On $e^{5x}+e^{4x}+e^{3x}+e^{2x}+e^{x}+1$

Question

Define the following,

$$F_2(x) := \frac{1}{2}+\frac{(2x)}{1!} B_2\Big(\tfrac{1}{2}\Big)+\frac{(2x)^2}{2!}B_3\Big(\tfrac{1}{2}\Big)+\frac{(2x)^3}{3!}B_4\Big(\tfrac{1}{2}\Big)+\dots $$

$$\color{brown}{F_3(x)} := \frac{1}{3}+\frac{(3x)}{1!} B_2\Big(\tfrac{1}{3}\Big)+\frac{(3x)^2}{2!}B_3\Big(\tfrac{1}{3}\Big)+\frac{(3x)^3}{3!}B_4\Big(\tfrac{1}{3}\Big)+\dots $$

$$F_4(x) := \frac{1}{4}+\frac{(4x)}{1!} B_2\Big(\tfrac{1}{4}\Big)+\frac{(4x)^2}{2!}B_3\Big(\tfrac{1}{4}\Big)+\frac{(4x)^3}{3!}B_4\Big(\tfrac{1}{4}\Big)+\dots $$

$$\color{brown}{F_6(x)} := \frac{1}{6}+\frac{(6x)}{1!} B_2\Big(\tfrac{1}{6}\Big)+\frac{(6x)^2}{2!}B_3\Big(\tfrac{1}{6}\Big)+\frac{(6x)^3}{3!}B_4\Big(\tfrac{1}{6}\Big)+\dots $$

From Summation of Series (2nd ed) by L.B Jolley, page 26, it seems that,

$$\color{brown}{F_3(x)} \overset{?}= \frac{1}{1+e^x+e^{2x}}\\ \color{brown}{F_6(x)} \overset{?}= \frac{1}{1+e^{x}+e^{2x}+e^{3x}+e^{4x}+e^{5x}}\tag1$$

I assume that $B_n(x)$ are the Bernoulli polynomials. However, when I try to numerically evaluate those two $F_k(x)$ using Mathematica, the LHS does not agree with the RHS.

Questions:

1. How to interpret $B(n)$ such that $(1)$ holds true?
2. And do $F_2(x)$ and $F_4(x)$ evaluate similar to $(1)$?

Answer 1

Ad Question 1: The polynomials $B_n(x)$ defined in L.B.W. Jolley's book Summation of Series are close relatives of the Bernoulli polynomials defined e.g. in L. Comtet's classic Advanced Combinatorics but not the same.

To differentiate the polynomials defined in L.B.W. Jolley's book from the more commonly defined Bernoulli polynomials we use, we denote them from now on with $\widetilde{B}_n$.

In number (1128) on p. 228 in Jolley's book and in the subsequent section (1129) we find a representation of $\widetilde{B}_n(x)$ as generating series. \begin{align*} t\frac{e^{xt}-1}{e^x-1}=\sum_{n\geq 1}n\widetilde{B}_n(x)\frac{t^n}{n!}\tag{1} \end{align*}

A generating function of the Bernoulli polynomials $B_n(x)$ is given as \begin{align*} t\frac{e^{xt}}{e^t-1}=\sum_{n\geq 0}B_n(x)\frac{t^n}{n!} \end{align*} Since a generating function of the Bernoulli numbers $B_n=B_n(0)$ is \begin{align*} \frac{t}{e^t-1}=\sum_{n\geq 0}B_n(0)\frac{t^n}{n!} \end{align*}

we obtain the following relationship between Bernoulli polynomials $B_n(x)$ and $\widetilde{B}_n(x)$

\begin{align*} \widetilde{B}_n(x)=\frac{B_n(x)-B_n(0)}{n}\qquad\qquad n\geq 1 \end{align*}

Ad question 2: The answer is affirmative. The nice formulas of $F_t(x)$ are valid for all $t\in \mathbb{N}$.

On one hand we obtain for $t\geq 1$ by expanding numerator and denominator of $F_t(x)$ with $1-e^x$

\begin{align*} F_t(x)&=\frac{1}{1+e^x+e^{2x}+\cdots+e^{(t-1)x}}\\ &=\frac{1-e^x}{1-e^{xt}} \end{align*}

On the other hand note, that the general expression of $F_t(x)$ is \begin{align*} F_t(x)=\frac{1}{t}+\sum_{n\geq 1}\widetilde{B}_{n+1}\left(\frac{1}{t}\right)\frac{(tx)^n}{n!}\tag{2} \end{align*}

We obtain

\begin{align*} F_t(x)&=\frac{1}{t}+\sum_{n\geq 1}\widetilde{B}_{n+1}\left(\frac{1}{t}\right)\frac{(tx)^n}{n!}\\ &=\frac{1}{t}+\sum_{n\geq 2}\widetilde{B}_n\left(\frac{1}{t}\right)\frac{(tx)^{n-1}}{(n-1)!}\\ &=\frac{1}{t}+\frac{1}{tx}\sum_{n\geq 2}n\widetilde{B}_n\left(\frac{1}{t}\right)\frac{(tx)^{n}}{n!}\tag{3}\\ &=\frac{1}{t}+\frac{1}{tx}\left(tx\frac{e^{\frac{1}{t}tx}-1}{e^{tx}-1}-\widetilde{B_1}\left(\frac{1}{t}\right)tx\right)\tag{4}\\ &=\frac{1}{t}+\frac{1}{tx}\left(tx\frac{e^x-1}{e^{tx}-1}-x\right)\\ &=\frac{e^x-1}{e^{tx}-1}\\ \end{align*}

and the claim follows.

Comment:

• In (3) we use the generating series for $n\widetilde{B}_n(x)$ according to (1)

• In (4) note that $\widetilde{B}_1(x)=x$

Epilogue:

Symmetry

The Function $F_t(x)$ and its reciprocal $\frac{1}{F_t(x)}$ admit a nice nearly symmetrical representation with respect to Bernoulli polynomials and their arguments. The following is valid

\begin{align*} F_t(x)-1& =\sum_{n\geq 0}\frac{B_{n+1}\left(\frac{1}{t}\right)-B_{n+1}(0)}{n+1}\frac{(tx)^n}{n!}\\ \frac{1}{F_t(x)}& =\sum_{n\geq 0}\frac{B_{n+1}(t)-B_{n+1}(0)}{n+1}\frac{x^n}{n!} \end{align*}

This holds, since according to (2) we get \begin{align*} F_t(x)&=\frac{1}{\sum_{k=0}^{t-1}e^{kx}} =\frac{1}{t}+\sum_{n\geq 1}\frac{B_{n+1}\left(\frac{1}{t}\right)-B_{n+1}(0)}{n+1}\frac{(tx)^n}{n!}\\ &=1+\sum_{n\geq 0}\frac{B_{n+1}\left(\frac{1}{t}\right)-B_{n+1}(0)}{n+1}\frac{(tx)^n}{n!}\\ \end{align*} and \begin{align*} \frac{1}{F_t(x)}&=\sum_{k=0}^{t-1}e^{kx} =\sum_{k=0}^{t-1}\sum_{n\geq 0}\frac{(kx)^n}{n!} =\sum_{n\geq 0}\left(\sum_{k=0}^{t-1}k^n\right)\frac{x^n}{n!}\\ &=\sum_{n\geq 0}\frac{B_{n+1}(t)-B_{n+1}(0)}{n+1}\frac{x^n}{n!} \end{align*}

Historical Note

The definition of the Bernoulli numbers $\widetilde{B}_n$ in L.B.W. Jolley's book is taken from the British Association Report from 1877. The first few numbers are

\begin{array}{cccccccccc} n&0&1&2&3&4&5&6&7&8\\ \hline\\ \widetilde{B}_n&-1&\frac{1}{6}&\frac{1}{30}&\frac{1}{42}&\frac{1}{30} &\frac{5}{66}&\frac{691}{2730}&\frac{7}{6}&\frac{3617}{510}\\ \\ B_n&0&-\frac{1}{2}&\frac{1}{6}&0&-\frac{1}{30}&0&\frac{1}{42}&0&-\frac{1}{30} \end{array}

This corresponds for $n\geq 1$ to Comtet's statement in section 1.14

(notation somewhat adapted) ... Bernoulli numbers, denoted by $b_n$ by Bourbaki, are sometimes also defined by: \begin{align*} \frac{t}{e^t-1}=1-\frac{1}{2}t+\sum_{n\geq 1}(-1)^{n+1}B_n\frac{t^{2n}}{(2n)!} \end{align*}

Each $B_n$ is then $>0$ and equals $(-1)^{n+1}B_{2n}$ as a function of our Bournoulli numbers.