# 朗伯W函数

W0(x)的图像，−1/ex ≤ 4

$z = W(z)e^{W(z)}.$

## 微分和积分

$W(x)=\frac{x}{\pi}\int_0^{\pi} \frac{\left(1-v\cot v\right)^2+v^2}{x+v\csc v \cdot e^{-v\cot v}} {\rm{d}}v,|\arg\left(x\right)|<\pi\,$
$W(x)=\int_{-\infty}^{-\frac{1}{e}}{-\frac{1}{\pi}}\Im \left[\frac{{\rm{d}}}{{\rm{d}}x}W(x)\right]\ln \left(1-\frac{z}{x}\right){\rm{d}}x\,$

$x\not\in\left[-\frac{1}{e},0\right],k\in{\mathbb{Z}}\,$,若$x\in\left(-\frac{1}{e},0\right),k=1,\pm2,\pm3,...\,$

$W_k(x)=1+\left(\ln x-1+2k\pi {{\rm{i}}}\right)e^{\frac{{\rm{i}}}{2\pi}\int_0^{\infty}\ln \frac{t-\ln t+\ln x+(2k+1)\pi{\rm{i}}}{t-\ln t+\ln x+(2k-1)\pi{\rm{i}}}\cdot\frac{{\rm{d}}t}{t+1}}=1+\left(\ln x-1+2k\pi {{\rm{i}}}\right)e^{\frac{{\rm{i}}}{2\pi}\int_0^{\infty}\ln \frac{\left(t-\ln t+\ln x\right)^2+\left(4k^2-1\right)\pi^2+2\pi\left(t-\ln t+\ln x\right){\rm{i}}}{\left(t-\ln t+\ln x\right)^2+\left(2k-1\right)^2\pi^2}\cdot\frac{{\rm{d}}t}{t+1}}\,$

$W_k(x)=1+\left(\ln x-1+2k\pi {{\rm{i}}}\right)e^{\frac{{\rm{i}}}{2\pi}\int_0^\infty\left[\frac{1}{2}\ln\frac{\left(t-\ln t+\ln x\right)^2+\left(2k+1\right)^2\pi^2}{\left(t-\ln t+\ln x\right)^2+\left(2k-1\right)^2\pi^2}+{\rm{i}}\arctan\frac{2\pi\left(t-\ln t+\ln x\right)}{\left(t-\ln t+\ln x\right)^2+\left(4k^2-1\right)\pi^2}\right]\cdot\frac{{\rm{d}}t}{t+1}}$
${}_{W_k(x)=1+\frac{\left(\ln x-1\right)\cos\frac{1}{4\pi}\int_0^{\infty}\ln\frac{\left(t-\ln t+\ln x\right)^2+\left(2k+1\right)^2\pi^2}{\left(t-\ln t+\ln x\right)^2+\left(2k-1\right)^2\pi^2}\cdot\frac{{\rm{d}}t}{t+1}-2k\pi\sin\frac{1}{4\pi}\int_0^{\infty}\ln\frac{\left(t-\ln t+\ln x\right)^2+\left(2k+1\right)^2\pi^2}{\left(t-\ln t+\ln x\right)^2+\left(2k-1\right)^2\pi^2}\cdot\frac{{\rm{d}}t}{t+1}+{\rm{i}}\left[\left(\ln x-1\right)\sin\frac{1}{4\pi}\int_0^{\infty}\ln\frac{\left(t-\ln t+\ln x\right)^2+\left(2k+1\right)^2\pi^2}{\left(t-\ln t+\ln x\right)^2+\left(2k-1\right)^2\pi^2}\cdot\frac{{\rm{d}}t}{t+1}+2k\pi\cos\frac{1}{4\pi}\int_0^{\infty}\ln\frac{\left(t-\ln t+\ln x\right)^2+\left(2k+1\right)^2\pi^2}{\left(t-\ln t+\ln x\right)^2+\left(2k-1\right)^2\pi^2}\cdot\frac{{\rm{d}}t}{t+1}\right]}{e^{\frac{1}{2\pi}\int_0^\infty\arctan\frac{2\pi\left(t-\ln t+\ln x\right)}{\left(t-\ln t+\ln x\right)^2+\left(4k^2-1\right)\pi^2}\cdot\frac{\rm{d}t}{t+1}}}}$

${}_{W_k(x)=u+v{\rm{i}},x=t+s{\rm{i}}}$,则有${}_{\left(u+v{\rm{i}}\right)e^{u+v{\rm{i}}}=t+s{\rm{i}}}$,展开分离出实部和虚部，

${}_{e^u\left(u\cos v-v\sin v\right)=t,e^u\left(u\sin v+v\cos v\right)=s}$,当${}_{s=0}$时，易知${}_{u=-v\cot v}$

${}_{W_k(x)=\frac{\left(1-\ln x\right)\sin\frac{1}{4\pi}\int_0^{\infty}\ln\frac{\left(t-\ln t+\ln x\right)^2+\left(2k+1\right)^2\pi^2}{\left(t-\ln t+\ln x\right)^2+\left(2k-1\right)^2\pi^2}\cdot\frac{{\rm{d}}t}{t+1}-2k\pi\cos\frac{1}{4\pi}\int_0^{\infty}\ln\frac{\left(t-\ln t+\ln x\right)^2+\left(2k+1\right)^2\pi^2}{\left(t-\ln t+\ln x\right)^2+\left(2k-1\right)^2\pi^2}\cdot\frac{{\rm{d}}t}{t+1}}{e^{\frac{1}{2\pi}\int_0^\infty\arctan\frac{2\pi\left(t-\ln t+\ln x\right)}{\left(t-\ln t+\ln x\right)^2+\left(4k^2-1\right)\pi^2}\cdot\frac{\rm{d}t}{t+1}}}\cot\frac{\left(\ln x-1\right)\sin\frac{1}{4\pi}\int_0^{\infty}\ln\frac{\left(t-\ln t+\ln x\right)^2+\left(2k+1\right)^2\pi^2}{\left(t-\ln t+\ln x\right)^2+\left(2k-1\right)^2\pi^2}\cdot\frac{{\rm{d}}t}{t+1}+2k\pi\cos\frac{1}{4\pi}\int_0^{\infty}\ln\frac{\left(t-\ln t+\ln x\right)^2+\left(2k+1\right)^2\pi^2}{\left(t-\ln t+\ln x\right)^2+\left(2k-1\right)^2\pi^2}\cdot\frac{{\rm{d}}t}{t+1}}{e^{\frac{1}{2\pi}\int_0^\infty\arctan\frac{2\pi\left(t-\ln t+\ln x\right)}{\left(t-\ln t+\ln x\right)^2+\left(4k^2-1\right)\pi^2}\cdot\frac{\rm{d}t}{t+1}}} +\frac{\left(\ln x-1\right)\sin\frac{1}{4\pi}\int_0^{\infty}\ln\frac{\left(t-\ln t+\ln x\right)^2+\left(2k+1\right)^2\pi^2}{\left(t-\ln t+\ln x\right)^2+\left(2k-1\right)^2\pi^2}\cdot\frac{{\rm{d}}t}{t+1}+2k\pi\cos\frac{1}{4\pi}\int_0^{\infty}\ln\frac{\left(t-\ln t+\ln x\right)^2+\left(2k+1\right)^2\pi^2}{\left(t-\ln t+\ln x\right)^2+\left(2k-1\right)^2\pi^2}\cdot\frac{{\rm{d}}t}{t+1}}{e^{\frac{1}{2\pi}\int_0^\infty\arctan\frac{2\pi\left(t-\ln t+\ln x\right)}{\left(t-\ln t+\ln x\right)^2+\left(4k^2-1\right)\pi^2}\cdot\frac{\rm{d}t}{t+1}}}{\rm{i}},}$
$W_0(x)=1+\left(\ln x-1\right)e^{-\frac{1}{\pi}\int_0^\infty\arg\left(t-\ln t+\ln x+\pi{\rm{i}}\right)\cdot\frac{\rm{d}t}{t+1}},x>0$

$x>\frac{1}{e}$，上式还可化为$W_0(x)=1+\left(\ln x-1\right)e^{-\frac{1}{\pi}\int_0^\infty\arctan\frac{\pi}{t-\ln t+\ln x}\cdot\frac{\rm{d}t}{t+1}}$

$z\left[1+W(z)\right]\frac{{\rm{d}}}{{\rm{d}}z}W(z)=W(z)$$z\neq -\frac{1}{e}\,,$

$\frac{{\rm{d}}}{{\rm{d}}z}W(z)=\frac{ W(z) }{z\left[1 + W(z)\right]}$$z\neq -\frac{1}{e} \,.$

$\int W(x) {\rm{d}}x = x \left[ W(x)+ \frac{1}{W (x) }-1 \right] + C$
$\int_0^1 W(x) {\rm{d}}x = \Omega+\frac{1}{\Omega} -2\approx 0.330336$

## 性质

$1\,$、函数$y=z^{z^{z^{z^{z^{z^{z^{\cdots}}}}}}} \,$

$2\,$、若$z>0 \,$，则$\ln W(z)=\ln z-W(z)\,$

## 泰勒级数

$W_0 \,$$x=0 \,$的泰勒级数如下：

$W_0 (x) = \sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!}\ x^n = x - x^2 + \frac{3}{2}x^3 - \frac{8}{3}x^4 + \frac{125}{24}x^5 - \cdots$

## 加法定理

$W(x)+W(y)=W\left[\frac{xy}{W(x)}+\frac{xy}{W(y)}\right]\,$
$x>0,y>0\,$

## 複數值

$\Re\left[W(x+y{\rm{i}})\right]=\sum_{k=1}^{\infty}\frac{(-k)^{k-1}}{k!}\sqrt{(x^2+y^2)^k}\cos \left(k\arctan\frac{x}{y}\right)\,$ , $x^2+y^2<\frac{1}{e^2}\,$

$\Im\left[W(x+y{\rm{i}})\right]=\sum_{k=1}^{\infty}\frac{(-k)^{k-1}}{k!}\sqrt{(x^2+y^2)^k}\sin \left(k\arctan\frac{x}{y}\right)\,$, $x^2+y^2<\frac{1}{e^2}\,$

$|W(x+y{\rm{i}})|=W(\sqrt{x+y})\,$

$\arg\left[W(x+y{\rm{i}})\right]=\sum_{k=1}^{\infty}\frac{(-k)^{k-1}}{k!}\arctan\left[\cot(k\arctan\frac{x}{y})\right]\,$, $x^2+y^2<\frac{1}{e^2}\,$

$\overline{W(x+y{\rm{i}})}=\sum_{k=1}^{\infty}\frac{(-k)^{k-1}}{k!}\sqrt{(x^2+y^2)^k}\left[\cos \left(k\arctan\frac{x}{y}\right)-{\rm{i}}\sin \left(k\arctan\frac{x}{y}\right)\right]\,$, $x^2+y^2<\frac{1}{e^2}\,$

## 特殊值

${}_{W\left(-\frac{\pi}{2}\right) = \frac{\pi}{2}{\rm{i}}}$
${}_{W\left(-\frac{\ln 2}{2}\right)= -\ln 2}$
${}_{W\left(-{1\over e}\right) = -1}$
${}_{W\left(1\right) = \Omega=\frac{1}{\int_{-\infty}^{\infty}\frac{{\rm{d}}x}{(e^x-x)^2+\pi^2}}-1\approx 0.56714329\dots}\,$欧米加常数
${}_{W(e) = 1}\,$
${}_{W(e^{e+1}) = e}\,$
${}_{W\left(\frac{1}{e^{1- \frac{1}{e}}}\right)= \frac{1}{e}}\,$
${}_{W(-\frac{1}{e})=-1}\,$
${}_{W({\pi}e^{\pi})=\pi}\,$
${}_{W(k{\ln k})={\ln k}}\,$ ${}_{(k>0)}\,$
${}_{W({\rm{i}}\pi)=-{\rm{i}}\pi}\,$
${}_{W(-{\rm{i}}\pi)={\rm{i}}\pi}\,$
${}_{W({\rm{i}}\cos1-\sin1)={\rm{i}}}\,$
${}_{W(-\frac{3}{2}{\pi})=-\frac{3}{2}{\pi}{\rm{i}}}\,$
${}_{W(-\frac{\sqrt[7]{8}}{7}{\ln 2})=-\frac{32}{7}{\ln 2}}\,$
${}_{W(-\frac{\sqrt{3}}{54}{\ln 3})=-\frac{9}{2}{\ln 3}}\,$
${}_{W(-\frac{\ln 2}{4})=-4{\ln 2}}\,$
${}_{W\left(-1\right)=\frac{e^{\frac{1}{2\pi}\int_0^\infty{1\over t+1}\arctan{2\pi\over t-\ln t}{\rm{d}}t}-\cos\left[\frac{1}{4\pi}\int_0^\infty{1\over t+1}\ln{\left(t-\ln t\right)^2\over 4\pi^2+\left(t-\ln t\right)^2}{\rm{d}}t\right]+\pi\sin\left[\frac{1}{4\pi}\int_0^\infty{1\over t+1}\ln{\left(t-\ln t\right)^2\over 4\pi^2+\left(t-\ln t\right)^2}{\rm{d}}t\right]-{\rm{i}}\left\{\pi\cos\left[\frac{1}{4\pi}\int_0^\infty{1\over t+1}\ln{\left(t-\ln t\right)^2\over 4\pi^2+\left(t-\ln t\right)^2}{\rm{d}}t\right]+\sin\left[\frac{1}{4\pi}\int_0^\infty{1\over t+1}\ln{\left(t-\ln t\right)^2\over 4\pi^2+\left(t-\ln t\right)^2}{\rm{d}}t\right]\right\}}{e^{\frac{1}{2\pi}\int_0^\infty{1\over t+1}\arctan{2\pi\over t-\ln t}{\rm{d}}t}}\approx -0.31813-1.33723{\rm{i}}} \,$
${}_{W(-\frac{\ln k}{k})=-\ln k}\,$
${}_{W\left[-\frac{\ln (x+1)}{x(x+1)^{\frac{1}{x}}}\right]=-\frac{x+1}{x}\ln (x+1)>,-1

## 应用

### 例子

$2^t = 5 t\,$
$\Rightarrow 1 = \frac{5 t}{2^t}\,$
$\Rightarrow 1 = 5 t \, e^{-t \ln 2}\,$
$\Rightarrow \frac{1}{5} = t \, e^{-t \ln 2}\,$
$\Rightarrow -\frac{\ln 2}{5} = ( - \, t \, \ln 2 ) \, e^{-t \ln 2}\,$
$\Rightarrow -t \ln 2 = W_k \left (-\frac{\ln 2}{5} \right )\,$
$\Rightarrow t = -\frac{ W_k \left ( -\frac{\ln 2}{5} \right )}{\ln 2}\,$

$Q^{a x + b} = c x + d \,$

$Q > 0 \and Q \neq 1\and c \neq 0$

$\frac{a}{c} Q^{b-\frac{ad}{c}}= \left(ax + \frac{ad}{c}\right)Q^{-\left(ax+\frac{ad}{c}\right)} \,$

$t = a x + \frac{a d}{c}$

$t Q^{-t} = \frac{a}{c} Q^{b-\frac{a d}{c}}$

$t{\ln Q} \cdot e^{-t \ln Q}= \frac{a}{c} Q^{b-\frac{a d}{c}}$

$t{\ln Q}=-W_k\left(-\frac{a\ln Q}{c}\,Q^{b-\frac{a d}{c}}\right)$

$\left(ax+\frac{ad}{c}\right){\ln Q}=-W_k\left(-\frac{a\ln Q}{c}\,Q^{b-\frac{a d}{c}}\right)$

$x = -\frac{W_k\left(-\frac{a\ln Q}{c}\,Q^{b-\frac{a d}{c}}\right)}{a\ln Q} - \frac{d}{c}$

$-\frac{a\ln Q}{c}\,Q^{b-\frac{a d}{c}} \in \left(-\infty ,-\frac{1}{e} \right)$,

$x = -\frac{W_k\left(-\frac{a\ln Q}{c}\,Q^{b-\frac{a d}{c}}\right)}{a\ln Q} - \frac{d}{c}$

${\rm{W}}_{-1}\left(-\frac{a\ln Q}{c}\,Q^{b-\frac{a d}{c}}\right)$

$x_1=-\frac{W\left(-\frac{a\ln Q}{c}\,Q^{b-\frac{a d}{c}}\right)}{a\ln Q} - \frac{d}{c}$

$x_2=-\frac{{\rm{W}}_{-1}\left(-\frac{a\ln Q}{c}\,Q^{b-\frac{a d}{c}}\right)}{a\ln Q} - \frac{d}{c}$

$x^x={\mathrm{t}}\, ,$

$x=\frac{\ln{\rm{t}}}{W(\ln {\rm{t}})}\,$

$x=\exp\left(W_k\left[\ln({\rm{t}})\right]\right).$

$x \log_b {x} = a \,$

$x = \frac{a {\ln b}}{W_k\left(a {\ln b}\right)}$

$x^a-b^x=0\,$
$a > 0 \,$ : $b > 0 \,$ : $x > 0 \,$

$a \ln x=x \ln b \,$
$\frac{\ln x}{x}=\frac{\ln b}{a}\,$
$e^{\frac{\ln x}{x}}=e^{\frac{\ln b}{a}} \,$
$x^{\frac{1}{x}}=b^{\frac{1}{a}}\,$

$\left(\frac{1}{x} \right)^{\frac{1}{x}}=b^{-\frac{1}{a}}\,$
$\frac{1}{x} =-\frac{\ln b}{aW\left(-\frac{1}{a} \ln b\right)}\,$

$(ax+b)^n=u^{cx+d} \,$

$x+\frac{b}{a}=\frac{u^{\frac{c}{n}x+\frac{d}{n}}}{a}\left(\cos\frac{2k\pi}{n}+{\rm{i}}\sin\frac{2k\pi}{n}\right)\,$

$u=e^{\ln u}\,$

$x+\frac{b}{a}=\frac{e^{\frac{c\ln u}{n}x+\frac{d\ln u}{n}}}{a}\left(\cos\frac{2k\pi}{n}+{\rm{i}}\sin\frac{2k\pi}{n}\right)\,$

$-\frac{c\ln u}{n}u^{-\frac{c}{n}x-\frac{cb}{na}}\,$

$\left(-\frac{c\ln u}{n}x-\frac{cb\ln u}{na}\right)e^{-\frac{c\ln u}{n}x-\frac{cb\ln u}{na}}=-\frac{c\ln u}{na}u^{\frac{d}{n}-\frac{cb}{na}}\left(\cos\frac{2k\pi}{n}+{\rm{i}}\sin\frac{2k\pi}{n}\right)\,$

$x_k=-\frac{n}{c\ln u}W_k\left[-\frac{c\ln u}{na}u^{\frac{d}{n}-\frac{cb}{na}}\left(\cos\frac{2k\pi}{n}+{\rm{i}}\sin\frac{2k\pi}{n}\right)\right]-\frac{b}{a}\,$

$k\in{\mathbb{Z}}\,$

## 一般化

$e^{-c x} = a_o (x-r) ~~\quad\qquad\qquad\qquad\qquad(1)$

Lambert W 函數之一般化[1][2][3] 包括:

• 一項在低維空間內廣義相對論量子力學的應用（量子引力），實際上一種以前未知的 連結 於此二區域中，如 “Journal of Classical and Quantum Gravity”[4] 所示其 (1) 的右邊式現為二維多項式 x：
$e^{-c x} = a_o (x-r_1 ) (x-r_2 ) ~~\qquad\qquad(2)$

• 量子力學的一特例特徵能的分析解三體問題，亦即（三維）氢分子離子[5]於此 (1)（或 (2)）的右手邊現為無限級數多項式之比於 x
$e^{-c x} = a_o \frac{\prod_{i=1}^{\infty} (x-r_i )}{ \prod_{i=1}^{\infty} (x-s_i)} \qquad \qquad\qquad(3)$

Lambert "W" 函數於基礎物理問題之應用並未完全即使標準情況如 (1) 最近在原子，分子，與光學物理領域可見。[6]

## 计算

W函数可以用以下的递推关系算出：

$w_{j+1}=w_j-\frac{w_j e^{w_j}-z}{e^{w_j}(w_j+1)-\frac{(w_j+2)(w_je^{w_j}-z)} {2w_j+2}}$

## 参考来源

1. ^ T.C. Scott and R.B. Mann, General Relativity and Quantum Mechanics: Towards a Generalization of the Lambert W Function, AAECC (Applicable Algebra in Engineering, Communication and Computing), vol. 17, no. 1, (April 2006), pp.41-47, [1]; Arxiv [2]
2. ^ T.C. Scott, G. Fee and J. Grotendorst, "Asymptotic series of Generalized Lambert W Function", SIGSAM, vol. 47, no. 3, (September 2013), pp. 75-83
3. ^ T.C. Scott, G. Fee, J. Grotendorst and W.Z. Zhang, "Numerics of the Generalized Lambert W Function", SIGSAM, vol. 48, no. 2, (June 2014), pp. 42-56
4. ^ P.S. Farrugia, R.B. Mann, and T.C. Scott, N-body Gravity and the Schrödinger Equation, Class. Quantum Grav. vol. 24, (2007), pp. 4647-4659, [3]; Arxiv [4]
5. ^ T.C. Scott, M. Aubert-Frécon and J. Grotendorst, New Approach for the Electronic Energies of the Hydrogen Molecular Ion, Chem. Phys. vol. 324, (2006), pp. 323-338, [5]; Arxiv [6]
6. ^ T.C. Scott, A. Lüchow, D. Bressanini and J.D. Morgan III, The Nodal Surfaces of Helium Atom Eigenfunctions, Phys. Rev. A 75, (2007), p. 060101, [7]