How can I prove my conjecture for the coefficients in $t(x)=\log(1+\exp(x)) $?

Question

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?

Answer 1

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.

Answer 2

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

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

Answer 3

For $n\ge0$, by virtue of the Faa di Bruno formula and some properties of the Bell polynomials of the second kind $B_{n,k}$, we obtain \begin{align*} t^{(n)}(x)&=\sum_{k=0}^n(\ln u)^{(k)} B_{n,k}\bigl(u'(x),u''(x),\dotsc,u^{(n-k+1)}(x)\bigr), \quad u=u(x)=1+\operatorname{e}^x\\ &=(\ln u)B_{n,0}\bigl(\operatorname{e}^x,\operatorname{e}^x,\dotsc,\operatorname{e}^x\bigr) +\sum_{k=1}^n(\ln u)^{(k)} B_{n,k}\bigl(\operatorname{e}^x,\operatorname{e}^x,\dotsc,\operatorname{e}^x\bigr)\\ &=\begin{cases} \ln u, &n=0\\ \displaystyle\sum_{k=1}^n(\ln u)^{(k)} B_{n,k}\bigl(\operatorname{e}^x,\operatorname{e}^x,\dotsc,\operatorname{e}^x\bigr), & n>0 \end{cases}\\ &=\begin{cases} \ln(1+\operatorname{e}^x), &n=0\\ \displaystyle\sum_{k=1}^n\frac{(-1)^{k-1}(k-1)!}{(1+\operatorname{e}^x)^{k}} \operatorname{e}^{kx}B_{n,k}(1,1,\dotsc,1), & n>0 \end{cases}\\ &=\begin{cases} \ln(1+\operatorname{e}^x), &n=0\\ \displaystyle\sum_{k=1}^n\frac{(-1)^{k-1}(k-1)!}{(1+\operatorname{e}^x)^{k}} \operatorname{e}^{kx}S(n,k), & n>0 \end{cases}\\ &\to \begin{cases} \ln2, &n=0\\ \displaystyle\sum_{k=1}^n(-1)^{k-1}\frac{(k-1)!}{2^{k}}S(n,k), & n>0 \end{cases} \end{align*} as $x\to0$, where $S(n,k)$ denotes the Stirling numbers of the second kind. Consequently, we have \begin{equation*} t(x)=\ln2+\sum_{n=1}^\infty \Biggl[\sum_{k=1}^n(-1)^{k-1}\frac{(k-1)!}{2^{k}}S(n,k)\Biggr]\frac{x^n}{n!}, \quad x<0. \end{equation*} Since \begin{equation*} \sum_{k=1}^n(-1)^{k-1}\frac{(k-1)!}{2^{k}}S(n,k) =(-1)^{n+1} (2^n-1) \zeta (1-n), \quad n\ge1, \end{equation*} we arrive at \begin{equation*} t(x)=\ln2+\sum_{n=1}^\infty(-1)^{n+1} \frac{(2^n-1) \zeta (1-n)}{n!}x^n, \quad x<0. \end{equation*} Since \begin{equation*} \eta (1-n)=(-1)^{n+1}(2^n-1) \zeta (1-n), \quad n\ge2, \end{equation*} we finally conclude that \begin{equation*} t(x)=\ln2+\frac{1}2x+\sum_{n=2}^\infty\frac{\eta(1-n)}{n!}x^n =\sum_{n=0}^\infty\frac{\eta(1-n)}{n!}x^n, \quad x<0. \end{equation*} For details of the notations and notions used above, please read the following references.

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