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

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.

1. Wei-Shih Du, Dongkyu Lim, and Feng Qi, Several recursive and closed-form formulas for some specific values of partial Bell polynomials, Advances in the Theory of Nonlinear Analysis and its Applications 6 (2022), no. 4, 528--537; available online at https://doi.org/10.31197/atnaa.1170948.
2. Bai-Ni Guo, Dongkyu Lim, and Feng Qi, Maclaurin's series expansions for positive integer powers of inverse (hyperbolic) sine and tangent functions, closed-form formula of specific partial Bell polynomials, and series representation of generalized logsine function, Applicable Analysis and Discrete Mathematics 16 (2022), no. 2, 427--466; available online at https://doi.org/10.2298/AADM210401017G.
3. Siqintuya Jin, Bai-Ni Guo, and Feng Qi, Partial Bell polynomials, falling and rising factorials, Stirling numbers, and combinatorial identities, Computer Modeling in Engineering & Sciences 132 (2022), no. 3, 781--799; available online at https://doi.org/10.32604/cmes.2022.019941.
4. Dongkyu Lim and Feng Qi, Increasing property and logarithmic convexity of two functions involving Dirichlet eta function, Journal of Mathematical Inequalities 16 (2022), no. 2, 463--469; available online at https://doi.org/10.7153/jmi-2022-16-33.
5. Feng Qi, Taylor's series expansions for real powers of two functions containing squares of inverse cosine function, closed-form formula for specific partial Bell polynomials, and series representations for real powers of Pi, Demonstratio Mathematica 55 (2022), no. 1, 710--736; available online at https://doi.org/10.1515/dema-2022-0157.
6. Feng Qi, Da-Wei Niu, Dongkyu Lim, and Bai-Ni Guo, Closed formulas and identities for the Bell polynomials and falling factorials, Contributions to Discrete Mathematics 15 (2020), no. 1, 163--174; available online at https://doi.org/10.11575/cdm.v15i1.68111.
7. Feng Qi, Da-Wei Niu, Dongkyu Lim, and Yong-Hong Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, Journal of Mathematical Analysis and Applications 491 (2020), no. 2, Paper No. 124382, 31 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124382.

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

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.