]] (indeed, a ∈ K[[x]]1 shows that x0
a =1; thus, x0
ais invertible inK; therefore, Proposition 3.3.7 entails that a is invertible inK[[x]]). Hence, f
have f = f
g ·gand thus h
x0i
f =hx0i f g ·g
=hx0i f
g ·hx0i g
| {z }
=1
(by (22))
=hx0i f g
and thus x0 f
g = x0
f = 1, so that f
g ∈ K[[x]]1). This shows that K[[x]]1 is closed under division. Thus, Proposition 3.7.10(b)is proven.
The two groups in Proposition 3.7.10 can now be connected through Exp and Log:
Theorem 3.7.11. The maps
Exp :(K[[x]]0,+, 0)→(K[[x]]1,·, 1) and
Log : (K[[x]]1,·, 1) →(K[[x]]0,+, 0) are mutually inverse group isomorphisms.
Proof of Theorem 3.7.11 (sketched). Lemma 3.7.9 yields that these two maps are group homomorphisms25. Lemma 3.7.8 shows that they are mutually inverse.
Combining these results, we conclude that these two maps are mutually inverse group isomorphisms. This proves Theorem 3.7.11.
Theorem 3.7.11 helps us turn addition into multiplication and vice versa when it comes to FPSs, at least if the constant terms are the right ones. This will come useful rather soon.