Closed-Form Formulas for the nth Derivative of the Power-Exponential Function x^x

Abstract

In this paper, the authors give a simple review of closed-form, explicit, and recursive formulas and related results for the nth derivative of the power-exponential function $x^x$, establish two closed-form and explicit formulas for partial Bell polynomials at some specific arguments, and present several new closed-form and explicit formulas for the nth derivative of the power-exponential function $x^x$ and for related functions and integer sequences.

1. Motivations

The power-exponential function

$$f\left(x\right)=\left\{\begin{array}{cc}{x}^x,\hfill & x>0\\ 1,\hfill & x=0\end{array}\right.$$

is elementary and typical in calculus for undergraduates. Its first few derivatives are often taught in calculus. However, its integrations and higher derivatives are not easy to be computed. Therefore, many mathematicians asked for, discussed, looked for, answered, and investigated derivatives and integrations of the function $f\left(x\right)$ and similar ones on the internet such as

• https://math.stackexchange.com/q/107870,
• https://math.stackexchange.com/q/137583,
• https://math.stackexchange.com/q/141347,
• https://math.stackexchange.com/q/146502,
• https://math.stackexchange.com/q/167780,
• https://math.stackexchange.com/q/177774,
• https://math.stackexchange.com/q/311903,
• https://math.stackexchange.com/q/376527,
• https://math.stackexchange.com/q/401122,
• https://math.stackexchange.com/q/407069,
• https://math.stackexchange.com/q/408503,
• https://math.stackexchange.com/q/410609,
• https://math.stackexchange.com/q/676641,
• https://math.stackexchange.com/q/740933,
• https://math.stackexchange.com/q/802256,
• https://math.stackexchange.com/q/848214,
• https://math.stackexchange.com/q/934896,
• https://math.stackexchange.com/q/942263,
• https://math.stackexchange.com/q/1006310,
• https://math.stackexchange.com/q/1015334,
• https://math.stackexchange.com/q/1041559,
• https://math.stackexchange.com/q/1404682,
• https://math.stackexchange.com/q/1480837,
• https://math.stackexchange.com/q/1710894,
• https://math.stackexchange.com/q/1755360,
• https://math.stackexchange.com/q/2061024,
• https://math.stackexchange.com/q/2064295,
• https://math.stackexchange.com/q/2083894,
• https://math.stackexchange.com/q/2165776,
• https://math.stackexchange.com/q/2214448,
• https://math.stackexchange.com/q/2286600,
• https://math.stackexchange.com/q/2287305,
• https://math.stackexchange.com/q/2445700,
• https://math.stackexchange.com/q/2598301,
• https://math.stackexchange.com/q/2599143,
• https://math.stackexchange.com/q/2963092,
• https://math.stackexchange.com/q/3106102,
• https://math.stackexchange.com/q/3337308,
• https://math.stackexchange.com/q/3413313,
• https://math.stackexchange.com/q/3459088,
• https://math.stackexchange.com/q/3631914,
• https://math.stackexchange.com/q/3860199,
• https://math.stackexchange.com/q/3961192,
• https://math.stackexchange.com/q/4185645,
• https://math.stackexchange.com/q/4463428,
• https://math.stackexchange.com/q/4568680.

Surprisingly, at so many web sites above, we do not find any general formula, including explicit and closed-form formula, for the nth derivative of the power-exponential function $f\left(x\right)$.

Does the general formula for the nth derivative of the power-exponential function $f\left(x\right)$ exist somewhere? What is the general formula for the nth derivative of the power-exponential function $f\left(x\right)$? We curiously posed these two questions at the site https://mathoverflow.net/q/437097 (accessed on 23 December 2022).

In this paper, we have two aims.

The first aim is to simply review several known results on closed-form, explicit, and recursive formulas and related results for the nth derivative of the power-exponential function $f\left(x\right)$, including Comtet’s numbers discussed in the monograph [1] (pp. 139–140), Kulkarni’s recursive formula obtained and mentioned in [2,3] (see also https://math.stackexchange.com/a/803045 (accessed on 20 May 2014), and Lehmer’s investigation of Comtet’s numbers and several other integer sequences carried out in the paper [4].

The second aim of this paper is to present several new closed-form and explicit formulas for the nth derivative of the power-exponential function $x^x$ and for related functions and integer sequences. We derive these formulas by virtue of the Faà di Bruno formula for derivatives of composite functions, by establishing two closed-form and explicit formulas for partial Bell polynomials at some specific arguments, and with the help of some properties of the Bell polynomials of the second kind.

2. Preliminaries

For fluently and smoothly proceeding, we prepare some notions and notations.

The Stirling numbers of the first kind $s(n,k)$ for $n\ge k\ge 0$ can be analytically generated [5] (p. 20, (1.30)) by

$$\frac{{[ln(1+t)]}^k}{k!}=\sum _{n=k}^{\infty }s(n,k)\frac{t^n}{n!},\left|t\right|<1,$$

which can be rearranged as Maclaurin’s series expansions of the power function

$${\left[\frac{ln(1+t)}{t}\right]}^k=\sum _{n=0}^{\infty }\frac{s(n+k,k)}{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}\frac{t^n}{n!},\left|t\right|<1,k\ge 0.$$

(1)

The falling factorial of a complex number $w\in \mathbb{C}$ is defined [6] (p. 7497) by

$${\langle w\rangle }_m=\prod _{\ell =0}^{m-1}(w-\ell )=\left\{\begin{array}{cc}w(w-1)\dots (w-m+1),\hfill & m\in \mathbb{N};\hfill \\ 1,\hfill & m=0.\hfill \end{array}\right.$$

The Faà di Bruno formula, see [7] (Theorem 11.4) and [1] (p. 139, Theorem C), can be described in terms of partial Bell polynomials $B_{n,k}(x_1,x_2,\dots ,x_{n-k+1})$ for $n\ge k\ge 0$ by

$$\frac{d^n}{d{x}^n}f\circ h\left(x\right)=\sum _{k=0}^n{f}^{\left(k\right)}\left(h\left(x\right)\right)B_{n,k}\left(h^{\prime }\left(x\right),h^{″}\left(x\right),\dots ,h^{(n-k+1)}\left(x\right)\right).$$

(2)

The partial Bell polynomials $B_{n,k}(x_1,x_2,\dots ,x_{n-k+1})$ for $n\ge k\ge 0$ satisfy the identities

$$\begin{array}{cc}\hfill B_{n,k}\left(ab{x}_1,a{b}^2{x}_2,\dots ,a{b}^{n-k+1}{x}_{n-k+1}\right)& =a^k{b}^n{B}_{n,k}(x_1,x_2,\dots ,x_{n-k+1}),\hfill \end{array}$$

(3)

$$\begin{array}{cc}\hfill B_{n,k}(0!,1!,2!,\dots ,(n-k)!)& ={(-1)}^{n-k}s(n,k),\hfill \end{array}$$

(4)

and

$$B_{n,k}\left(x_1,x_2,\dots ,x^{n-k+1}\right)=\sum _{\ell =0}^k\left(\genfrac{}{}{0pt}{}{n}{\ell }\right)x_1^{\ell }{B}_{n-\ell ,k-\ell }\left(0,x_2,\dots ,x^{n-k+1}\right).$$

(5)

These identities can be found in [7] (p. 412) and [1] (pp. 135–136).

3. A Simple Review of the nth Derivative of $x^x$

Basing on comments and answers made by several mathematicians at the sites https://mathoverflow.net/q/437097 (accessed on 28 December 2022) and https://math.stackexchange.com/q/802256 (accessed on 28 December 2022), we simply review some works by four mathematicians on the nth derivative of the power-exponential function $f\left(x\right)=x^x$ for $x>0$.

3.1. Comtet’s Numbers

What is the nth derivative of $x^{ax}$ for $x>0$ and any fixed real number $a\ne 0$? In [1] (pp. 139–140, Example), Comtet introduced the integer sequence $b(n,k)$ by

$$\frac{{[(1+t)ln(1+t)]}^k}{k!}=\sum _{n=k}^{\infty }b(n,k)\frac{t^n}{n!},$$

(6)

presented the explicit formula

$$b(n,k)=\left\{\begin{array}{cc}1,\hfill & (n,k)=(0,0);\\ 0,\hfill & n\in \mathbb{N},k=0;\\ {\displaystyle \sum _{\ell =k}^n}\left(\genfrac{}{}{0pt}{}{\ell }{k}\right)k^{\ell -k}s(n,\ell ),\hfill & n\ge k>0,\end{array}\right.$$

(7)

and obtained the closed-form formula

$$\frac{d^n\left(x^{ax}\right)}{d{x}^n}=a^n{x}^{ax}\sum _{j=0}^n\left(\genfrac{}{}{0pt}{}{n}{j}\right){(lnx)}^j\sum _{k=0}^{n-j}\frac{b(n-j,n-k-j)}{{\left(ax\right)}^k},n\in {\mathbb{N}}_0.$$

As in [4], we call $b(n,k)$ Comtet’s numbers.

3.2. Kulkarni’s Recursive Formula

In 1984, observing that $f^{(n+1)}\left(x\right)$ is the nth derivative of $f^{\prime }\left(x\right)=x^x(1+lnx)$ for $x>0$, writing down Leibniz’s rule for the nth derivative of a product of two n-time differentiable functions, and using the fact that the kth derivative of $1+lnx$ is ${(-1)}^{k-1}\frac{(k-1)!}{x^k}$, Kulkarni [2] gave the recursive formula

$$f^{(n+1)}\left(x\right)=(1+lnx)f^{\left(n\right)}\left(x\right)+\sum _{k=1}^n{(-1)}^{k-1}\left(\genfrac{}{}{0pt}{}{n}{k}\right)(k-1)!\frac{f^{(n-k)}\left(x\right)}{x^k}.$$

See also [3] or click the site https://math.stackexchange.com/a/803045 (accessed on 20 May 2014).

3.3. Lehmer’s Investigation

In the sense that $F\left(F^{-1}\left(x\right)\right)=F^{-1}\left(F\left(x\right)\right)=x$, the function

$$\phi \left(x\right)=\sum _{k=1}^{\infty }{(-1)}^{k-1}{(k-1)}^{k-1}\frac{x^k}{k!},$$

which is the inverse function of $(1+x)ln(1+x)$, was considered in the paper [4]. Via the function $\phi \left(x\right)$, Lehmer introduced the notion $B(n,k)$ by

$$\frac{{\left[\phi \left(x\right)\right]}^k}{k!}=\sum _{n=1}^{\infty }B(n,k)\frac{x^n}{n!}.$$

In the paper [4], Lehmer treated $b(n,k)$, $B(n,k)$, $s(n,k)$, and $S(n,k)$ together and presented numerous properties and relations.

4. A Lemma

The following lemma plays a key role in this paper.

Lemma 1. 

For $n\ge k\ge 0$, partial Bell polynomials $B_{n,k}$ satisfy

$$\begin{array}{c}{B}_{n,k}(0,0!,1!,2!,\dots ,(n-k-2)!,(n-k-1)!)\hfill \\ \hfill ={(-1)}^{n-k}n!\sum _{j=0}^k\frac{{(-1)}^j}{(k-j)!}\sum _{\ell =0}^{n-k}\frac{s(\ell +j,j)}{(\ell +j)!}\left(\genfrac{}{}{0pt}{}{j}{n-k-\ell }\right)\end{array}$$

(8)

and

$$\begin{array}{c}{B}_{n,k}(0,0!,1!,2!,\dots ,(n-k-2)!,(n-k-1)!)\hfill \\ \hfill ={(-1)}^{n-k}\sum _{j=0}^k{(-1)}^j\left(\genfrac{}{}{0pt}{}{n}{k-j}\right)b(n-k+j,j).\end{array}$$

(9)

Proof. 

Making use of the formula

$$\frac{1}{k!}{\left(\sum _{m=1}^{\infty }{x}_m\frac{t^m}{m!}\right)}^k=\sum _{n=k}^{\infty }{B}_{n,k}(x_1,x_2,\dots ,x_{n-k+1})\frac{t^n}{n!},k\in {\mathbb{N}}_0$$

listed in [1] (p. 133) yields

$$B_{n+k,k}(x_1,x_2,\dots ,x_{n+1})=\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)\underset{t\to 0}{lim}\frac{d^n}{d{t}^n}{\left[\sum _{m=0}^{\infty }{x}_{m+1}\frac{t^m}{(m+1)!}\right]}^k.$$

Taking

$$(x_1,x_2,x_3,x_4,\dots ,x_n,\dots )=(0,0!,1!,2!,\dots ,(n-2)!,\dots )$$

results in

$$\begin{array}{cc}\hfill & B_{n+k,k}(0,0!,1!,2!,\dots ,(n-2)!,(n-1)!)\hfill \\ \hfill & =\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)\underset{t\to 0}{lim}\frac{d^n}{d{t}^n}{\left[\sum _{m=1}^{\infty }(m-1)!\frac{t^m}{(m+1)!}\right]}^k\hfill \\ \hfill & =\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)\underset{t\to 0}{lim}\frac{d^n}{d{t}^n}{\left[1+\frac{(1-t)ln(1-t)}{t}\right]}^k\hfill \\ \hfill & =\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)\underset{t\to 0}{lim}\frac{d^n}{d{t}^n}\sum _{j=0}^k\left(\genfrac{}{}{0pt}{}{k}{j}\right){(t-1)}^j{\left[\frac{ln(1-t)}{-t}\right]}^j\hfill \\ \hfill & =\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)\underset{t\to 0}{lim}\frac{d^n}{d{t}^n}\sum _{j=0}^k\left(\genfrac{}{}{0pt}{}{k}{j}\right){(t-1)}^j\sum _{\ell =0}^{\infty }\frac{s(\ell +j,j)}{\left(\genfrac{}{}{0pt}{}{\ell +j}{j}\right)}\frac{{(-t)}^{\ell }}{\ell !}\hfill \\ \hfill & =\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)\sum _{j=0}^k\left(\genfrac{}{}{0pt}{}{k}{j}\right)\sum _{\ell =0}^{\infty }\frac{s(\ell +j,j)}{\left(\genfrac{}{}{0pt}{}{\ell +j}{j}\right)}\frac{{(-1)}^{\ell }}{\ell !}\underset{t\to 0}{lim}\frac{d^n}{d{t}^n}\left[{(t-1)}^j{t}^{\ell }\right]\hfill \\ \hfill & =\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)\sum _{j=0}^k{(-1)}^j\left(\genfrac{}{}{0pt}{}{k}{j}\right)\sum _{\ell =0}^{\infty }\frac{s(\ell +j,j)}{\left(\genfrac{}{}{0pt}{}{\ell +j}{j}\right)}\frac{{(-1)}^{\ell }}{\ell !}\underset{t\to 0}{lim}\frac{d^n}{d{t}^n}\sum _{q=0}^j{(-1)}^q\left(\genfrac{}{}{0pt}{}{j}{q}\right)t^{\ell +q}\hfill \\ \hfill & =\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)\sum _{j=0}^k{(-1)}^j\left(\genfrac{}{}{0pt}{}{k}{j}\right)\sum _{\ell =0}^{\infty }\frac{s(\ell +j,j)}{\left(\genfrac{}{}{0pt}{}{\ell +j}{j}\right)}\frac{{(-1)}^{\ell }}{\ell !}\underset{t\to 0}{lim}\sum _{q=0}^j{(-1)}^q\left(\genfrac{}{}{0pt}{}{j}{q}\right){\langle \ell +q\rangle }_n{t}^{\ell +q-n}\hfill \\ \hfill & ={(-1)}^{n}n!\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)\sum _{j=0}^k{(-1)}^j\left(\genfrac{}{}{0pt}{}{k}{j}\right)\sum _{\ell =0}^{\infty }\frac{s(\ell +j,j)}{\left(\genfrac{}{}{0pt}{}{\ell +j}{j}\right)}\frac{1}{\ell !}\left(\genfrac{}{}{0pt}{}{j}{n-\ell }\right)\hfill \\ \hfill & ={(-1)}^n(n+k)!\sum _{j=0}^k\frac{{(-1)}^j}{(k-j)!}\sum _{\ell =0}^n\frac{s(\ell +j,j)}{(\ell +j)!}\left(\genfrac{}{}{0pt}{}{j}{n-\ell }\right).\hfill \end{array}$$

In conclusion, we arrive at

$$\begin{array}{c}{B}_{n+k,k}(0,0!,1!,2!,\dots ,(n-2)!,(n-1)!)\hfill \\ \hfill ={(-1)}^n(n+k)!\sum _{j=0}^k\frac{{(-1)}^j}{(k-j)!}\sum _{\ell =0}^n\frac{s(\ell +j,j)}{(\ell +j)!}\left(\genfrac{}{}{0pt}{}{j}{n-\ell }\right).\end{array}$$

Replacing $n+k$ by n leads to the closed-form Formula (8).

The Equation (6) can be rearranged as

$${\left[\frac{(1+t)ln(1+t)}{t}\right]}^k=\sum _{n=0}^{\infty }\frac{b(n+k,k)}{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}\frac{t^n}{n!}.$$

(10)

Basing on the arguments in the proof of (8) and making use of (10), we obtain

$$\begin{array}{cc}\hfill & B_{n+k,k}(0,0!,1!,2!,\dots ,(n-2)!,(n-1)!)\hfill \\ \hfill & =\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)\underset{t\to 0}{lim}\frac{d^n}{d{t}^n}{\left[1-\frac{(1-t)ln(1-t)}{-t}\right]}^k\hfill \\ \hfill & =\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)\underset{t\to 0}{lim}\frac{d^n}{d{t}^n}\sum _{j=0}^k{(-1)}^j\left(\genfrac{}{}{0pt}{}{k}{j}\right){\left[\frac{(1-t)ln(1-t)}{-t}\right]}^j\hfill \\ \hfill & =\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)\underset{t\to 0}{lim}\frac{d^n}{d{t}^n}\sum _{j=0}^k{(-1)}^j\left(\genfrac{}{}{0pt}{}{k}{j}\right)\sum _{m=0}^{\infty }\frac{b(m+j,j)}{\left(\genfrac{}{}{0pt}{}{m+j}{j}\right)}\frac{{(-t)}^m}{m!}\hfill \\ \hfill & =\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)\sum _{j=0}^k{(-1)}^j\left(\genfrac{}{}{0pt}{}{k}{j}\right)\underset{t\to 0}{lim}\sum _{m=n}^{\infty }{(-1)}^m\frac{b(m+j,j)}{\left(\genfrac{}{}{0pt}{}{m+j}{j}\right)}\frac{t^{m-n}}{(m-n)!}\hfill \\ \hfill & ={(-1)}^n\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)\sum _{j=0}^k{(-1)}^j\left(\genfrac{}{}{0pt}{}{k}{j}\right)\frac{b(n+j,j)}{\left(\genfrac{}{}{0pt}{}{n+j}{j}\right)}\hfill \\ \hfill & ={(-1)}^n\sum _{j=0}^k{(-1)}^j\left(\genfrac{}{}{0pt}{}{n+k}{k-j}\right)b(n+j,j).\hfill \end{array}$$

Replacing $n+k$ by n gives (9). The proof of Lemma 1 is thus complete. □

5. Two New Formulas for the nth Derivative of $x^x$

By virtue of the closed-form Formulas (8) and (9) in Lemma 1, we now present two new closed-form and explicit formulas for the nth derivative of $x^x$ as follows.

Theorem 1. 

For $n\in {\mathbb{N}}_0=\{0,1,2,\dots \}$, we have

$${\left(x^x\right)}^{\left(n\right)}=n!x^{x-n}\sum _{k=0}^n{x}^k\sum _{j=0}^k\left[\sum _{q=0}^{n-k}\frac{s(q+j,j)}{(q+j)!}\left(\genfrac{}{}{0pt}{}{j}{n-k-q}\right)\right]\frac{{(lnx)}^{k-j}}{(k-j)!}$$

(11)

and

$${\left(x^x\right)}^{\left(n\right)}=x^{x-n}\sum _{k=0}^n{x}^k\sum _{j=0}^k{(-1)}^j\left(\genfrac{}{}{0pt}{}{j-n-1}{j}\right)b(n-j,k-j){(lnx)}^j.$$

(12)

Consequently, Taylor’s series expansion around $x=1$ is

$$x^x=\sum _{n=0}^{\infty }\left[\sum _{k=0}^n\sum _{q=k}^n\frac{s(q,k)}{q!}\left(\genfrac{}{}{0pt}{}{k}{n-q}\right)\right]{(x-1)}^n,|x-1|<1$$

(13)

and

$$x^x=\sum _{n=0}^{\infty }\left[\sum _{k=0}^{n}b(n,k)\right]\frac{{(x-1)}^n}{n!},|x-1|<1.$$

(14)

Proof. 

For $n\in {\mathbb{N}}_0$, we have

$$\begin{array}{cc}\hfill {\left(x^x\right)}^{\left(n\right)}={\left(e^{xlnx}\right)}^{\left(n\right)}& =\sum _{k=0}^n{\left(e^u\right)}^{\left(k\right)}{B}_{n,k}\left(u^{\prime }\left(x\right),u^{″}\left(x\right),\dots ,u^{(n-k)}\left(x\right),u^{(n-k+1)}\left(x\right)\right)\hfill \\ \hfill & =\sum _{k=0}^n{e}^{x ln x}{B}_{n,k}(1+lnx,\frac{1}{x},-\frac{1}{x^2},\frac{2}{x^3},-\frac{6}{x^4},\dots ,\hfill \\ \hfill & {(-1)}^{n-k}\frac{(n-k-2)!}{x^{n-k-1}},{(-1)}^{n-k+1}\frac{(n-k-1)!}{x^{n-k}}),\hfill \end{array}$$

where $u=u\left(x\right)=xlnx$.

By the Formulas (3) and (5) in sequence, we acquire

$$\begin{array}{cc}\hfill {\left(x^x\right)}^{\left(n\right)}& =\sum _{k=0}^n{x}^x\sum _{\ell =0}^k\left(\genfrac{}{}{0pt}{}{n}{\ell }\right){(1+lnx)}^{\ell }{B}_{n-\ell ,k-\ell }(0,\frac{1}{x},-\frac{1}{x^2},\frac{2}{x^3},-\frac{6}{x^4},\dots ,\hfill \\ \hfill & {(-1)}^{n-k}\frac{(n-k-2)!}{x^{n-k-1}},{(-1)}^{n-k+1}\frac{(n-k-1)!}{x^{n-k}})\hfill \\ \hfill & ={(-1)}^n{x}^{x-n}\sum _{k=0}^n{x}^k\sum _{\ell =0}^k{(-1)}^{\ell }\left(\genfrac{}{}{0pt}{}{n}{\ell }\right){(1+lnx)}^{\ell }\hfill \\ \hfill & \times B_{n-\ell ,k-\ell }(0,0!,1!,2!,\dots ,(n-k-2)!,(n-k-1)!).\hfill \end{array}$$

By virtue of the Formula (8) in Lemma 1, we conclude the explicit formula

$$\begin{array}{cc}\hfill {\left(x^x\right)}^{\left(n\right)}& =n!x^{x-n}\sum _{k=0}^n{(-1)}^k{x}^k\sum _{\ell =0}^k{(-1)}^{\ell }[\sum _{j=0}^{k-\ell }\frac{{(-1)}^j}{(k-\ell -j)!}\hfill \\ \hfill & \times \sum _{q=0}^{n-k}\frac{s(q+j,j)}{(q+j)!}\left(\genfrac{}{}{0pt}{}{j}{n-k-q}\right)]\frac{{(1+lnx)}^{\ell }}{\ell !}\hfill \\ \hfill & =n!x^{x-n}\sum _{k=0}^n{(-1)}^k{x}^k\sum _{j=0}^k{(-1)}^j\sum _{q=0}^{n-k}\frac{s(q+j,j)}{(q+j)!}\left(\genfrac{}{}{0pt}{}{j}{n-k-q}\right)\hfill \\ \hfill & \times \sum _{\ell =0}^{k-j}\frac{{(-1)}^{\ell }}{(k-j-\ell )!}\frac{{(1+lnx)}^{\ell }}{\ell !}\hfill \\ \hfill & =n!x^{x-n}\sum _{k=0}^n{x}^k\sum _{j=0}^k\frac{{(lnx)}^{k-j}}{(k-j)!}\sum _{q=0}^{n-k}\frac{s(q+j,j)}{(q+j)!}\left(\genfrac{}{}{0pt}{}{j}{n-k-q}\right).\hfill \end{array}$$

Accordingly, the explicit Formula (11) follows.

By virtue of the Formula (9) in Lemma 1, we arrive at

$$\begin{array}{cc}\hfill {\left(x^x\right)}^{\left(n\right)}& =x^{x-n}\sum _{k=0}^n{(-1)}^k{x}^k\sum _{\ell =0}^k{(-1)}^{\ell }\left(\genfrac{}{}{0pt}{}{n}{\ell }\right)\hfill \\ \hfill & \times \left[\sum _{j=0}^{k-\ell }{(-1)}^j\left(\genfrac{}{}{0pt}{}{n-\ell }{n-k+j}\right)b(n-k+j,j)\right]{(1+lnx)}^{\ell }\hfill \\ \hfill & =x^{x-n}\sum _{k=0}^n{(-1)}^k{x}^k\sum _{j=0}^k{(-1)}^{j}b(n-k+j,j)\hfill \\ \hfill & \times \sum _{\ell =0}^{k-j}{(-1)}^{\ell }\left(\genfrac{}{}{0pt}{}{n}{\ell }\right)\left(\genfrac{}{}{0pt}{}{n-\ell }{n-(k-j)}\right){(1+lnx)}^{\ell }.\hfill \end{array}$$

Since

$$\sum _{\ell =0}^m{(-1)}^{\ell }\left(\genfrac{}{}{0pt}{}{n}{\ell }\right)\left(\genfrac{}{}{0pt}{}{n-\ell }{n-m}\right)x^{\ell }=\left(\genfrac{}{}{0pt}{}{m-n-1}{m}\right){(x-1)}^m$$

(15)

for $m,n\in {\mathbb{N}}_0$, we finally obtain

$${\left(x^x\right)}^{\left(n\right)}=x^{x-n}\sum _{k=0}^n{(-1)}^k{x}^k\sum _{j=0}^k{(-1)}^j\left(\genfrac{}{}{0pt}{}{k-j-n-1}{k-j}\right)b(n-k+j,j){(lnx)}^{k-j}.$$

The Formula (12) follows.

Taylor’s series expansions (13) and (14) can be derived from Taylor’s theorem and taking the limit $x\to 1$ in (11) and (12) respectively. The proof of Theorem 1 is thus complete. □

Remark 1. 

The variants of the results (11) and (13) in Theorem 1 have been announced at the sites https://mathoverflow.net/q/437097 (accessed on 23 December 2022) and https://math.stackexchange.com/a/4605027 (accessed on 20 May 2014) as a comment and an answer without proof.

The form of Taylor’s series expansion (14) is very nice.

Remark 2. 

In the papers [8,9], among other things, several Maclaurin and Taylor series expansions have been discovered.

6. Two New Formulas for Comtet’s Numbers

Reformulating the generating function (6) of Comtet’s numbers $b(n,k)$, employing the series expansion (1) of the Stirling numbers of the first kind $s(n,k)$, applying the Faà di Bruno Formula (2), and using the identities (3) and (4), we now present two new closed-form and explicit formulas for the integer sequence $b(n,k)$, as follows. In next section, we will derive the third new closed-form and explicit formula for the integer sequence $b(n,k)$.

Theorem 2. 

For $n\ge k\ge 0$, we have

$$b(n,k)=\frac{n!}{(n-k)!}\sum _{j=0}^k\frac{1}{j!}\sum _{\ell =0}^{n-k}\frac{\left(\genfrac{}{}{0pt}{}{n-k}{\ell }\right)}{\left(\genfrac{}{}{0pt}{}{\ell +j}{j}\right)}s(\ell +j,j)s(n-k-\ell ,k-j)$$

(16)

and

$$b(n,k)=n!\sum _{j=n-k}^n\left(\genfrac{}{}{0pt}{}{k}{n-j}\right)\frac{s(j,k)}{j!}.$$

(17)

Proof. 

The series expansion (10) means that

$$\begin{array}{cc}\hfill \frac{b(n+k,k)}{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}& =\underset{t\to 0}{lim}\frac{d^n}{d{t}^n}{\left[\frac{(1+t)ln(1+t)}{t}\right]}^k\hfill \\ \hfill & =\underset{t\to 0}{lim}\frac{d^n}{d{t}^n}{\left[\frac{ln(1+t)}{t}+ln(1+t)\right]}^k\hfill \\ \hfill & =\underset{t\to 0}{lim}\frac{d^n}{d{t}^n}\sum _{j=0}^k\left(\genfrac{}{}{0pt}{}{k}{j}\right){\left[\frac{ln(1+t)}{t}\right]}^j{[ln(1+t)]}^{k-j}\hfill \\ \hfill & =\underset{t\to 0}{lim}\sum _{j=0}^k\left(\genfrac{}{}{0pt}{}{k}{j}\right)\sum _{\ell =0}^n\left(\genfrac{}{}{0pt}{}{n}{\ell }\right){\left\{{\left[\frac{ln(1+t)}{t}\right]}^j\right\}}^{(\ell )}{\left\{{[ln(1+t)]}^{k-j}\right\}}^{(n-\ell )},\hfill \end{array}$$

where, by the series expansion (1) of the Stirling numbers of the first kind $s(n,k)$,

$$\begin{array}{cc}\hfill {\left\{{\left[\frac{ln(1+t)}{t}\right]}^j\right\}}^{(\ell )}& ={\left[\sum _{m=0}^{\infty }\frac{s(m+j,j)}{\left(\genfrac{}{}{0pt}{}{m+j}{j}\right)}\frac{t^m}{m!}\right]}^{(\ell )}\hfill \\ \hfill & =\sum _{m=\ell }^{\infty }\frac{s(m+j,j)}{\left(\genfrac{}{}{0pt}{}{m+j}{j}\right)}\frac{t^{m-\ell }}{(m-\ell )!}\hfill \\ \hfill & \to \frac{s(\ell +j,j)}{\left(\genfrac{}{}{0pt}{}{\ell +j}{j}\right)},t\to 0\hfill \end{array}$$

(18)

and, by the Faà di Bruno Formula (2) with $f\left(u\right)=u^{k-j}$ and $u=h\left(t\right)=ln(1+t)$, and by the identities (3) and (4),

$$\begin{array}{c}{\left\{{[ln(1+t)]}^{k-j}\right\}}^{(n-\ell )}=\sum _{m=0}^{n-\ell }{\left(u^{k-j}\right)}^{\left(m\right)}{B}_{n-\ell ,m}(\frac{1}{1+t},\\ -\frac{1}{{(1+t)}^2},\dots ,{(-1)}^{n-\ell -m}\frac{(n-\ell -m)!}{{(1+t)}^{n-\ell -m+1}})\\ =\sum _{m=0}^{n-\ell }{\langle k-j\rangle }_m{[ln(1+t)]}^{k-j-m}\frac{{(-1)}^{n-\ell +m}}{{(1+t)}^{n-\ell }}{B}_{n-\ell ,m}\left(0!,1!,\dots ,(n-\ell -m)!\right)\\ \begin{array}{cc}& \to {(-1)}^{n-\ell +k-j}(k-j)!B_{n-\ell ,k-j}\left(0!,1!,\dots ,(n-\ell -k+j)!\right),t\to 0\hfill \\ & ={(-1)}^{n-\ell +k-j}(k-j)!{(-1)}^{n-\ell -k+j}s(n-\ell ,k-j)\hfill \\ & =(k-j)!s(n-\ell ,k-j).\hfill \end{array}\end{array}$$

Accordingly, we arrive at

$$\frac{b(n+k,k)}{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}=\sum _{j=0}^k\left(\genfrac{}{}{0pt}{}{k}{j}\right)\sum _{\ell =0}^n\left(\genfrac{}{}{0pt}{}{n}{\ell }\right)\frac{s(\ell +j,j)}{\left(\genfrac{}{}{0pt}{}{\ell +j}{j}\right)}(k-j)!s(n-\ell ,k-j).$$

Replacing $n+k$ by n leads to (16).

From (10), we can also compute by

$$\begin{array}{cc}\hfill \frac{b(n+k,k)}{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}& =\underset{t\to 0}{lim}\frac{d^n}{d{t}^n}{\left[\frac{(1+t)ln(1+t)}{t}\right]}^k\hfill \\ \hfill & =\underset{t\to 0}{lim}\frac{d^n}{d{t}^n}\left\{{(1+t)}^k{\left[\frac{ln(1+t)}{t}\right]}^k\right\}\hfill \\ \hfill & =\underset{t\to 0}{lim}\sum _{j=0}^n\left(\genfrac{}{}{0pt}{}{n}{j}\right){\left\{{\left[\frac{ln(1+t)}{t}\right]}^k\right\}}^{\left(j\right)}{\left[{(1+t)}^k\right]}^{(n-j)}.\hfill \end{array}$$

Making use of (18), we arrive at

$$\begin{array}{cc}\hfill \frac{b(n+k,k)}{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}& =\sum _{j=0}^n\left(\genfrac{}{}{0pt}{}{n}{j}\right)\underset{t\to 0}{lim}{\left\{{\left[\frac{ln(1+t)}{t}\right]}^k\right\}}^{\left(j\right)}\underset{t\to 0}{lim}{\left[{(1+t)}^k\right]}^{(n-j)}\hfill \\ \hfill & =\sum _{j=n-k}^n\left(\genfrac{}{}{0pt}{}{n}{j}\right)\frac{s(j+k,k)}{\left(\genfrac{}{}{0pt}{}{j+k}{k}\right)}{\langle k\rangle }_{n-j}\hfill \\ \hfill & =\sum _{j=n-k}^n\left(\genfrac{}{}{0pt}{}{n}{j}\right)\frac{s(j+k,k)}{\left(\genfrac{}{}{0pt}{}{j+k}{k}\right)}\frac{k!}{(k-n+j)!}.\hfill \end{array}$$

Replacing $n+k$ by n and simplifying lead to

$$\begin{array}{cc}\hfill b(n,k)& =\left(\genfrac{}{}{0pt}{}{n}{k}\right)\sum _{j=n-2k}^{n-k}\left(\genfrac{}{}{0pt}{}{n-k}{j}\right)\frac{s(j+k,k)}{\left(\genfrac{}{}{0pt}{}{j+k}{k}\right)}\frac{k!}{(k-(n-k)+j)!}\hfill \\ \hfill & =n!\sum _{j=n-2k}^{n-k}\left(\genfrac{}{}{0pt}{}{k}{n-k-j}\right)\frac{s(j+k,k)}{(j+k)!}\hfill \\ \hfill & =n!\sum _{j=n-k}^n\left(\genfrac{}{}{0pt}{}{k}{n-j}\right)\frac{s(j,k)}{j!}.\hfill \end{array}$$

The Formula (17) is thus proved. The proof of Theorem 2 is complete. □

Remark 3. 

The form of the Formula (17) is better and simpler than the one of the Formula (7) for Comtet’s numbers $b(n,k)$.

7. The Third New Formula for Comtet’s Numbers

In [4] (p. 467, Theorem 1), Lehmer defined the function ${\sigma }_n\left(t\right)$ by

$${\sigma }_n\left(t\right)=\sum _{k=0}^{n}b(n,k)t^k$$

(19)

and proved that the function $\sigma \left(t\right)$ can be generated by

$${(1+x)}^{t(1+x)}=\sum _{n=0}^{\infty }{\sigma }_n\left(t\right)\frac{x^n}{n!}.$$

In this section, we discover closed-form formulas for the nth derivative of the power-exponential function ${(1+x)}^{t(1+x)}$, for the function $\sigma \left(t\right)$, and for the integer sequence $b(n,k)$.

Theorem 3. 

For $n\in {\mathbb{N}}_0$, we have

$$\begin{array}{c}{\left[{(1+x)}^{t(1+x)}\right]}^{\left(n\right)}=n!{(1+x)}^{t(1+x)-n}\sum _{k=0}^n{t}^k{(1+x)}^k\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\times \sum _{j=0}^k\left[\sum _{q=0}^{n-k}\frac{s(q+j,j)}{(q+j)!}\left(\genfrac{}{}{0pt}{}{j}{n-k-q}\right)\right]\frac{{[ln(1+x)]}^{k-j}}{(k-j)!}\end{array}$$

(20)

and

$$\begin{array}{c}{\left[{(1+x)}^{t(1+x)}\right]}^{\left(n\right)}={(1+x)}^{t(1+x)-n}\sum _{k=0}^n{t}^k{(1+x)}^k\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\times \sum _{j=0}^k{(-1)}^j\left(\genfrac{}{}{0pt}{}{j-n-1}{j}\right)b(n-j,k-j){[ln(1+x)]}^j.\end{array}$$

(21)

Consequently, we acquire the closed-form formulas

$${\sigma }_n\left(t\right)=n!\sum _{k=0}^n\left[\sum _{j=k}^n\left(\genfrac{}{}{0pt}{}{k}{n-j}\right)\frac{s(j,k)}{j!}\right]t^k,n\ge 0$$

(22)

and

$$b(n,k)=n!\sum _{j=k}^n\left(\genfrac{}{}{0pt}{}{k}{n-j}\right)\frac{s(j,k)}{j!},n\ge k\ge 0.$$

(23)

Proof. 

For $n\in {\mathbb{N}}_0$, we have

$$\begin{array}{cc}\hfill {\left[{(1+x)}^{t(1+x)}\right]}^{\left(n\right)}& ={\left[e^{t(1+x)ln(1+x)}\right]}^{\left(n\right)}\hfill \\ \hfill & =\sum _{k=0}^n\frac{d^k{e}^{tu}}{d{u}^k}{B}_{n,k}\left(u^{\prime }\left(x\right),u^{″}\left(x\right),\dots ,u^{(n-k)}\left(x\right),u^{(n-k+1)}\left(x\right)\right)\hfill \\ \hfill & =\sum _{k=0}^n{t}^k{e}^{t(1+x)ln(1+x)}{B}_{n,k}(1+ln(1+x),\frac{0!}{1+x},\frac{-1!}{{(1+x)}^2},\hfill \\ \hfill & \frac{2!}{{(1+x)}^3},\dots ,\frac{{(-1)}^{n-k}(n-k-2)!}{{(1+x)}^{n-k-1}},\frac{{(-1)}^{n-k+1}(n-k-1)!}{{(1+x)}^{n-k}}),\hfill \end{array}$$

where $u=u\left(x\right)=(1+x)ln(1+x)$. By the Formulas (5) and (3) in sequence, we acquire

$$\begin{array}{c}{\left[{(1+x)}^{t(1+x)}\right]}^{\left(n\right)}={(1+x)}^{t(1+x)}\sum _{k=0}^n{t}^k\sum _{\ell =0}^k\left(\genfrac{}{}{0pt}{}{n}{\ell }\right){[1+ln(1+x)]}^{\ell }{B}_{n-\ell ,k-\ell }(0,\\ \begin{array}{cc}& \frac{0!}{1+x},\frac{-1!}{{(1+x)}^2},\dots ,\frac{{(-1)}^{n-k}(n-k-2)!}{{(1+x)}^{n-k-1}},\frac{{(-1)}^{n-k+1}(n-k-1)!}{{(1+x)}^{n-k}})\hfill \\ & ={(-1)}^n{(1+x)}^{t(1+x)-n}\sum _{k=0}^n{t}^k{(1+x)}^k\sum _{\ell =0}^k{(-1)}^{\ell }\left(\genfrac{}{}{0pt}{}{n}{\ell }\right){[1+ln(1+x)]}^{\ell }\hfill \\ & \times B_{n-\ell ,k-\ell }(0,0!,1!,2!,\dots ,(n-k-2)!,(n-k-1)!).\hfill \end{array}\end{array}$$

By virtue of the Formula (8) in Lemma 1, we conclude

$$\begin{array}{c}{\left[{(1+x)}^{t(1+x)}\right]}^{\left(n\right)}={(1+x)}^{t(1+x)-n}\sum _{k=0}^n{(-1)}^k{t}^k{(1+x)}^k\sum _{\ell =0}^k{(-1)}^{\ell }\left(\genfrac{}{}{0pt}{}{n}{\ell }\right)\\ \times {[1+ln(1+x)]}^{\ell }(n-\ell )!\sum _{j=0}^{k-\ell }\frac{{(-1)}^j}{(k-\ell -j)!}\sum _{q=0}^{n-k}\frac{s(q+j,j)}{(q+j)!}\left(\genfrac{}{}{0pt}{}{j}{n-k-q}\right)\\ =n!{(1+x)}^{t(1+x)-n}\sum _{k=0}^n{(-1)}^k{t}^k{(1+x)}^k\sum _{j=0}^k\frac{{(-1)}^j}{(k-j)!}\\ \times \left[\sum _{\ell =0}^{k-j}{(-1)}^{\ell }\left(\genfrac{}{}{0pt}{}{k-j}{\ell }\right){[1+ln(1+x)]}^{\ell }\right]\sum _{q=0}^{n-k}\frac{s(q+j,j)}{(q+j)!}\left(\genfrac{}{}{0pt}{}{j}{n-k-q}\right)\\ =n!{(1+x)}^{t(1+x)-n}\sum _{k=0}^n{t}^k{(1+x)}^k\sum _{j=0}^k\frac{{[ln(1+x)]}^{k-j}}{(k-j)!}\sum _{q=0}^{n-k}\frac{s(q+j,j)}{(q+j)!}\left(\genfrac{}{}{0pt}{}{j}{n-k-q}\right).\end{array}$$

The explicit Formula (20) follows.

By virtue of the Formula (9) in Lemma 1, we conclude

$$\begin{array}{c}{\left[{(1+x)}^{t(1+x)}\right]}^{\left(n\right)}={(1+x)}^{t(1+x)-n}\sum _{k=0}^n{(-1)}^k{t}^k{(1+x)}^k\\ \times \sum _{\ell =0}^k{(-1)}^{\ell }\left(\genfrac{}{}{0pt}{}{n}{\ell }\right){[1+ln(1+x)]}^{\ell }\sum _{j=0}^{k-\ell }{(-1)}^j\left(\genfrac{}{}{0pt}{}{n}{k-\ell -j}\right)b(n-k+j,j)\\ ={(1+x)}^{t(1+x)-n}\sum _{k=0}^n{(-1)}^k{t}^k{(1+x)}^k\\ \times \sum _{j=0}^k{(-1)}^j\left[\sum _{\ell =0}^{k-j}{(-1)}^{\ell }\left(\genfrac{}{}{0pt}{}{n}{\ell }\right)\left(\genfrac{}{}{0pt}{}{n}{k-j-\ell }\right){[1+ln(1+x)]}^{\ell }\right]b(n-k+j,j)\\ ={(1+x)}^{t(1+x)-n}\sum _{k=0}^n{(-1)}^k{t}^k{(1+x)}^k\\ \times \sum _{j=0}^k{(-1)}^j\left(\genfrac{}{}{0pt}{}{k-j-n-1}{k-j}\right)b(n-k+j,j){[ln(1+x)]}^{k-j}\\ ={(1+x)}^{t(1+x)-n}\sum _{k=0}^n{t}^k{(1+x)}^k\sum _{j=0}^k{(-1)}^j\left(\genfrac{}{}{0pt}{}{j-n-1}{j}\right)b(n-j,k-j){[ln(1+x)]}^j,\end{array}$$

where we interchanged the order of double summations and used the identity (15). The explicit Formula (21) follows.

Taking $x\to 0$ in (20) gives

$$\begin{array}{cc}\hfill {\sigma }_n\left(t\right)& =\underset{x\to 0}{lim}{\left[{(1+x)}^{t(1+x)}\right]}^{\left(n\right)}\hfill \\ \hfill & =n!\sum _{k=0}^n\left[\sum _{q=0}^{n-k}\frac{s(q+k,k)}{(q+k)!}\left(\genfrac{}{}{0pt}{}{k}{n-k-q}\right)\right]t^k\hfill \\ \hfill & =n!\sum _{k=0}^n\left[\sum _{j=k}^n\frac{s(j,k)}{j!}\left(\genfrac{}{}{0pt}{}{k}{n-j}\right)\right]t^k.\hfill \end{array}$$

The Formula (22) is thus proved. Taking $x\to 0$ in (21) results in (19).

Comparing (19) with (22) and equating coefficients of the terms $t^k$ lead to the nice Formula (23). The proof of Theorem 3 is thus complete. □

Remark 4. 

The form of the Formula (23) is also better and simpler than the one of the Formula (7) for Comtet’s numbers.

8. Conclusions

The identity (9) in Lemma 1 connects partial Bell polynomials with Comtet’s numbers. Those two identities in Lemma 1 are applicable.

The closed-form and explicit Formulas (11) and (12) in Theorem 1 are our main results. Taylor’s series expansions (13) and (14) demonstrate the beauty and symmetry of our main results.

The Formulas (16) and (17) in Theorem 2 and the Formula (23) in Theorem 3 once again demonstrate the beauty and symmetry of our main results.

Comparing the Formula (17) in Theorem 2 and the Formula (23) in Theorem 3 shows some symmetry in form.