Let $F$ be a formal Dirichlet series $F(s) = \sum_{n=1}^\infty a_n n^{-s}$ and $1/F(s)$ its multiplicative inverse. The formal logarithm of $F$ is defined as $$\ln F(s) = \ln(a_1) + \sum_{n=2}^\infty \frac{[n^{-s}] (F'(s) / F(s))}{-\ln n} n^{-s}$$ with $$F'(s) = \sum_{n=1}^\infty -\ln n \cdot a_n n^{-s}$$ so essentially, $\ln F(s)$ is the formal antiderivative of $F'(s) / F(s)$, which makes sense.

Apparently, it is the case that if the logarithm of a Dirichlet series has non-negative real coefficients, the original series also has non-negative real coefficients (cf. e.g. Exponentiation of a Dirichlet series).

The SE question I linked uses an analytic argument to show that, but I am wondering whether there is also a nice and simple formal argument that does not take the detour of considering the actual function $F(s)$ in the complex plane. I somehow feel there should be one, but I cannot see it.

  • $\begingroup$ Similarly to Daniel's answer, if $a_1 = 1$ then $\log F(s) = \log (1-(1-F(s))) = - \sum_{k=1}^\infty \frac{(1-F(s))^k}{k}$ and everything is well-defined as formal Dirichlet series. As analytic series, we need to find $\sigma$ such that $|1-F(s)| < 1$ for $\Re(s) > \sigma$, then the Dirichlet series for $\log F(s)$ converges for $\Re(s) > \sigma$. $\endgroup$ – reuns Dec 7 '17 at 0:32
  • $\begingroup$ Sorry, but how does that help me with showing that $F$ has non-negative coefficients? $\endgroup$ – Manuel Eberl Dec 7 '17 at 18:27
  • $\begingroup$ ?? What do you not understand in this : $G(s) = \sum_{n=1}^\infty a_n n^{-s}, G(s)^k = \sum_{n=1}^\infty a_n(k) n^{-s}$ where $a_n(k+1) = \sum_{d | n} a_d a_{n/d}(k) $ (prove it !). If the $a_n$ are non-negative then so are the $a_n(k)$ and $\exp(G(s)) = \sum_{k=0}^\infty \frac{G(s)^k}{k!}=\sum_{k=0}^\infty \frac{1}{k!}\sum_{n=1}^\infty a_n(k) n^{-s}$ has non-negative coeffecients. $\endgroup$ – reuns Dec 7 '17 at 18:31
  • $\begingroup$ Okay, it is clear to me that $\exp(G(s))$, as you define it, has non-negative coefficients. What is not clear to me is how what you wrote shows that $\exp (\log F(s)) = F(s)$. If you want to use an analytic argument, I think that at the very least, you will require interchanging of the two infinite sums. In case of non-negative coefficients, that should be unproblematic, but I'm wondering whether there is some general way to show $\log (\exp F(s)) = F(s)$ for any Dirichlet series $F(s)$. $\endgroup$ – Manuel Eberl Dec 7 '17 at 21:36
  • $\begingroup$ The coefficient of $N^{-s}$ in $F(s)$ only affects the coefficients $n^{-s}, n \ge N$ in $F(s)^k,(1-F(s))^k, \log F(s), \exp(\log(F(s))$.. Thus it suffices to prove it for $F(s) = \sum_{n=1}^N a_n n^{-s}$ and to let $N \to \infty$. That's the point of all the formal series theory. $\endgroup$ – reuns Dec 7 '17 at 21:39

The formal calculation is essentially identical to the analytic argument in the convergent case. If the formal logarithm of $F$ is

$$\ln a_1 + \underbrace{\sum_{n = 2}^{\infty} \frac{b_n}{n^s}}_{Q(s)}$$

we can look at the formal Dirichlet series

$$a_1\cdot \exp Q(s) = a_1 \Biggl( 1 + \sum_{k = 1}^{\infty} \frac{1}{k!} Q(s)^k\Biggr).$$

Since the coefficient of $n^{-s}$ in $Q(s)^k$ is zero for $n < 2^k$, each coefficient in the resulting formal series is determined by a finite partial sum, so no convergence issue arises. And if the coefficients of $Q(s)$ are non-negative, so are the coefficients of $Q(s)^k$ and hence of the formal exponential.

It remains to see that the formal exponential is the inverse of the formal logarithm. That can be done by explicit formal calculations, but I find it easier to use the analytic identity. To determine the coefficient of $n^{-s}$ in the formal exponential, we only need the coefficients of $m^{-s}$ for $m \leqslant n$ in the formal logarithm. To determine these coefficients, we only need the coefficients of $m^{-s}$ for $m \leqslant n$ in the formal derivative $F'$ and the formal inverse $1/F$. But all of these are determined by the coefficients of $m^{-s}$ for $m \leqslant n$ in $F$.

Thus truncating the series of $F$ after $a_n n^{-s}$, computing the logarithm of the - now trivially convergent - truncated series, and finally the exponential of that logarithm leads back to the original truncated series by the analytic identities. Since the formal computation agrees with this analytic one for all coefficients of $m^{-s}$ for $m \leqslant n$, the formal exponential of the formal logarithm agrees with the original formal series up to and including the coefficient of $n^{-s}$. But $n$ was arbitrary, so we have the equality of the full formal series.

We can identify a Dirichlet series - formal or convergent - with its coefficient sequence. Thus the space of all formal Dirichlet series can be identified with the sequence space $S = \mathbb{C}^{\mathbb{N}\setminus \{0\}}$. On $S$, we consider its natural $\mathbb{C}$-vector space structure, and two internal multiplications,

  1. the pointwise muliplication, given by $(f \odot g)(n) = f(n)\cdot g(n)$, and
  2. the Dirichlet convolution, given by $$(f \ast g)(n) = \sum_{k \mid n} f(k)g\biggl(\frac{n}{k}\biggr)\,.$$

Both internal multiplications are $\mathbb{C}$-bilinear, commutative and associative. This is straightforwardly verified using the corresponding properties of the multiplication in $\mathbb{C}$ and divisibility properties in $\mathbb{N}\setminus \{0\}$. Since we have two internal multiplications, the power notation $f^k$ is a priori ambiguous, we shall use it for the power with respect to the Dirichlet convolution, so $f^2 = f\ast f$ etc., and if $f(1) \neq 0$, then $f^{-1}$ denotes the Dirichlet inverse of $f$.

We denote the multiplicative unit of the Dirichlet convolution by $\varepsilon$, and introduce some linear operators on $S$:

  • the trunction operators $T_n$, given by $(T_nf)(k) = f(k)$ for $k \leqslant n$ and $(T_nf)(k) = 0$ for $k > n$,
  • the tail operator, given by $Zf = f - T_1f$,
  • the formal differentiation, given by $(Df)(n) = -\ln n\cdot f(n)$, or $Df = (-\ln) \odot f$,
  • for $f\in S$ with $f(1) = 0$, the formal integral $If = \bigl(-\frac{1}{\ln}\bigr)\odot f$, where we use the convention $0\cdot \text{undefined} = 0$ for $If(1)$.

We note the following easily verified properties:

  • $D$ and $I$ commute with truncations, $DT_nf = T_nDf$ and $IT_n f = T_n If$ whenever $If$ is defined (more generally, $g\odot T_n f = T_n (g \odot f)$ for all $f,g\in S$),
  • $T_n (f \ast g) = T_n (T_n f \ast T_n g)$ for all $f, g \in S$, in particular $T_n (f^k) = T_n \bigl((T_ n f)^k\bigr)$,
  • $DIf = f$, whenever $If$ is defined,
  • $IDf = Zf$ for every $f\in S$,
  • $Df = 0 \iff Zf = 0$, i.e. a formal Dirichlet series has zero derivative if and only if it is constant,
  • $D$ satisfies the product rule (i.e., it is a derivation) with respect to Dirichlet convolution, $D(f\ast g) = (Df) \ast g + f \ast (Dg)$, and by induction the power rule $D(f^k) = k f^{k-1} \ast Df$ for $k \in \mathbb{N}$ (for $k \in \mathbb{Z}$ if $f$ has a Dirichlet inverse).

Now we introduce a topology on $S$ that allows us to work with infinite constructions. For every $f \in S$, a neighbourhood basis of $f$ is given by the family $\mathscr{U}(f) = \{ U_n(f) : n \in \mathbb{N}\}$, where

$$U_n(f) = \{ g \in S : T_n g = T_n f\}.$$

This topology is induced by the (ultra)metric

$$d(f,g) = \inf\: \bigl\{ 2^{-n} : T_n f = T_n g\},$$

it is the product topology on $\mathbb{C}^{\mathbb{N}\setminus \{0\}}$ if we endow $\mathbb{C}$ with the discrete topology instead of the standard topology.

A sequence $(f_m)$ in $S$ converges in this topology if and only if for every $k\in \mathbb{N}\setminus \{0\}$ there is an $M_k$ such that $f_m(k) = f_n(k)$ for all $m, n \geqslant M_k$. The feature we need is that a series $\sum_m f_m$ converges in $S$ if and only if $f_m \to 0$.

We note that the operators $T_n, D, I$, and the Dirichlet convolution are continuous in this topology, consequently a convergent series can always be differentiated (and integrated, when the constant term vanishes) term by term. Also, $$\lim_{n \to \infty} T_n f = f\,.$$ The continuity of the Dirichlet convolution yields another representation of the Dirichlet inverse, as a geometric series. If $f(1) \neq 0$, we can write

$$f = f(1)\cdot \bigl(\varepsilon + f(1)^{-1}Zf\bigr),$$

and since $(Zf)^k \to 0$ - $T_n\bigl((Zf)^k\bigr) = 0$ for $n < 2^k$ - the series

$$Rf = \sum_{k = 0}^{\infty} (-1)^k\frac{(Zf)^k}{f(1)^{k+1}}$$

converges. Then

\begin{align} f \ast Rf &= \bigl(f(1)\varepsilon + Zf\bigr) \ast Rf \\ &= f(1)Rf + Zf \ast Rf \\ &= \sum_{k = 0}^{\infty} \frac{(-1)^k(Zf)^k}{f(1)^k} + \sum_{k = 0}^{\infty} \frac{(-1)^k (Zf)^{k+1}}{f(1)^{k+1}} \\ &= \frac{(-1)^0(Zf)^0}{f(1)^0} \\ &= \varepsilon \end{align}

shows that $Rf = f^{-1}$.

Now we introduce the formal logarithm and exponential. For the formal logarithm, we require $f(1) \neq 0$. Then we define

$$Lf = (\ln f(1))\varepsilon + \sum_{k = 1}^{\infty} \frac{(-1)^{k-1}}{k}\biggl(\frac{Zf}{f(1)}\biggr)^k$$

using the familiar Taylor expansion of $\ln (1+z)$. The formal exponential is defined as

$$Ef = e^{f(1)}\cdot \sum_{k = 0}^{\infty} \frac{(Zf)^k}{k!}.$$

We cannot - in this setting - define the exponential as

$$\sum_{k = 0}^{\infty} \frac{f^k}{k!}$$

since that series doesn't converge in $S$ if $f(1) \neq 0$. To obtain a convergent series, we must separate the constant term from the rest. This is similar to determining the Maclaurin expansion of the composition of (sufficiently smooth) functions. If $f(0) \neq 0$, we must use the Taylor expansion of $g$ about $f(0)$, not the Maclaurin expansion of $g$, to find the Maclaurin expansion of $g\circ f$.

To see that $E$ is a left inverse of $L$, we note the identity

$$\exp(\ln (1+X)) = 1 + X$$

of formal power series. This follows from the identity $\exp(\ln (1+z)) \equiv 1 + z$ for $\lvert z\rvert < 1$ and the identity theorem for convergent power series. Since this is a formal identity, we can substitute an arbitrary element of a $\mathbb{C}$-algebra for the indeterminate $X$ for which the intermediate series make sense. This is the case for elements of $S$ without constant term, hence

$$\exp\bigl(L \bigl(\varepsilon + f(1)^{-1} Zf\bigr)\bigr) = \varepsilon + f(1)^{-1} Z f$$

and consequently

$$E(Lf) = e^{\ln f(1)}\bigl( \varepsilon + f(1)^{-1} Zf\bigr) = f(1)\varepsilon + Zf = f$$

for all $f\in S$ with $f(1) \neq 0$.

Since $g^k$ is nonnegative if $g$ is nonnegative, it is evident that $Ef$ is nonnegative if $Zf$ is nonnegative and $f(1)$ real.

However, that does not quite settle the question because you used a different definition for the formal logarithm. We need to see that these two definitions yield the same formal Dirichlet series. Your definition was

$$\tilde{L} f = \bigl(\ln f(1)\bigr)\varepsilon + I\bigl((Df)\ast f^{-1}\bigr)\,.$$

Using $D f = D(Zf)$ and $f^{-1} = Rf$, we compute

\begin{align} (Df) \ast f^{-1} &= D(Zf) \ast Rf \\ &= D(Zf) \ast \sum_{k = 0}^{\infty} (-1)^k \frac{(Zf)^k}{f(1)^{k+1}} \\ &= \sum_{k = 0}^{\infty} (-1)^k \frac{(Zf)^k \ast D (Zf)}{f(1)^{k+1}} \\ &= \sum_{k = 1}^{\infty} (-1)^{k-1} \frac{(Zf)^{k-1} D(Zf)}{f(1)^k} \\ &= \sum_{k = 1}^{\infty} \frac{(-1)^{k-1}}{k} \cdot \frac{D\bigl((Z f)^k\bigr)}{f(1)^k} \\ &= D\sum_{k = 1}^{\infty} \frac{(-1)^{k-1}}{k} \cdot \frac{(Zf)^k}{f(1)^k} \\ &= D\bigl(L f - (\ln f(1))\varepsilon\bigr) \\ &= D(L f) \end{align}

and hence see that

$$\tilde{L} f = (\ln f(1))\varepsilon + ID(Lf) = (\ln f(1))\varepsilon + Z(Lf) = Lf\,,$$

thus completing the proof that a formal Dirichlet series whose (formal) logarithm has nonnegative coefficients itself has nonnegative coefficients.

  • $\begingroup$ Thanks for that answer, but I'm still struggling a lot with understanding how this actually works. I first tried to prove $(\exp(F))' = \exp(F) F'$ purely formally by computing the coefficients, because I thought I could use that to show $\exp(\ln F) = F$ via $(\exp(-\ln F)F)' = 0$, but I failed miserably (I ended up with two double sums that didn't match up somehow). I then tried to understand your analytic argument, but I'm still not quite sure how these truncations work. Could you please go into some more detail? $\endgroup$ – Manuel Eberl Dec 7 '17 at 18:26
  • $\begingroup$ Regarding the truncations, let us define the truncation operator $T_n$ on formal Dirichlet series by $$T_n\Biggl(\sum_{k = 1}^{\infty} \frac{a_k}{k^s}\Biggr) = \sum_{k = 1}^n \frac{a_k}{k^s}.$$ Then the point is that these truncation operators are compatible with the other operations, $$T_n(1/F) = T_n(1/T_n(F)),\; T_n(F\ast G) = T_n\bigl(T_n(F) \ast T_n(G)\bigr),\; T_n(F') = (T_n(F))',\; T_n (\ln F) = T_n \bigl(\ln (T_n(F))\bigr),$$ and $T_n\bigl(\exp(F)\bigr) = T_n\bigl(\exp(T_n(F))\bigr)$, where however for $\exp$ we separate the constant term from the others. Writing $F(s) = a_1 + G(s)$ … $\endgroup$ – Daniel Fischer Dec 7 '17 at 21:41
  • $\begingroup$ where $G$ contains the terms $a_k k^{-s}$ for $k \geqslant 2$, we define $\exp(F) = e^{a_1}\cdot \exp (G)$, with $\exp(G(s)) = \sum_{m = 0}^{\infty} \frac{G(s)^m}{m!}$. This is analogous to computations with Taylor polynomials, to compute the Taylor polynomial of order $n$ of the product of two functions, we multiply the Taylor polynomials of order $n$ of the two factors and ignore the terms of order $> n$. And when computing the Taylor polynomial of $e^{f(x)}$ we also must separate the constant term from the rest, so that isn't a quirk of formal Dirichlet series either. … $\endgroup$ – Daniel Fischer Dec 7 '17 at 21:42
  • $\begingroup$ If you take that on faith for the moment, maybe you can see how the argument works. By truncating, we obtain a convergent Dirichlet series (actually a Dirichlet polynomial), call it $\tilde{F}$. Then we can write $\ln \tilde{F}$ in the given form, and compute $\exp(\ln \tilde{F})$, and since we're dealing with nice convergent series, we know that indeed $\exp(\ln \tilde{F}(s)) = \tilde{F}(s)$ in some half-plane. Then by the uniqueness theorem for the coefficients, we know the coefficients are those we started with. $\endgroup$ – Daniel Fischer Dec 7 '17 at 21:42
  • $\begingroup$ By the compatibility, it follows that the first $n$ coefficients of $\exp(\ln F(s))$ are equal to the first $n$ coefficients of $F(s)$. But $n$ was arbitrary. It's too late today to start a proper write-up, but I should find some time tomorrow. $\endgroup$ – Daniel Fischer Dec 7 '17 at 21:42

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