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?

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update: Moreover, if we take the/a "complementary" function $$ u(x)=\log(-1+\exp(x)) $$ then we seem to get the similar looking $$ u(x) = \log(x) - \sum_{k=1}^\infty { \zeta(1-k) \over k!}x^k$$

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

It is not difficult to see that \begin{equation*} \ln\frac{\operatorname{e}^x-1}{x}=\frac{x}{2}+\ln\frac{\sinh(x/2)}{x/2} =\frac{x}{2}+\ln\frac{\sin(x\operatorname{i}/2)}{x\operatorname{i}/2}, \end{equation*} where $\operatorname{i}=\sqrt{-1}\,$ is the imaginary unit in complex analysis. On Page 75 in the handbook [1] and on Page 55 in the handbook [2] listed below, we find \begin{equation*} \ln\frac{\sin z}{z}=-\sum_{k=1}^{\infty}\frac{2^{2k-1}}{k}|B_{2k}|\frac{z^{2k}}{(2k)!}, \quad |z|<\pi, \end{equation*} where $B_{2k}$ denotes the Bernoulli numbers. Thus, it follows that \begin{equation*} \ln\frac{\sin(x\operatorname{i}/2)}{x\operatorname{i}/2} =\sum_{k=1}^{\infty}\frac{B_{2k}}{2k}\frac{x^{2k}}{(2k)!}, \quad |x|<2\pi. \end{equation*} Accordingly, we derive \begin{equation*} \boxed{\ln\frac{\operatorname{e}^x-1}{x}=\frac{x}{2}+\sum_{k=1}^{\infty}\frac{B_{2k}}{2k}\frac{x^{2k}}{(2k)!} =\frac{x}{2}-\sum_{k=1}^{\infty}\zeta(1-2k)\frac{x^{2k}}{(2k)!}, \quad |x|<2\pi,} \end{equation*} where we used the identity \begin{equation*} \zeta(1-2n)=-\frac{B_{2n}}{2n}, \quad n\in\mathbb{N} \end{equation*} found on Page 807 in the handbook [1] below.

References

1. M. Abramowitz and I. A. Stegun (Eds), Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, National Bureau of Standards, Applied Mathematics Series 55, 10th printing, Dover Publications, New York and Washington, 1972.

2. I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, Translated from the Russian, Translation edited and with a preface by Daniel Zwillinger and Victor Moll, Eighth edition, Revised from the seventh edition, Elsevier/Academic Press, Amsterdam, 2015; available online at https://doi.org/10.1016/B978-0-12-384933-5.00013-8.

Answer 4

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 5

The question posed by Gottfried Helms at How can I prove my conjecture for the coefficients in $t(x)=\log(1+\exp(x)) $? is equivalent to the known Maclaurin power series expansion \begin{equation}\label{log-cosine-series-expansion} \ln\cos x=-\sum_{n=1}^{\infty}\frac{2^{2n-1}(2^{2n}-1)}{n(2n)!}|B_{2n}|x^{2n},\quad |x|<\frac{\pi}2; \end{equation} see the second power series expansion on Page 55 in the reference [2] below.

By the above expansion, it is not difficult to see that \begin{align*} \ln\frac{\operatorname{e}^x+1}{2}&=\frac{x}{2}+\ln\frac{\operatorname{e}^{x/2}+\operatorname{e}^{-x/2}}{2}\\ &=\frac{x}{2}+\ln\cosh\frac{x}{2}\\ &=\frac{x}{2}+\ln\cos\frac{x\operatorname{i}}{2}\\ &=\frac{x}{2}+\sum_{k=1}^{\infty}\bigl(2^{2k}-1\bigr)\frac{B_{2k}}{2k}\frac{x^{2k}}{(2k)!}\\ &=\frac{x}{2}-\sum_{k=1}^{\infty}\bigl(2^{2k}-1\bigr)\zeta(1-2k)\frac{x^{2k}}{(2k)!}\\ &=\frac{x}{2}+\sum_{k=1}^{\infty}\eta(1-2k)\frac{x^{2k}}{(2k)!} \end{align*} for $|x|<\pi$, where $\operatorname{i}=\sqrt{-1}\,$ is the imaginary unit in complex analysis and we used the identity \begin{equation}\label{abram-23.2.15} \zeta(1-2k)=-\frac{B_{2k}}{2k}, \quad k\in\mathbb{N}, \end{equation} on Page 807 in the handbook [1] below.

Reference

1. M. Abramowitz and I. A. Stegun (Eds), Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, National Bureau of Standards, Applied Mathematics Series 55, 10th printing, Dover Publications, New York and Washington, 1972.
2. I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, Translated from the Russian, Translation edited and with a preface by Daniel Zwillinger and Victor Moll, Eighth edition, Revised from the seventh edition, Elsevier/Academic Press, Amsterdam, 2015; available online at https://doi.org/10.1016/B978-0-12-384933-5.00013-8.
3. Hong-Chao Zhang, Bai-Ni Guo, and Wei-Shih Du, On Qi's normalized remainder of Maclaurin power series expansion of logarithm of secant function, Axioms 13 (2024), no. 12, Article 860, 11 pages; available online at https://doi.org/10.3390/axioms13120860.

Answer 6

The following answer was motivated by the idea in Section 1.1 of the paper

Zhi-Hua Bao, Ravi Prakash Agarwal, Feng Qi, and Wei-Shih Du, Some properties on normalized tails of Maclaurin power series expansion of exponential function, Symmetry 16 (2024), no. 8, Article 989, 15 pages; available online at https://doi.org/10.3390/sym16080989.

It is well known that the Bernoulli numbers $B_n$ are generalized by \begin{equation*} \frac{x}{\operatorname{e}^x-1}=\sum_{n=0}^\infty B_n\frac{x^n}{n!}=1-\frac{x}2+\sum_{n=1}^\infty B_{2n}\frac{x^{2n}}{(2n)!}, \quad |x|<2\pi. \end{equation*} Differentiating and making use of the above equation give \begin{align*} \biggl(\ln\frac{\operatorname{e}^x-1}{x}\biggr)'&=1-\frac{1}{x}+\frac{1}{x}\frac{x}{\operatorname{e}^x-1}\\ &=1-\frac{1}{x}+\frac{1}{x}\Biggl[1-\frac{x}2+\sum_{j=1}^\infty B_{2j}\frac{x^{2j}}{(2j)!}\Biggr]\\ &=\frac{1}{2}+\sum_{j=1}^\infty B_{2j}\frac{x^{2j-1}}{(2j)!}, \quad |x|<2\pi. \end{align*} Integrating over the interval $(0,x)$ on both sides of the above equality at the very ends yields \begin{equation*} \ln\frac{\operatorname{e}^x-1}{x}=\frac{x}{2}+\sum_{j=1}^\infty\frac{B_{2j}}{2j}\frac{x^{2j}}{(2j)!}, \quad |x|<2\pi. \end{equation*} Further in view of the identity \begin{equation*} \zeta(1-2k)=-\frac{B_{2k}}{2k}, \quad k\in\mathbb{N}, \end{equation*} we obtain \begin{equation*} \boxed{\ln\frac{\operatorname{e}^x-1}{x}=\frac{x}{2}-\sum_{k=1}^{\infty}\zeta(1-2k)\frac{x^{2k}}{(2k)!}, \quad |x|<2\pi.} \end{equation*}