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A Brief Look into the Lambert W Function
对朗伯 W 函数的简要探讨
Thomas P. Dence
Ashland University, Ashland, USA
Email: tdence@ashland.edu
Received December 4, 2012; revised March 14, 2013; accepted March 22, 2013
Copyright © 2013 Thomas P. Dence. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
ABSTRACT
摘要
The Lambert W function has its origin traced back 250 years, but it's just been in the past several decades when some of the real usefulness of the function has been brought to the attention of the scientific community.
朗伯 W 函数的起源可追溯至 250 年前，但直到近几十年，该函数的部分实际应用价值才引起科学界的关注。
Keywords: Lagrange Inversion Theorem; Infinite Tower of Exponents
关键词：拉格朗日反演定理；无穷指数塔
1. Introduction
1. 引言
The Lambert W function, named after Johann Heinrich Lambert [1], is a standard function in both Mathematica, where it's called Product log (x), and in Maple, where you can use both Lambert w ( x ) w(x) w(x) or Lambert W ( 0 , x ) W(0, x) W(0,x). The zero in this latter expression denotes the principal branch of the inverse of x e x x e^{x} xex. The actual usage of the letter W has a rather vague origin. One source [2] attributes it to some earlier papers on the subject that wrote the function, since Euler had studied the equation x e x = α x e^{x} = \alpha xex=α [4] (although Euler credits Lambert as studying the equation first [5]), but they decided Euler had enough items attached to his name! Another source [3] attributes the W to honor the British mathematician Sir Edward M. Wright (famous co-author with G. H. Hardy of An Introduction to the Theory of Numbers ) who did a lot of pioneering work with the function. Finally, Robert Corless and David Jeffrey of the University of Western Ontario have written, during the past several decades, a number of journal articles on the function. Their paper in 1996, in collaboration with Gaston Gonnet, David Hare, and Donald Knuth, was where Lambert's name got attached to the function [2]. It could have been coined the Euler W function, since Euler had studied the equation x e x = α x e^{x} = \alpha xex=α [4] (although Euler credits Lambert as studying the equation first [5]), but they decided Euler had enough items attached to his name!
朗伯 W 函数以约翰·海因里希·朗伯命名[1]，是两款数学软件中的标准函数：在 Mathematica 中，它被称为 Product log(x)；在 Maple 中，可表示为 Lambert w ( x ) w(x) w(x) 或 Lambert W ( 0 , x ) W(0, x) W(0,x)。后一种表达式中的"0"表示 x e x x e^{x} xex 反函数的主分支。字母"W"的实际使用起源相当模糊。有文献[2] 指出，这一符号源于早期研究该函数的论文------欧拉曾研究过方程 x e x = α x e^{x} = \alpha xex=α [4]（尽管欧拉认为朗伯是首个研究该方程的学者[5]），但研究者们认为欧拉名下已关联足够多的数学概念，故未将该函数命名为"欧拉函数"；另有文献[3] 认为，"W"是为纪念英国数学家爱德华·M·赖特爵士（他与 G·H·哈代合著《数论导论》，且对朗伯 W 函数开展了大量开创性研究）。此外，加拿大西安大略大学的罗伯特·科莱斯与戴维·杰弗里在近几十年间发表了多篇关于该函数的期刊论文：1996 年，他们与加斯东·冈内特、戴维·黑尔及唐纳德·克努特合作撰写的论文，首次将"朗伯"之名与该函数正式关联[2]。
2. Definition
2. 定义
The exponential function y = x e x y = x e^{x} y=xex is defined for all real x, but has a codomain of y ≥ − 1 e y ≥ -\frac{1}{e} y≥−e1. This function (Figure 1(a)) is the product of two elementary functions, each defined on the entire real line, and each being one-to-one; but the product is not injective. Consequently, if we restrict the domain to x ≥ − 1 x ≥ -1 x≥−1, then x e x x e^{x} xex will possess an inverse, which is a function, and it's this function that is now known as the (principal) Lambert W function (Figure 1(b)), written as W ( x ) W(x) W(x). An alternative branch for W would be defined for that portion of x e x x e^{x} xex when x ≤ − 1 x ≤ -1 x≤−1. We won't consider that situation in this article.
指数函数 y = x e x y = x e^{x} y=xex 对所有实数 x x x均有定义，但其值域为 y ≥ − 1 e y ≥ -\frac{1}{e} y≥−e1。该函数（图 1(a)）是两个基本函数的乘积：这两个基本函数均在全体实数域上有定义且均为一一映射，但它们的乘积并非单射。因此，若将定义域限制为 x ≥ − 1 x ≥ -1 x≥−1，则 x e x x e^{x} xex 会存在一个反函数，这个反函数便是如今的（主分支）朗伯 W 函数（图 1(b)），记为 W ( x ) W(x) W(x)。W 函数的另一分支可由 x ≤ − 1 x ≤ -1 x≤−1 时的 x e x x e^{x} xex 定义，但本文不对此分支展开讨论。
Figure 1. (a) Graph of f ( x ) = x e x f (x) = x e^x f(x)=xex; (b) Graph of W ( x ) W (x) W(x), x ∈ [ − 1 / e , + ∞ ) x \in [-1/e, +\infty) x∈[−1/e,+∞).
图 1. (a) 函数 f ( x ) = x e x f (x) = x e^x f(x)=xex 的图像；(b) 函数 W ( x ) W (x) W(x) 的图像（ x ∈ [ − 1 / e , + ∞ ) x \in [-1/e, +\infty) x∈[−1/e,+∞)）。
Several function values of W are easy to compute, such as W ( − 1 e ) = − 1 W(-\frac{1}{e}) = -1 W(−e1)=−1, W ( 0 ) = 0 W(0) = 0 W(0)=0, W ( − 1 2 ln 2 ) = − ln 2 W(-\frac{1}{2} \ln 2) = -\ln 2 W(−21ln2)=−ln2 and W ( e ) = 1 W(e) = 1 W(e)=1. The value of W ( 1 ) W(1) W(1), known as the omega constant, has the approximate value 0.567143. The number W ( 1 ) W(1) W(1) is, in some sense, a distant cousin of the golden ratio ϕ \phi ϕ, since 1 ϕ \frac{1}{\phi} ϕ1 is a solution to 1 x = x + 1 \frac{1}{x} = x + 1 x1=x+1, and W ( 1 ) W(1) W(1) is the solution to 1 x = e x \frac{1}{x} = e^{x} x1=ex, and x + 1 x + 1 x+1 is the linear Maclaurin approximation to e x e^{x} ex (Figure 2). Since W is the inverse of x e x x e^{x} xex, it follows that W ( 1 ) e W ( 1 ) = 1 W(1) e^{W(1)} = 1 W(1)eW(1)=1 and that the slope of the curve in Figure 1(b) at the point ( 1 , W ( 1 ) ) (1, W(1)) (1,W(1)) is [ 1 + e W ( 1 ) ] − 1 ≈ 0.6381 [1 + e^{W(1)}]^{-1} ≈ 0.6381 [1+eW(1)]−1≈0.6381.
W 函数的部分函数值易于计算，例如 W ( − 1 e ) = − 1 W(-\frac{1}{e}) = -1 W(−e1)=−1、 W ( 0 ) = 0 W(0) = 0 W(0)=0、 W ( − 1 2 ln 2 ) = − ln 2 W(-\frac{1}{2} \ln 2) = -\ln 2 W(−21ln2)=−ln2 及 W ( e ) = 1 W(e) = 1 W(e)=1。其中 W ( 1 ) W(1) W(1) 被称为"欧米伽常数"，近似值为 0.567143。从某种意义上看， W ( 1 ) W(1) W(1) 是黄金比例 ϕ \phi ϕ 的"远亲"：因为 1 ϕ \frac{1}{\phi} ϕ1 是方程 1 x = x + 1 \frac{1}{x} = x + 1 x1=x+1 的解，而 W ( 1 ) W(1) W(1) 是方程 1 x = e x \frac{1}{x} = e^{x} x1=ex 的解，且 x + 1 x + 1 x+1 恰为 e x e^{x} ex 的线性麦克劳林近似（图 2）。由于 W 函数是 x e x x e^{x} xex 的反函数，由此可推出 W ( 1 ) e W ( 1 ) = 1 W(1) e^{W(1)} = 1 W(1)eW(1)=1，且图 1(b) 中曲线在点 ( 1 , W ( 1 ) ) (1, W(1)) (1,W(1)) 处的斜率为 [ 1 + e W ( 1 ) ] − 1 ≈ 0.6381 [1 + e^{W(1)}]^{-1} ≈ 0.6381 [1+eW(1)]−1≈0.6381。
3. Computation
3. 计算方法
A natural question is how to compute arbitrary values of W ( x ) W(x) W(x). One result, from the Lagrange inversion theorem, asserts that the Lambert W function has the Taylor series expansion [6,7]
一个自然的问题是：如何计算 W ( x ) W(x) W(x) 的任意取值？根据拉格朗日反演定理，朗伯 W 函数具有如下泰勒级数展开式[6,7]：
W ( x ) = ∑ n = 1 ∞ ( − n ) n − 1 n ! x n , (1) W(x) = \sum_{n=1}^{\infty} \frac{(-n)^{n-1}}{n!} x^{n},\tag{1} W(x)=n=1∑∞n!(−n)n−1xn,(1)
which, unfortunately, has a radius of convergence of merely 1 e \frac{1}{e} e1. Since the denominator n ! n! n! grows rapidly, it's advantageous to write the series with the coefficients defined recursively as
但遗憾的是，该级数的收敛半径仅为 1 e \frac{1}{e} e1。由于分母 n ! n! n! 增长迅速，将级数改写为系数递归定义的形式会更有利：即
W ( x ) = ∑ n = 1 ∞ c n x n W(x) = \sum_{n=1}^{\infty} c_{n} x^{n} W(x)=n=1∑∞cnxn,
with c 1 = 1 c_{1} = 1 c1=1, c n = − n − 1 n c n − 1 c_{n} = -\frac{n - 1}{n} c_{n - 1} cn=−nn−1cn−1 for all n ≥ 2 n ≥ 2 n≥2.
其中 c 1 = 1 c_{1} = 1 c1=1，且对所有 n ≥ 2 n ≥ 2 n≥2 有 c n = − n − 1 n c n − 1 c_{n} = -\frac{n - 1}{n} c_{n - 1} cn=−nn−1cn−1。
This recursion lends itself to easy programming evaluation. Testing this, with say a series of 150 terms (which is plenty, considering that ∣ x ∣ ≤ 1 e |x| ≤ \frac{1}{e} ∣x∣≤e1), with x = − 1 2 ln 2 ≈ − 0.3465736 x = -\frac{1}{2} \ln 2 ≈ -0.3465736 x=−21ln2≈−0.3465736, we obtain a partial sum value of S = − 0.69314684 S = -0.69314684 S=−0.69314684, which differs from the exact value of − ln 2 -\ln 2 −ln2 by 0.0000003. We also note that ∣ x ∣ < 1 e |x| < \frac{1}{e} ∣x∣<e1, so the use of the series is justified. On the other hand, a TI-graphing calculator returns "overflow error" if we try to determine W ( 1 ) W(1) W(1), primarily since the coefficients grow rapidly.
这种递归形式便于通过编程计算。例如，取 150 项级数（考虑到 ∣ x ∣ ≤ 1 e |x| ≤ \frac{1}{e} ∣x∣≤e1，150 项已足够），当 x = − 1 2 ln 2 ≈ − 0.3465736 x = -\frac{1}{2} \ln 2 ≈ -0.3465736 x=−21ln2≈−0.3465736 时，计算得到部分和 S = − 0.69314684 S = -0.69314684 S=−0.69314684，与精确值 − ln 2 -\ln 2 −ln2 的差值仅为 0.0000003。同时，由于 ∣ x ∣ < 1 e |x| < \frac{1}{e} ∣x∣<e1，使用该级数计算是合理的。反之，若尝试用 TI 绘图计算器计算 W ( 1 ) W(1) W(1)，会出现"溢出错误"，这主要是因为级数系数增长过快。
Suppose that ∣ x ∣ > 1 e |x| > \frac{1}{e} ∣x∣>e1 and we wish to compute W ( x ) W(x) W(x). One possibility is the series
若 ∣ x ∣ > 1 e |x| > \frac{1}{e} ∣x∣>e1 且需计算 W ( x ) W(x) W(x)，可考虑如下级数：
W ( x ) = ln x − ln ( ln x ) + ∑ n = 1 ∞ ( − 1 ) n ( ln x ) n a n ( x ) , (2) W(x) = \ln x - \ln (\ln x) + \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(\ln x)^{n}} a_{n}(x),\tag{2} W(x)=lnx−ln(lnx)+n=1∑∞(lnx)n(−1)nan(x),(2)
where 其中
a n ( x ) = ∑ m = 1 n ( − 1 ) m S ( n , n − m + 1 ) [ ln ( ln x ) ] m m ! a_{n}(x) = \sum_{m=1}^{n} (-1)^{m} S(n, n - m + 1) \frac{[\ln (\ln x)]^{m}}{m!} an(x)=m=1∑n(−1)mS(n,n−m+1)m![ln(lnx)]m
and S ( n , k ) S(n, k) S(n,k) denotes a Stirling number of the first kind [3].
且 S ( n , k ) S(n, k) S(n,k) 表示第一类斯特林数[3]。
The series (2) is somewhat impractical to use because of the difficulty in determining a n ( x ) a_{n}(x) an(x); it turns out to be more useful to employ some standard numerical schemes for approximating W ( x ) W(x) W(x).
由于确定 a n ( x ) a_{n}(x) an(x) 存在困难，该级数（2）的实用性有限；事实上，采用标准数值方法近似计算 W ( x ) W(x) W(x) 会更有效。
First, setting W ( x ) = y W(x) = y W(x)=y, we need to solve y e y = x y e^{y} = x yey=x. Defining the function g g g by g ( y ) = y e y − x g(y) = y e^{y} - x g(y)=yey−x, we use Newton's method to approximate y y y in g ( y ) = 0 g(y) = 0 g(y)=0. This gives
首先，令 W ( x ) = y W(x) = y W(x)=y，则需求解方程 y e y = x y e^{y} = x yey=x。定义函数 g ( y ) = y e y − x g(y) = y e^{y} - x g(y)=yey−x，采用牛顿法近似求解 g ( y ) = 0 g(y) = 0 g(y)=0 的根，迭代公式为：
y n + 1 = y n − g ( y n ) g ′ ( y n ) = y n 2 + x e − y n y n + 1 . y_{n+1} = y_{n} - \frac{g(y_{n})}{g'(y_{n})} = \frac{y_{n}^{2} + x e^{-y_{n}}}{y_{n} + 1}. yn+1=yn−g′(yn)g(yn)=yn+1yn2+xe−yn..
To determine W ( 2 ) W(2) W(2), for example, starting with an initial approximate of y 0 = 0 y_{0} = 0 y0=0, after 7 more iterations we get y 8 = 0.852605502 y_{8} = 0.852605502 y8=0.852605502, which is an excellent approximation to W ( 2 ) W(2) W(2) because y 8 e y 8 y_{8} e^{y_{8}} y8ey8 returns 2 on the calculator. If x x x is a relatively small number, then an initial approximate of 0 will suffice for the algorithm; but if x x x is large, then ln x \ln x lnx can be chosen for y 0 y_{0} y0. For instance, if x = 10 x = 10 x=10, choose y 0 = ln 10 ≈ 2.302585 y_{0} = \ln 10 ≈ 2.302585 y0=ln10≈2.302585, and after 5 iterations we get W ( 10 ) ≈ 1.745528003 W(10) ≈ 1.745528003 W(10)≈1.745528003.
以计算 W ( 2 ) W(2) W(2) 为例：取初始近似值 y 0 = 0 y_{0} = 0 y0=0，经过 7 次迭代后得到 y 8 = 0.852605502 y_{8} = 0.852605502 y8=0.852605502------该值是 W ( 2 ) W(2) W(2) 的优良近似，因为计算器计算 y 8 e y 8 y_{8} e^{y_{8}} y8ey8 的结果为 2。若 x x x 是较小的数，取初始近似值 0 即可满足算法需求；若 x x x 较大，则可选择 y 0 = ln x y_{0} = \ln x y0=lnx。例如，当 x = 10 x = 10 x=10 时，取 y 0 = ln 10 ≈ 2.302585 y_{0} = \ln 10 ≈ 2.302585 y0=ln10≈2.302585，经过 5 次迭代后得到 W ( 10 ) ≈ 1.745528003 W(10) ≈ 1.745528003 W(10)≈1.745528003。
Newton's method is a favorite iteration scheme for many because of its simplicity, though the convergence, quadratic in general, is typically relatively slow. A faster choice is furnished by Halley's method (of Halley's comet fame), which produces cubic convergence, and happens to be the choice implemented by the software Maple; this scheme gives [8]
牛顿法因原理简单成为许多人的首选迭代方法，但其收敛性通常为二次收敛，速度相对较慢。更快的选择是哈雷法（因哈雷彗星而闻名），该方法具有三次收敛性，且恰为软件 Maple 所采用的计算方案[8]，其迭代公式为：
y n + 1 = y n − y n e y n − x ( y n + 1 ) e y n − ( y n + 2 ) ( y n e y n − x ) 2 y n + 2 . y_{n+1} = y_n - \frac{y_n e^{y_n} - x}{(y_n + 1)e^{y_n} - \frac{(y_n + 2)(y_n e^{y_n} - x)}{2y_n + 2}}. yn+1=yn−(yn+1)eyn−2yn+2(yn+2)(yneyn−x)yneyn−x.
Employing this gives W ( 10 ) ≈ 1.745528003 W(10) ≈ 1.745528003 W(10)≈1.745528003 after 3 iterations. This complex looking scheme is actually what you get when you apply Newton's method to the function g ( y ) g ′ ( y ) \frac{g(y)}{\sqrt{g'(y)}} g′(y) g(y) [9]. An alternative root-finding scheme, using continued fraction expansion, is described in [10].
采用该方法计算 W ( 10 ) W(10) W(10) 时，仅需 3 次迭代即可得到 W ( 10 ) ≈ 1.745528003 W(10) ≈ 1.745528003 W(10)≈1.745528003。这一看似复杂的方案，本质上是对函数 g ( y ) g ′ ( y ) \frac{g(y)}{\sqrt{g'(y)}} g′(y) g(y) 应用牛顿法的结果[9]。文献[10] 还介绍了另一种基于连分数展开的求根方法。
4. Calculus
4. 微积分性质
We know that since y = x e x y = x e^{x} y=xex is an increasing and differentiable function for all x > − 1 x > -1 x>−1, then its inverse y = W ( x ) y = W(x) y=W(x) is likewise increasing and differentiable for all x > − 1 e x > -\frac{1}{e} x>−e1. Differentiating x = y e y x = y e^{y} x=yey with respect to y y y, we obtain
已知函数 y = x e x y = x e^{x} y=xex 在所有 x > − 1 x > -1 x>−1 的范围内是单调递增且可微的，因此其反函数 y = W ( x ) y = W(x) y=W(x) 在所有 x > − 1 e x > -\frac{1}{e} x>−e1 的范围内也同样单调递增且可微。对 x = y e y x = y e^{y} x=yey 关于 y y y 求导，可得
1 = d W d x d x d y = d W d x [ y e y + e y ] , 1 = \frac{dW}{dx} \frac{dx}{dy} = \frac{dW}{dx} \left[ y e^y + e^y \right], 1=dxdWdydx=dxdW[yey+ey],
so
因此
W ′ ( x ) = 1 ( y + 1 ) e y = y ( y + 1 ) y e y = y ( y + 1 ) x = W ( x ) [ W ( x ) + 1 ] x = e − W ( x ) 1 + W ( x ) , x ≠ 0. (3) \begin{aligned} W'(x) &= \frac{1}{(y+1)e^y} = \frac{y}{(y+1)ye^y}\\ &= \frac{y}{(y+1)x} = \frac{W(x)}{[W(x) + 1]x} = \frac{e^{-W(x)}}{1 + W(x)}, \quad x \neq 0. \end{aligned} \tag{3} W′(x)=(y+1)ey1=(y+1)yeyy=(y+1)xy=[W(x)+1]xW(x)=1+W(x)e−W(x),x=0.(3)
In particular,
特别地
W ′ ( e ) = 1 e ( 1 + 1 ) = 1 2 e W'(e) = \frac{1}{e(1 + 1)} = \frac{1}{2e} W′(e)=e(1+1)1=2e1,
and similarly,
同理，
lim x → ( − 1 e ) + W ′ ( x ) = ∞ \lim_{x \to (-\frac{1}{e})^+} W'(x) = \infty limx→(−e1)+W′(x)=∞.
What about W ′ ( 0 ) W'(0) W′(0)? The right-hand side of ( 3 ) (3) (3) is indeterminant at x = 0 x = 0 x=0, but division of both sides of ( 1 ) (1) (1) by x x x and taking the limit as x → 0 x \to 0 x→0 give
那么 W ′ ( 0 ) W'(0) W′(0) 如何计算？ ( 3 ) (3) (3) 在 x = 0 x = 0 x=0 处为不定式，但将 ( 1 ) (1) (1) 两边同除以 x x x 并取 x → 0 x \to 0 x→0 的极限，可得
lim x → 0 W ( x ) e W ( x ) x = 1. \lim_{x \to 0} \frac{W(x) e^{W(x)}}{x} = 1. x→0limxW(x)eW(x)=1.
Since lim x → 0 e W ( x ) = e 0 = 1 \lim_{x \to 0} e^{W(x)} = e^{0} = 1 limx→0eW(x)=e0=1, this yields lim x → 0 W ( x ) x = 1 \lim_{x \to 0} \frac{W(x)}{x} = 1 limx→0xW(x)=1. Thus,
由于 lim x → 0 e W ( x ) = e 0 = 1 \lim_{x \to 0} e^{W(x)} = e^{0} = 1 limx→0eW(x)=e0=1，因此 lim x → 0 W ( x ) x = 1 \lim_{x \to 0} \frac{W(x)}{x} = 1 limx→0xW(x)=1，进而有：
W ′ ( 0 ) = lim x → 0 W ( x ) x [ 1 + W ( x ) ] = lim x → 0 W ( x ) / x 1 + W ( x ) = 1 1 = 1. W'(0) = \lim_{x \to 0} \frac{W(x)}{x \left[ 1 + W(x) \right]} = \lim_{x \to 0} \frac{W(x)/x}{1 + W(x)} = \frac{1}{1}= 1. W′(0)=x→0limx[1+W(x)]W(x)=x→0lim1+W(x)W(x)/x=11=1.
For large x x x, the graph of W ( x ) W(x) W(x) bears strong resemblance to ln x \ln x lnx, since from (2) we have lim x → ∞ W ( x ) ln x = 1 \lim_{x \to \infty} \frac{W(x)}{\ln x} = 1 limx→∞lnxW(x)=1 although we have to be careful here because the difference ∣ W ( x ) − ln x ∣ |W(x) - \ln x| ∣W(x)−lnx∣ increases without bound as x → ∞ x \to \infty x→∞ [7]. The graph of W ( x ) W(x) W(x), like that of ln x \ln x lnx, is concave downward for all x x x since x e x xe^x xex is concave upward. If we differentiate (3), and omit the argument x x x for brevity, then
对于较大的 x x x， W ( x ) W(x) W(x) 的图像与 ln x \ln x lnx 非常相似，因为从公式 (2) 我们有 lim x → ∞ W ( x ) ln x = 1 \lim_{x \to \infty} \frac{W(x)}{\ln x} = 1 limx→∞lnxW(x)=1，尽管我们必须小心，因为差值 ∣ W ( x ) − ln x ∣ |W(x) - \ln x| ∣W(x)−lnx∣ 随着 x → ∞ x \to \infty x→∞ 无限增加 [7]。 W ( x ) W(x) W(x) 的图像，像 ln x \ln x lnx 一样，对于所有的 x x x 都是向下凹的，因为 x e x xe^x xex 是向上凹的。如果我们对公式 (3) 求导，并为了简洁省略参数 x x x，那么
W ′ ′ = x ( 1 + W ) W ′ − W [ x W ′ + 1 + W ] x 2 ( 1 + W ) 2 = − 2 W 2 − W 3 x 2 ( 1 + W ) 3 < 0. W'' = \frac{x(1+W)W' - W[xW' + 1 + W]}{x^2(1+W)^2} = \frac{-2W^2 - W^3}{x^2(1+W)^3} < 0. W′′=x2(1+W)2x(1+W)W′−W[xW′+1+W]=x2(1+W)3−2W2−W3<0.
Rewriting W ′ ′ W'' W′′ as e − 2 W [ − ( W + 2 ) ] ( 1 + W ) 3 \frac{e^{-2W}\left[-(W+2)\right]}{(1+W)^3} (1+W)3e−2W[−(W+2)] puts this into the form which fits the general case for W ( n ) W^{(n)} W(n) [5]. In fact, from this form, we readily see that there is a point of inflection on the curve when W = − 2 W = -2 W=−2, which actually falls on the other branch of the W W W function.
将 W ′ ′ W'' W′′ 重写为 e − 2 W [ − ( W + 2 ) ] ( 1 + W ) 3 \frac{e^{-2W}\left[-(W+2)\right]}{(1+W)^3} (1+W)3e−2W[−(W+2)] 使其符合 W ( n ) W^{(n)} W(n) 的一般情况 [5]。实际上，从这种形式，我们很容易看出当 W = − 2 W = -2 W=−2 时，曲线上有一个拐点，这实际上落在 W W W 函数的另一个分支上。
Continuing along the calculus vein, we should examine, if possible, the integral of W ( x ) W(x) W(x). To this end, recall that y = W ( x ) y = W(x) y=W(x) iff x = y e y x = y e^{y} x=yey. Thus, d x = e y ( 1 + y ) d y dx = e^{y}(1 + y) dy dx=ey(1+y)dy, and
接下来从微积分角度进一步分析，若可能，还需研究 W ( x ) W(x) W(x) 的积分。为此，回顾 y = W ( x ) y = W(x) y=W(x) 当且仅当 x = y e y x = y e^{y} x=yey，因此 d x = e y ( 1 + y ) d y dx = e^{y}(1 + y) dy dx=ey(1+y)dy，进而有：
∫ W ( x ) d x = ∫ y d y W ′ ( x ) = ∫ y d y W ( x ) x ( 1 + W ( x ) ) = ∫ [ y e y + y 2 e y ] d y \begin{aligned} \int W(x) \, dx &= \int y \frac{dy}{W'(x)} = \int \frac{y \, dy}{W(x)} x(1 + W(x))\\ &= \int \left[ ye^y + y^2 e^y \right] \, dy \end{aligned} ∫W(x)dx=∫yW′(x)dy=∫W(x)ydyx(1+W(x))=∫[yey+y2ey]dy
and integrating this last integral by parts, we obtain
并使用分部积分法计算最后一个积分，我们得到
x [ y − 1 + 1 y ] , x \left[ y - 1 + \frac{1}{y} \right], x[y−1+y1],
which now gives
这现在给出了
∫ W ( x ) d x = x [ W ( x ) − 1 + 1 W ( x ) ] + C . (4) \int W(x) \, dx = x \left[ W(x) - 1 + \frac{1}{W(x)} \right] + C. \tag{4} ∫W(x)dx=x[W(x)−1+W(x)1]+C.(4)
In particular, the area of the region bounded by the curve W ( x ) W(x) W(x), the x-axis, and the line x = ε x = \varepsilon x=ε is, therefore,
特别地，由曲线 W ( x ) W(x) W(x)，x轴和直线 x = ε x = \varepsilon x=ε 围成的区域的面积因此为，
∫ 0 ε W ( x ) d x = x [ W ( x ) − 1 + 1 W ( x ) ] ∣ 0 ε = ε [ 1 − 1 + 1 ] − lim ε → 0 ε [ W ( ε ) − 1 + 1 W ( ε ) ] = e − 1. \begin{aligned} \int_{0}^{\varepsilon} W(x) \, dx &= x \left[ W(x) - 1 + \frac{1}{W(x)} \right] \bigg|{0}^{\varepsilon}\\&= \varepsilon \left[ 1 - 1 + 1 \right] - \lim{\varepsilon \to 0} \varepsilon \left[ W(\varepsilon) - 1 + \frac{1}{W(\varepsilon)} \right]\\& = e - 1.\end{aligned} ∫0εW(x)dx=x[W(x)−1+W(x)1] 0ε=ε[1−1+1]−ε→0limε[W(ε)−1+W(ε)1]=e−1.
We note this result agrees with evaluating the integral via inverse functions [11], because then
我们注意到这个结果与通过反函数[11]计算积分是一致的，因为
∫ 0 e W ( x ) d x = e W ( e ) − 0 W ( 0 ) − ∫ 0 W ( e ) x e x d x = e − 1. \int_{0}^{e} W(x) \, dx = e \, W(e) - 0 \, W(0) - \int_{0}^{W(e)} x e^x \, dx = e - 1. ∫0eW(x)dx=eW(e)−0W(0)−∫0W(e)xexdx=e−1.
Other integrals, involving functions containing W W W, can be computed, some just with a special change of variable [6]. For instance,
涉及 W W W 的其他积分可以计算，有些只需通过变量的特殊变换[6]。例如，
∫ x W ( x ) d x = 1 2 e 2 W ( x ) [ W ( x ) − 1 2 ] [ W 2 ( x ) + 1 2 ] + C . \int x W(x) \, dx = \frac{1}{2} e^{2 W(x)} \left[ W(x) - \frac{1}{2} \right] \left[ W^2(x) + \frac{1}{2} \right] + C. ∫xW(x)dx=21e2W(x)[W(x)−21][W2(x)+21]+C.
The function x W ( x ) xW(x) xW(x) is concave up, connecting ( 0 , 0 ) (0,0) (0,0)
函数 x W ( x ) xW(x) xW(x) 是向上凹的，连接 ( 0 , 0 ) (0,0) (0,0)
and ( e , e ) (e,e) (e,e), hence its area ( ∫ 0 e x W ( x ) d x = ( 3 e 2 + 1 ) / 8 ) \left(\int_0^e x W(x) dx = \left(3e^2 + 1\right)/8\right) (∫0exW(x)dx=(3e2+1)/8) is less than ∫ 0 e x d x = e 2 / 2 \int_0^e x dx = e^2/2 ∫0exdx=e2/2. Similarly we find
以及 ( e , e ) (e,e) (e,e)，因此其面积 ( ∫ 0 e x W ( x ) d x = ( 3 e 2 + 1 ) / 8 ) \left(\int_0^e x W(x) dx = \left(3e^2 + 1\right)/8\right) (∫0exW(x)dx=(3e2+1)/8) 小于 ∫ 0 e x d x = e 2 / 2 \int_0^e x dx = e^2/2 ∫0exdx=e2/2。同样我们发现
∫ 0 e x W ( x ) d x = 1 4 e 2 W ( x ) [ 1 + 2 W ( x ) ] ∣ 0 e = 3 e 2 / 4 , \int_0^e \frac{x}{W(x)} dx = \frac{1}{4} e^{2W(x)} \left[1 + 2W(x)\right] \Bigg|_0^e = 3e^2/4, ∫0eW(x)xdx=41e2W(x)[1+2W(x)] 0e=3e2/4,
and this is greater than e 2 / 2 e^2/2 e2/2 since x / W ( x ) x/W(x) x/W(x) is increasing and concave down from ( 0 , 1 ) (0,1) (0,1) to ( e , e ) (e,e) (e,e).
这大于 e 2 / 2 e^2/2 e2/2，因为 x / W ( x ) x/W(x) x/W(x) 从 ( 0 , 1 ) (0,1) (0,1) 到 ( e , e ) (e,e) (e,e) 是递增且向下凹的。
5. Applications
5. 应用领域
An article appeared in the February, 2000, issue of FOCUS , the newsletter of the Mathematical Association of America, touting the merits of the W function as a candidate for a new elementary function to be studied in schools and to be included in textbooks [12]. The rationale for this was that not only is W a radically different function from the traditional elementary functions of polynomials, rationals, exponentials, logarithmics, and trigonometrics, but its calculus provides a wealth of interesting, and powerful, applications. A number of these are mentioned in a paper by Corless et al., where they describe such applications as enumeration of trees, combustion, enzyme kinetics, linear delay equations, population growth, spread of disease, and the analysis of algorithms [3]. An article [13] by Packel and Yuen shows that W is instrumental in determining the maximum range for a projectile with linear resistance (problems of this type have certainly been important for several thousand years). The solution for the current in a series diode/resistor circuit can also be written in terms of W. Applications of W are found in complex cases involving atomic, nuclear, and optical physics. The first physics problem to be solved explicitly in terms of W was one in which the exchange forces between two nuclei within the hydrogen molecular ion ( H 2 + ) (H_2^+) (H2+) were calculated [14].
2000 年 2 月，美国数学协会通讯《聚焦》（FOCUS ）刊登了一篇文章，宣扬朗伯 W 函数作为"新基本函数"的优势，认为其应纳入学校教学内容并收录进教材[12]。理由如下：W 函数不仅与多项式、有理函数、指数函数、对数函数、三角函数等传统基本函数截然不同，其微积分性质还衍生出丰富有趣且实用性强的应用。科莱斯等人的论文中提及了多个应用场景，包括树的计数、燃烧学、酶动力学、线性时滞方程、种群增长、疾病传播及算法分析[3]。帕克尔与袁合著的文章[13] 指出，W 函数对确定"线性阻力下抛射体的最大射程"起关键作用（这类问题数千年来一直具有重要意义）；此外，二极管/电阻串联电路的电流解也可通过 W 函数表示。在复杂场景中，W 函数的应用还涉及原子物理、核物理与光学：首个明确用 W 函数求解的物理问题，是计算氢分子离子 ( H 2 + ) (H_2^+) (H2+) 内两个原子核之间的交换力[14]。
Several other cases involve generalized Gaussian noise, solar winds, black holes, general relativity, quantum chromodynamics, fuel consumption, Stirling's formula for n ! n! n!, cardiorespiratory control, water-wave heights in oceanography, enumeration of trees in combinatorics, and statistical mechanics [5,15-17]. A really interesting analog of W ( x ) W(x) W(x) is given by Dan Kalman [18], where he defines a function glog, similar to W, in that glog is the inverse to e x x \frac{e^{x}}{x} xex. The glog function bears a strong resemblance to W ( x ) W(x) W(x), possessing similar properties and useful common applications, such as solving exponential-linear equations. The two functions are intimately related by
其他应用场景还包括：广义高斯噪声、太阳风、黑洞、广义相对论、量子色动力学、燃料消耗、 n ! n! n! 的斯特林公式、心肺调节、海洋学中的水波高度、组合数学中的树计数及统计力学[5,15-17]。丹·卡尔曼[18] 提出了一个与 W ( x ) W(x) W(x) 十分相似的函数------glog 函数：glog 是 e x x \frac{e^{x}}{x} xex 的反函数，与 W ( x ) W(x) W(x) 性质相似且存在共同实用场景（如求解指数-线性方程）。两者通过以下关系式紧密关联： W ( x ) = − g log ( − 1 / x ) and g log ( x ) = − W ( − 1 / x ) . W(x) = -g \log(-1/x) \text{ and } g \log(x) = -W(-1/x). W(x)=−glog(−1/x) and glog(x)=−W(−1/x).
6. Algebra
6. 代数应用
In a high-school precalculus course one might be presented with the elementary equation 2 x = 10 2^x = 10 2x=10 to solve. Now, instead, let's solve a similar equation 2 x = 10 x 2^x = 10x 2x=10x, which means that it won't suffice to begin by taking the logarithm of both sides. Instead, we proceed as follows:
在高中的微积分先修课程中，可能会遇到一个基本的方程 2 x = 10 2^x = 10 2x=10 来求解。现在，让我们解决一个类似的方程 2 x = 10 x 2^x = 10x 2x=10x，这意味着仅仅对两边取对数是不够的。相反，我们按如下步骤进行：
2 x = 10 x ↔ 1 = 10 x 2 − x = 10 x e − x ln 2 ↔ − ln 2 10 = − x ln 2 e − x ln 2 2^x = 10x \leftrightarrow 1 = 10x 2^{-x} = 10x e^{-x \ln 2} \leftrightarrow \frac {-\ln 2}{10} = -x \ln 2 e^{-x \ln 2} 2x=10x↔1=10x2−x=10xe−xln2↔10−ln2=−xln2e−xln2
Since the right-hand side of this last equation is of the form z e z ze^z zez, and since we know y = z e z y = ze^z y=zez if z = W ( y ) z = W (y) z=W(y) then
由于最后一个方程的右边是 z e z ze^z zez 的形式，并且由于我们知道 y = z e z y = ze^z y=zez 时 z = W ( y ) z = W (y) z=W(y)，那么
− x ln 2 = W ( − ln 2 10 ) , or x = − 1 ln 2 W ( − 1 10 ln 2 ) . -x \ln 2 = W\left (-\frac {\ln 2}{10}\right), \text { or } x = -\frac {1}{\ln 2} W\left (-\frac {1}{10} \ln 2\right). −xln2=W(−10ln2), or x=−ln21W(−101ln2).
Using Kalman's glog function we can solve 2 x = 10 x 2^x = 10x 2x=10x
and get
使用 Kalman 的 glog 函数，我们可以求解 2 x = 10 x 2^x = 10x 2x=10x
得到
x = 1 log 2 g log ( 10 log 2 ) . x = \frac {1}{\log 2} g \log\left (\frac {10}{\log 2}\right). x=log21glog(log210).
Since
由于
∣ − 1 10 ln 2 ∣ ≈ 0.069 < 1 e , \left|-\frac {1}{10} \ln 2\right| \approx 0.069 < \frac {1}{e}, −101ln2 ≈0.069<e1,
we can use (1) to approximate W ( − 1 10 ln 2 ) W\left (-\frac {1}{10} \ln 2\right) W(−101ln2) and get -0.074690848, so x = 0.1077550149 x = 0.1077550149 x=0.1077550149.
我们可以使用 (1) 来近似 W ( − 1 10 ln 2 ) W\left (-\frac {1}{10} \ln 2\right) W(−101ln2) 并得到 -0.074690848，因此 x = 0.1077550149 x = 0.1077550149 x=0.1077550149。
Checking, we find
检查后，我们发现
2 x = 1.07755015 = 10 x 2^x = 1.07755015 = 10x 2x=1.07755015=10x.
The equation 2 x = 10 x 2^x = 10x 2x=10x is a special case of a more general setting a b x + c = d x + e a^{bx+c} = dx + e abx+c=dx+e, where we assume the base a > 0 a > 0 a>0 and where neither b b b nor d d d equals zero. The substitution − t = b x + b e d -t = bx + \frac {be}{d} −t=bx+dbe then gives
方程 2 x = 10 x 2^x = 10x 2x=10x 是更一般设置 a b x + c = d x + e a^{bx+c} = dx + e abx+c=dx+e 的一个特例，其中我们假设底数 a > 0 a > 0 a>0 并且 b b b 和 d d d 都不为零。替换 − t = b x + b e d -t = bx + \frac {be}{d} −t=bx+dbe 则给出
a b x + c = a − t − b e / d a c = d [ − t − b e d b ] + e = − t d b − b e b + e = − t d b , a^{bx+c} = a^{-t-be/d} a^c = d \left [ \frac {-t - \frac {be}{d}}{b} \right] + e = \frac {-td}{b} - \frac {be}{b} + e = \frac {-td}{b}, abx+c=a−t−be/dac=d[b−t−dbe]+e=b−td−bbe+e=b−td,
and, thus,
因此，
a − t a c − b e d = − t d b ↔ − b i d a c − b e d = a t ↔ − b d a c − b e d = t a t = t e t ln a . a^{-t} a^{c - \frac{be}{d}} = \frac{-td}{b} \leftrightarrow \frac{-b}{id} a^{c - \frac{be}{d}} = a^t \leftrightarrow \frac{-b}{d} a^{c - \frac{be}{d}} = t a^t = t e^{t \ln a}. a−tac−dbe=b−td↔id−bac−dbe=at↔d−bac−dbe=tat=tetlna.
Multiplication of both sides by ln a \ln a lna gives
两边同时乘以 ln a \ln a lna 得到
− b d ( ln a ) a c − b e d = t ( ln a ) e t ln a , \frac{-b}{d} (\ln a) a^{c - \frac{be}{d}} = t (\ln a) e^{t \ln a}, d−b(lna)ac−dbe=t(lna)etlna,
which now has the form
现在具有形式
u = w e w , u = we^w, u=wew, so w = W ( u ) , w = W(u), w=W(u), or
t ln a = W ( − b d ln a ⋅ a c − b e d ) t \ln a = W\left(-\frac{b}{d} \ln a \cdot a^{c - \frac{be}{d}}\right) tlna=W(−dblna⋅ac−dbe)
and, hence,
因此
− b x − b e d = t = 1 ln a W ( − b d ln a ⋅ a c − b e d ) , -bx - \frac{be}{d} = t = \frac{1}{\ln a} W\left(-\frac{b}{d} \ln a \cdot a^{c - \frac{be}{d}}\right), −bx−dbe=t=lna1W(−dblna⋅ac−dbe),
that is,
即
x = − 1 b ln a W ( − b d ln a ⋅ a c − b e d ) − e d . x = \frac{-1}{b \ln a} W\left(-\frac{b}{d} \ln a \cdot a^{c - \frac{be}{d}}\right) - \frac{e}{d}. x=blna−1W(−dblna⋅ac−dbe)−de.
Another interesting algebraic application involves the infinite tower of exponents x x x . . . x^{x^{x^{.^{.^{.}}}}} xxx..., which will be denoted by T ( x ) T(x) T(x). To solve the particular equation T ( x ) = 2 T(x) = 2 T(x)=2 one might argue that this is equivalent to x T ( x ) = 2 x^{T(x)} = 2 xT(x)=2, in which case we have x 2 = 2 x^2 = 2 x2=2, so x = 2 x = \sqrt{2} x=2 , which is the correct solution to T ( x ) = 2 T(x) = 2 T(x)=2. But what about T ( x ) = 3 T(x) = 3 T(x)=3, T ( x ) = 4 T(x) = 4 T(x)=4, or T ( x ) = y T(x) = y T(x)=y. It stands to reason that as y increases, so does x. But with T ( x ) = 4 T(x) = 4 T(x)=4, we can write this as x T ( x ) = 4 x^{T(x)} = 4 xT(x)=4, or x 4 = 4 x^4 = 4 x4=4, so x = 2 x = \sqrt{2} x=2 again! Something isn't right.
另一个有趣的代数应用涉及无限指数塔 x x x . . . x^{x^{x^{.^{.^{.}}}}} xxx...，这将被表示为 T ( x ) T(x) T(x)。为了解决特定的方程 T ( x ) = 2 T(x) = 2 T(x)=2，人们可能会认为这相当于 x T ( x ) = 2 x^{T(x)} = 2 xT(x)=2，在这种情况下我们有 x 2 = 2 x^2 = 2 x2=2，所以 x = 2 x = \sqrt{2} x=2 ，这是 T ( x ) = 2 T(x) = 2 T(x)=2 的正确解。但是 T ( x ) = 3 T(x) = 3 T(x)=3， T ( x ) = 4 T(x) = 4 T(x)=4 或 T ( x ) = y T(x) = y T(x)=y 呢？按理说，随着 y 的增加，x 也会增加。但是当 T ( x ) = 4 T(x) = 4 T(x)=4 时，我们可以将其写为 x T ( x ) = 4 x^{T(x)} = 4 xT(x)=4，或者 x 4 = 4 x^4 = 4 x4=4，所以 x = 2 x = \sqrt{2} x=2 再次出现！有些不对劲。
The problem lies with the domain of T T T. We find in [19] that the infinite tower of exponents is only defined (i.e., its interval of convergence) for 1 e e ≤ x ≤ e e \frac{1}{e^e} \leq x \leq \sqrt[e]{e} ee1≤x≤ee , or approximately 0.065988 < x < 1.444667 0.065988 < x < 1.444667 0.065988<x<1.444667. So if x x x is selected from this interval, what is T ( x ) T(x) T(x)? If we set T ( x ) = y T(x) = y T(x)=y then
问题在于 T T T 的定义域。我们在 [19] 中发现，无限指数塔仅在 1 e e ≤ x ≤ e e \frac{1}{e^e} \leq x \leq \sqrt[e]{e} ee1≤x≤ee ，或者大约 0.065988 < x < 1.444667 0.065988 < x < 1.444667 0.065988<x<1.444667 时定义（即其收敛区间）。因此，如果从这个区间中选择 x x x， T ( x ) T(x) T(x) 是什么？如果我们设 T ( x ) = y T(x) = y T(x)=y 那么
x T ( x ) = y ↔ ln x ⋅ e y ln x = y ln x ↔ W ( − ln x ) = − y ln x ↔ W ( − ln x ) − ln x = y . \begin{aligned} {x^{T(x)}} & = y \leftrightarrow \ln x \cdot {e^{y\ln x}} = y\ln x \leftrightarrow W( - \ln x)\\ & = - y\ln x \leftrightarrow \frac{{W( - \ln x)}}{{ - \ln x}} = y. \end{aligned} xT(x)=y↔lnx⋅eylnx=ylnx↔W(−lnx)=−ylnx↔−lnxW(−lnx)=y.
Note also that y = 1 y = 1 y=1 when x = 1 x = 1 x=1, and the above expression for y y y gives a function continuous at x = 1 x = 1 x=1 since lim u → 0 [ W ( u ) / u ] = 1 \lim_{u \to 0} [\text{W}(u)/u] = 1 limu→0[W(u)/u]=1. Hence, if 1 e e ≤ x ≤ e e \frac{1}{e^e} \leq x \leq \sqrt[e]{e} ee1≤x≤ee , then e ≥ − ln x ≥ − 1 e e \geq -\ln x \geq -\frac{1}{e} e≥−lnx≥−e1, so 1 e ≤ y ≤ e \frac{1}{e} \leq y \leq e e1≤y≤e, and this is why the equation T ( x ) = 2 \text{T}(x) = 2 T(x)=2 is solvable, but T ( x ) = 4 \text{T}(x) = 4 T(x)=4 is not.
另请注意，当 x = 1 x = 1 x=1 时 y = 1 y = 1 y=1，且上述 y y y 的表达式给出的函数在 x = 1 x = 1 x=1 处连续，因为 lim u → 0 [ W ( u ) / u ] = 1 \lim_{u \to 0} [\text{W}(u)/u] = 1 limu→0[W(u)/u]=1。因此，若 1 e e ≤ x ≤ e e \frac{1}{e^e} \leq x \leq \sqrt[e]{e} ee1≤x≤ee ，则 e ≥ − ln x ≥ − 1 e e \geq -\ln x \geq -\frac{1}{e} e≥−lnx≥−e1，故 1 e ≤ y ≤ e \frac{1}{e} \leq y \leq e e1≤y≤e，这就是方程 T ( x ) = 2 \text{T}(x) = 2 T(x)=2 可解而 T ( x ) = 4 \text{T}(x) = 4 T(x)=4 不可解的原因。
The graph of T is therefore an increasing function with domain [ 1 e e , e e ] \left[ \frac{1}{e^e}, \sqrt[e]{e} \right] [ee1,ee ] and range [ 1 e , e ] \left[ \frac{1}{e}, e \right] [e1,e]. It also passes through the two obvious points of ( 1 e , W ( 1 ) ) ≈ ( 1 e , 0.567143 ) \left( \frac{1}{e}, \text{W}(1) \right) \approx \left( \frac{1}{e}, 0.567143 \right) (e1,W(1))≈(e1,0.567143) and ( 1 , 1 ) (1, 1) (1,1). What else can we deduce? Checking for differentiability, we have from (3),
因此，T 的图像是一个单调递增函数，定义域为 [ 1 e e , e e ] \left[ \frac{1}{e^e}, \sqrt[e]{e} \right] [ee1,ee ]，值域为 [ 1 e , e ] \left[ \frac{1}{e}, e \right] [e1,e]。它还经过两个明显的点 ( 1 e , W ( 1 ) ) ≈ ( 1 e , 0.567143 ) \left( \frac{1}{e}, \text{W}(1) \right) \approx \left( \frac{1}{e}, 0.567143 \right) (e1,W(1))≈(e1,0.567143) 和 ( 1 , 1 ) (1, 1) (1,1)。我们还能推导出什么呢？检查可微性，由（3）式可得，
T ′ ( x ) = [ − ln x ] ⋅ W ( − ln x ) − ln x [ 1 + W ( − ln x ) ] ( − 1 x ) − W ( − ln x ) ( − 1 x ) = W ( − ln x ) ⋅ W ( − ln x ) x ln 2 x ( 1 + W ( − ln x ) ) = [ W ( − ln x ) ln x ] 2 ⋅ 1 x [ 1 − ( ln x ) T ( x ) ] \begin{array}{lll} T'(x) & = \frac{{[ - \ln x] \cdot W( - \ln x)}}{{ - \ln x[1 + W( - \ln x)]}}\left( { - \frac{1}{x}} \right) - W( - \ln x)\left( { - \frac{1}{x}} \right)\\[1em] & = \frac{{W( - \ln x) \cdot W( - \ln x)}}{{x{{\ln }^2}x(1 + W( - \ln x))}}\\[1em] & = {\left[ {\frac{{W( - \ln x)}}{{\ln x}}} \right]^2} \cdot \frac{1}{{x[1 - (\ln x)T(x)]}} \end{array} T′(x)=−lnx[1+W(−lnx)][−lnx]⋅W(−lnx)(−x1)−W(−lnx)(−x1)=xln2x(1+W(−lnx))W(−lnx)⋅W(−lnx)=[lnxW(−lnx)]2⋅x[1−(lnx)T(x)]1
and since the limit of this expression is 1 as x → 1 x \to 1 x→1, then T ′ ( 1 ) = 1 T'(1) = 1 T′(1)=1, and hence T ′ T' T′ is never 0, so T T T is always strictly increasing.
并且由于这个表达式的极限当 x → 1 x \to 1 x→1 时为 1，那么 T ′ ( 1 ) = 1 T'(1) = 1 T′(1)=1，因此 T ′ T' T′ 永远不会为 0，所以 T T T 总是严格递增的。
The following small table (Table 1) of values will prove helpful.
以下的小表格（表1）的数值将证明是有帮助的。
Alternatively, we could have found T ′ T' T′ by implicit differentiation of $y = x x . . . = T ( x ) = x y x^{x^{.^{.^{.}}}} = T(x) = x^y xx...=T(x)=xy.
或者，我们可以通过隐式微分 y = x x ⋯ = T ( x ) = x y y = x^{x^{\cdots}} = T(x) = x^y y=xx⋯=T(x)=xy 来找到 T ′ T' T′。
Alternatively, we could have found T ′ T' T′ by implicit differentiation of y = x x ⋅ ⋅ ⋅ = T ( x ) = x y y = x^{x^{\cdot^{\cdot^{\cdot}}}} = T(x) = x^y y=xx⋅⋅⋅=T(x)=xy.
或者，我们可以通过隐式微分 y = x x ⋯ = T ( x ) = x y y = x^{x^{\cdots}} = T(x) = x^y y=xx⋯=T(x)=xy 来找到 T ′ T' T′。
| x x x | T ′ ( x ) T'(x) T′(x) |
|---|---|
| e 1 / e e^{1/e} e1/e | ∞ \infty ∞ |
| 1 | 1 |
| 1 / e 1/e 1/e | ≈ 0.558 \approx 0.558 ≈0.558 |
| 1 / e e 1/e^e 1/ee | e e 2 e 2 ≈ 1.025 \frac{e^e}{2e^2} \approx 1.025 2e2ee≈1.025 |
| 0.394 | 0.557 |
| Table 1. Some derivative values of T ( x ) T(x) T(x) | |
| 表 1. T ( x ) T(x) T(x) 的部分导数数值 |
Thus 1 y y ′ = y x + y ′ ln x \frac{1}{y} y' = \frac{y}{x} + y' \ln x y1y′=xy+y′lnx, or y ′ = y 2 x − x y ln x y' = \frac{y^2}{x - x y \ln x} y′=x−xylnxy2; so again, y ′ = y 2 x ( 1 − y ln x ) y' = \frac{y^2}{x(1 - y \ln x)} y′=x(1−ylnx)y2. Using this form for y ′ y' y′ to compute y ′ ′ y'' y′′ (after some algebraic manipulations and cancellations) gives,
此外，也可通过对 y = T ( x ) = x y y = T(x) = x^y y=T(x)=xy 隐函数求导得到 T ′ ( x ) T'(x) T′(x)，结果同上： y ′ = y 2 x ( 1 − y ln x ) y' = \frac{y^2}{x(1 - y \ln x)} y′=x(1−ylnx)y2。利用该导数表达式进一步计算二阶导数 y ′ ′ y'' y′′（经代数运算与消元后）：
y ′ ′ = y 3 ( 2 − y ln x ) + y 2 ( y − 1 + y ln x ) ( 1 − y ln x ) x 2 ( 1 − y ln x ) 3 y'' = \frac{y^3(2 - y \ln x) + y^2(y - 1 + y \ln x)(1 - y \ln x)}{x^2(1 - y \ln x)^3} y′′=x2(1−ylnx)3y3(2−ylnx)+y2(y−1+ylnx)(1−ylnx)
This complex expression appears to yield negative values for all x < 0.394 x < 0.394 x<0.394 and positive values for all x > 0.394 x > 0.394 x>0.394, and
该复杂表达式在 x < 0.394 x < 0.394 x<0.394 时取值为负，在 x > 0.394 x > 0.394 x>0.394 时取值为正；且
y ′ ( 0.394 ) = 0.581 7 2 0.394 × [ 1 − 0.5817 × ln 0.394 ] ≈ 0.3384 0.394 × ( 1 + 0.5817 × 0.9314 ) ≈ 0.557 y'(0.394) = \frac{0.5817^2}{0.394 \times [1 - 0.5817 \times \ln 0.394]} ≈ \frac{0.3384}{0.394 \times (1 + 0.5817 \times 0.9314)} ≈ 0.557 y′(0.394)=0.394×[1−0.5817×ln0.394]0.58172≈0.394×(1+0.5817×0.9314)0.3384≈0.557.
Hence, we have an inflection point at x ≈ 0.394 x ≈ 0.394 x≈0.394 (where y ≈ 0.5817 y ≈ 0.5817 y≈0.5817). Putting all of these pieces of the puzzle together, we obtain a decent graph of T T T, as shown in Figure 3.
因此，曲线在 x ≈ 0.394 x ≈ 0.394 x≈0.394（对应 y ≈ 0.5817 y ≈ 0.5817 y≈0.5817）处存在拐点。整合上述所有信息，可得到 T ( x ) T(x) T(x) 的完整图像，如图 3 所示。
1 y y ′ = y ′ x + y ′ ln x or y ′ = y 2 x − x y ln x ; \frac{1}{y}y' = \frac{y'}{x} + y'\ln x \text{ or } y' = \frac{y^2}{x - xy\ln x}; y1y′=xy′+y′lnx or y′=x−xylnxy2;
so again
因此再次
y ′ ( 1 ) = 1 y'(1) = 1 y′(1)=1 .
Use of this form for easier access to y ′ ′ y'' y′′ then gives, after some algebraic manipulations and cancellations,
使用这种形式可以更容易地访问 y ′ ′ y'' y′′，然后经过一些代数操作和消去，
y ′ ′ = y 3 ( 2 − y ln x ) + y 2 ( y − 1 + y ln x ) ( 1 − y ln x ) x 2 ( 1 − y ln x ) 3 y'' = \frac{y^3(2 - y\ln x) + y^2(y - 1 + y\ln x)(1 - y\ln x)}{x^2(1 - y\ln x)^3} y′′=x2(1−ylnx)3y3(2−ylnx)+y2(y−1+ylnx)(1−ylnx) .
This complex expression appears to yield negative values for all x < 0.394 x < 0.394 x<0.394 and positive values for all x > 0.394 x > 0.394 x>0.394, and y ′ ′ ( 0.394 ) ≈ 0 y''(0.394) \approx 0 y′′(0.394)≈0 . Hence, we have an inflection point at x ≈ 0.394 , y ≈ 0.5817 x \approx 0.394, y \approx 0.5817 x≈0.394,y≈0.5817 . Also
这个复杂的表达式似乎在所有 x < 0.394 x < 0.394 x<0.394 时产生负值，在所有 x > 0.394 x > 0.394 x>0.394 时产生正值，并且 y ′ ′ ( 0.394 ) ≈ 0 y''(0.394) \approx 0 y′′(0.394)≈0 。因此，我们在 x ≈ 0.394 , y ≈ 0.5817 x \approx 0.394, y \approx 0.5817 x≈0.394,y≈0.5817 处有一个拐点。此外，
y ′ ( 0.394 ) = 0.581 7 2 0.394 × [ 1 + 0.9314 ( 0.5817 ) ] = 0.557 y'(0.394) = \frac{0.5817^2}{0.394 \times [1 + 0.9314(0.5817)]} = 0.557 y′(0.394)=0.394×[1+0.9314(0.5817)]0.58172=0.557 . Putting all of these puzzle pieces together, we obtain a decent graph of T T T, as shown in Figure 3.
将所有这些拼图拼在一起，我们得到了 T T T 的一个不错的图，如图3所示
y ′ ( 0.394 ) = 0.581 7 2 0.394 × [ 1 + 0.9314 ( 0.5817 ) ] = 0.557 y'(0.394) = \frac{0.5817^2}{0.394 \times [1 + 0.9314(0.5817)]} = 0.557 y′(0.394)=0.394×[1+0.9314(0.5817)]0.58172=0.557
The tower function T T T must necessarily possess an inverse T − 1 T^{-1} T−1. We note then that
塔函数 T T T 必须必然具有一个逆函数 T − 1 T^{-1} T−1。我们注意到，
T ( x ) = y → x T ( x ) = y → x y = y → x = y 1 / y T(x) = y \rightarrow x^{T(x)} = y \rightarrow x^y = y \rightarrow x = y^{1/y} T(x)=y→xT(x)=y→xy=y→x=y1/y
and, consequently, this inverse is T − 1 ( x ) = x 1 / x T^{-1}(x) = x^{1/x} T−1(x)=x1/x. Composition of the two functions give the interesting pair of identities,
因此，这个逆函数是 T − 1 ( x ) = x 1 / x T^{-1}(x) = x^{1/x} T−1(x)=x1/x。两个函数的组合给出了一对有趣的恒等式，
( x 1 / x ) ( x 1 / x ) ⋅ ⋅ ⋅ = x , 1 e ≤ x ≤ e and x x x ⋅ ⋅ ⋅ = x , 1 e e ≤ x ≤ e \left(x^{1/x}\right)^{\left(x^{1/x}\right)^{\cdot^{\cdot^{\cdot}}}} = x, \frac{1}{e} \leq x \leq e \quad \text{and} \quad \sqrt[{{x^{{x^{{ \cdot ^{{ \cdot ^ \cdot }}}}}}}}]{x} = x,\frac{1}{{{e^e}}} \le x \le e (x1/x)(x1/x)⋅⋅⋅=x,e1≤x≤eandxx⋅⋅⋅x =x,ee1≤x≤e .
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