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I'm considering the transfer-function $$ t(x) = \log(1 + \exp(x)) $$ and find the beginning of the power series (simply using Pari/GP) as $$ t(x) = \log(2) + 1/2 x + 1/8 x^2 – 1/192 x^4 + 1/2880 x^6 - \ldots $$ Examining the pattern of the coefficients I find the much likely composition $$ t(x) = \sum_{k=0}^\infty {\eta(1-k) \over k! }x^k $$ where $ \eta() $ is the Dirichlet eta-(or "alternating zeta") function.
I'm using this definition in further computations and besides the convincing simplicitiness of the pattern the results are always meaningful. However, I've no idea how I could prove this description of the coefficients.

Q: Does someone has a source or an idea, how to do such a proof on oneself?

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Are you familiar with Faadi Bruno's forlmula. – Mhenni Benghorbal Feb 18 '13 at 18:02
Yes, but I find it a much complicated one. When I study iteration of functions and can use a Carleman-matrix-epression (and the formal powers of that Carleman-matrices) then I could always avoid to involve into that formula. I know of a handful of articles which put Carleman/Bell-matrix and Faa-di-Bruno together, as for instance Aldrovandi/Freitas but I wouldn't see the solution for the proof for the decryption of the transfer-function $t(x)$ in question. – Gottfried Helms Feb 18 '13 at 18:07
up vote 12 down vote accepted

The Dirichlet eta function is given by $\eta(s)=\sum_{n=1}^{\infty}(-1)^{n-1}n^{-s}$, but this converges only for $s$ with positive real part, and you are proposing to use its behavior for negative integers. A globally convergent series for $\eta$ can be derived using the Riemann zeta function (cf. here): $$ \eta(s)=(1-2^{1-s})\zeta(s)=\sum_{n=0}^{\infty}2^{-(n+1)}\sum_{k=0}^{n}(-1)^{k}{{n}\choose{k}}(k+1)^{-s}. $$ Using this expansion allows us to write $\eta(1-k)$ as $$ \eta(1-k)=\sum_{n=0}^{\infty}2^{-(n+1)}\sum_{j=0}^{n}(-1)^{j}{{n}\choose{j}}(j+1)^{k-1}. $$

Your power series is then $$ \begin{eqnarray} \sum_{k=0}^{\infty}\frac{\eta(1-k)x^k}{k!} &=&\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}2^{-(n+1)}\sum_{j=0}^{n}(-1)^{j}{{n}\choose{j}}(j+1)^{k-1}\left(\frac{x^{k}}{k!}\right) \\ &=&\sum_{n=0}^{\infty}2^{-(n+1)}\sum_{j=0}^{n}\frac{(-1)^{j}}{j+1}{{n}\choose{j}}\sum_{k=0}^{\infty}\frac{\left(x(j+1)\right)^{k}}{k!} \\ &=&\sum_{n=0}^{\infty}2^{-(n+1)}\sum_{j=0}^{n}\frac{(-1)^{j}}{j+1}{{n}\choose{j}}\left(e^x\right)^{j+1} \\ &=&\sum_{n=0}^{\infty}2^{-(n+1)}\int_{-e^{x}}^{0} dy\sum_{j=0}^{n}{{n}\choose{j}}y^{j} \\ &=&\sum_{n=0}^{\infty}2^{-(n+1)}\int_{-e^{x}}^{0}\left(1+y\right)^{n}dy \\ &=&\sum_{n=0}^{\infty}2^{-(n+1)}\frac{\left(1+y\right)^{n+1}}{n+1}\Bigg\vert_{-e^x}^{0} \\ &=&\sum_{n=0}^{\infty}2^{-(n+1)}\frac{1-\left(1-e^x\right)^{n+1}}{n+1} \\ &=&f\left(\frac{1}{2}\right) - f\left(\frac{1-e^x}{2}\right), \end{eqnarray} $$ where $$f(z)=\sum_{n=0}^{\infty}\frac{z^{n+1}}{n+1}=-\log \left(1-z\right).$$ Putting this together, we find $$ \sum_{k=0}^{\infty}\frac{\eta(1-k)x^k}{k!} = -\log\left(\frac{1}{2}\right)+\log\left(\frac{1+e^x}{2}\right)=\log\left(1+e^x\right), $$ as you conjectured.

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Wow. Thanks a lot -this shall give me much to chew on and to study... to use this sequence of steps for other, but related, situations; it looks much helpful, indeed! – Gottfried Helms Feb 19 '13 at 20:38

Related problems: (I), (II), (III), (IV). Here is a formula for the nth derivative of the function $\ln(1+e^{x})$ at the point $x=0$

$$ \left( \ln(1+e^{x})\right)^{(n)}= \sum _{k=1}^{n}\begin{Bmatrix} n\\k \end{Bmatrix} \left( -1 \right)^{k+1}2^{-k}\, \Gamma\left( k \right),\quad n \in \mathbb{N}, $$

where $\begin{Bmatrix} n\\k \end{Bmatrix} $ are the Stirling numbers of the second kind. The above formula allows us to construct the Taylor series of the function as

$$ \ln(1+e^x) = \ln(2)+\sum_{n=1}^{\infty} \sum _{k=1}^{n}\begin{Bmatrix} n\\k \end{Bmatrix} \left( -1 \right)^{k+1}\, 2^{-k}\,\Gamma\left( k \right) \frac{x^n}{n!}. $$

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can you prove the first assertion? – nbubis Feb 19 '13 at 15:02
@nbubis: I'll try to do this later. – Mhenni Benghorbal Feb 19 '13 at 15:04
(where the curly braces mean the Stirling numbers 2'nd kind) – Gottfried Helms Feb 19 '13 at 16:05
@GottfriedHelms: Yes, I'll add this. – Mhenni Benghorbal Feb 19 '13 at 16:25
arrgh - that Stirling numbers 2'nd kind rang my alarm. It's just to use the Carleman-matrices for $x\to \exp(x)-1$, then $ x \to x+1 $ and the first column of that for $x \to \log(1+x) $. This gives the decomposition in Stirling-numbers 2nd and 1st kind and binomial-coefficients. Well, but after that: how is that the sequence of Dirichlet-etas? – Gottfried Helms Feb 19 '13 at 17:48

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