如何确定方程解？

极光永明

（MSE问题链接）
现我已经知晓方程$$\log_2\left(x+1\right)+x=2$$的解的唯一性.
我的问题是：除了猜测之外，有没有严谨的方法可以推导出$x=1$？（尽管这个猜测几乎是显然的）
在上面提供的MSE链接中，有人提到了Lambert-W函数，我在查阅之后仍然没有什么头绪？望大佬赐教。

12673zf

在实数范围内，用函数单调性+猜根（观察）就可以搞定吧（正常高一就会做了），左边递增，而且恰好1的时候取等。

极光永明

12673zf 发表于 2023-1-19 12:26
在实数范围内，用函数单调性+猜根（观察）就可以搞定吧（正常高一就会做了），左边递增，而且恰好1的时候取 ...

看问题，我需要一个不依赖于猜根的解答。

hbghlyj

Mathematica

WolframAlpha\[W(2^{x + 1} \log2) - \log2\over\log2\] inverse function Log2[x+1]+x

如何求Lambert W函数的级数?
zhuanlan.zhihu.com/p/216093102

$W(x)e^{W(x)}=x$ ，利用隐函数的求导法则，两边对 $x$ 求导有：
$$[W(x)+1]e^{W(x)}\cdot\frac{\mathrm dW(x)}{\mathrm dx}=1$$
$$\Rightarrow\frac{\mathrm dW(x)}{\mathrm dx}=\frac{1}{e^{W(x)}+W(x)\cdot e^{W(x)}}=\frac{1}{e^{W(x)}+x}=\frac{1}{x+\frac{x}{W(x)}}=\frac{W(x)}{x+xW(x)}$$
更一般地， $$\frac{\mathrm d^nW(x)}{\mathrm dx^n}=\frac{e^{-nW(x)}\cdot p_n[W(x)]}{[1+W(x)]^{2n+1}}$$其中多项式 $p_n(w)$ 满足递推式$$p_{n+1}(w)=(-nw+3n-1)p_n(w)+(1+w)p'_n(w)$$且 $p_1(w)=1$ ，对 $n\geq1$ 的整数有 $$p_n(0)=\frac{(-n)^{n-1}}{n!}$$利用 $W(0)=0$ 的便利性，我们在 $x=0$ 处展开 $W(x)$ 可以得到其对应的泰勒级数为$$\sum_{n=1}^{\infty}{\frac{(-n)^{n-1}}{n!}\cdot x^n}=x-x^2+\frac{3}{2}x^3-\cdot\cdot\cdot$$对于泰勒级数，我们还有一个有必要讨论的问题，那就是它的收敛半径，很显然 $$\frac{1}{R}=\lim_{n \rightarrow \infty}{\left| \frac{a_{n+1}}{a_n} \right|}=\lim_{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n-1}=e$$故 $R=\frac{1}{e}$

Lagrange inversion
The Lambert $W$ function is the function \(W(z)\) that is implicitly defined by the equation
\[W(z) e^{W(z)} = z.\]
We may use the theorem to compute the Taylor series of \(W(z)\) at \(z=0.\) We take \(f(w) = we^w\) and \(a = 0.\) Recognizing that
\[\frac{d^n}{dx^n} e^{\alpha x} = \alpha^n e^{\alpha x},\]
this gives
\begin{align*}
   W(z) &= \sum_{n=1}^{\infty} \left[\lim_{w \to 0} \frac{d^{n-1}}{dw^{n-1}} e^{-nw} \right] \frac{z^n}{n!} \\
   {} &= \sum_{n=1}^{\infty} (-n)^{n-1} \frac{z^n}{n!} \\
   {} &= z-z^2+\frac{3}{2}z^3-\frac{8}{3}z^4+O(z^5).
\end{align*}
The radius of convergence of this series is \(e^{-1}\) (giving the principal branch of the Lambert function).
A series that converges for larger $z$ (though not for all $z$) can also be derived by series inversion.  The function \(f(z) = W(e^z) - 1\) satisfies the equation
\[1 + f(z) + \ln (1 + f(z)) = z.\]
Then \(z + \ln (1 + z)\) can be expanded into a power series and inverted.  This gives a series for \(f(z+1) = W(e^{z+1})-1\text{:}\)
\[W(e^{1+z}) = 1 + \frac{z}{2} + \frac{z^2}{16} - \frac{z^3}{192} - \frac{z^4}{3072} + \frac{13 z^5}{61440} -  O(z^6).\]
\(W(x)\) can be computed by substituting \(\ln x - 1\) for $z$ in the above series. For example, substituting $-1$ for $z$ gives the value of \(W(1) \approx 0.567143.\)

hbghlyj

来自tigertooth4的博客
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2023-5-3 04:21 Upload

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A Lambert curve is a curve given by $C =\{(z,w) \; | \; z = w e^w\} \subset \mathbb{C}^* \times \mathbb{C}^*.$
Note that the above equation defines \(w\) as an implicit function of \(z\). Such functions are not unique, we call them the Lambert W function.
By inverse function theorem, \(w\) can be written locally as a function of \(z\). Note that there are two branches, the branches we care about is called \(w_0\). From wikipedia, one could find the asymptotic expansion of \(w_0\) w.r.t. \(z\), which is\[w_0(z) = \sum_{k\ge 1}\frac{(-k)^{k-1}}{k!}x^k\]Now if we redefine the Lambert curve to be \(x = y e^{-y}\), and define\[t = 1 + \sum_{k=1}^\infty \frac{k^k}{k!}x^k,\]how the expression \(x=y e^{-y}\) may be changed when we substitute \(y\) by \(t\)?

Notice that the Taylor expansions of \(w_0(z)\) and \(t(x)\) are alike, it is not so hard to establish a functional relationship between the two. So let's start by finding a Taylor series of \(y\) w.r.t. \(x\). It is easy to see two definitions of Lambert curves are identified if we introduce the transform \(y(x) = -w_0(x)\). So \(y\) can be expand to a Taylor series:\[y(x)=\sum_{k\ge 1}\frac{k^{k-1}}{k!}x^k\]By term-by-term differentiation, we know \(x y’(x) = t-1\). But from another point of view, by implicit differentiation, we find \(1 = xy’(y^{-1} - 1)\). So \(y = 1-t^{-1}\). And, if written in the new parameter \(t\), Lambert curve can be formatted as$$x = (1-t^{-1}) e^{-(1-t^{-1})}.$$

hbghlyj

极光永明 发表于 2023-1-19 04:21
（MSE问题链接）
现我已经知晓方程$$\log_2\left(x+1\right)+x=2$$的解的唯一性.

对于$|ϵ|\ll1$,可以求方程$$\log_2\left(x+1\right)+x=2+ϵ$$的渐近解:

1. InverseSeries[Series[Log[2, ϵ + 1] + ϵ - 2, {ϵ, 1, 3}]]

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\[x\sim1+\frac{\epsilon }{1+\frac{1}{2 \log (2)}}+\frac{\epsilon ^2}{4 \left(1+\frac{1}{2 \log (2)}\right)^2 (1+2 \log (2))}+\frac{\epsilon ^3 \left(\frac{1}{8 (1+2 \log (2))^2}-\frac{1}{12 (1+2 \log (2))}\right)}{\left(1+\frac{1}{2 \log (2)}\right)^3}+O\left(\epsilon ^4\right)\](当$ϵ=0$时是$x=1$.)
这个级数可以像Lambert W函数那样求出收敛半径吗

hbghlyj

极光永明 发表于 2023-1-19 04:21
现我已经知晓方程$\log_2\left(x+1\right)+x=2$的解的唯一性.

若改为\[\log_{x+1}2+x=2\]则有2个解$x_1≈0.715007$和$x_2=1$
函数$f(x)=\log_{x+1}2+x-2$不是单射,它有2个连续的反函数$f_+^{-1}$和$f_-^{-1}$
$f(x)$在$x=1$的级数为Series[Log[x+1,2]+x-2,{x,1,4}]\begin{multline*}f(x)=(x-1) \left(1-\frac{1}{2 \log (2)}\right)+\frac{(x-1)^2 (2+\log (2))}{8 \log ^2(2)}\\+\frac{(x-1)^3 \left(-3-\log ^2(2)-3 \log (2)\right)}{24 \log ^3(2)}\\+\frac{(x-1)^4 \left(12+3 \log ^3(2)+11 \log ^2(2)+18 \log (2)\right)}{192 \log ^4(2)}+O\left((x-1)^5\right)\end{multline*}反演的级数$f_+^{-1}$为\begin{multline*}1+\frac{x}{1-\frac{1}{2 \log (2)}}-\frac{x^2 (2+\log (2))}{4 \left(\left(1-\frac{1}{2 \log (2)}\right)^2 \log (2) (2 \log (2)-1)\right)}\\+\frac{x^3 \left(-\frac{-3-\log ^2(2)-3 \log (2)}{12 \log ^2(2) (2 \log (2)-1)}-\frac{(2+\log (2)) \left(-\frac{1}{\log (2) (2 \log (2)-1)}-\frac{1}{2 (2 \log (2)-1)}\right)}{4 \log (2) (2 \log (2)-1)}\right)}{\left(1-\frac{1}{2 \log (2)}\right)^3}\\+\frac{x^4 \left(-\frac{(2+\log (2)) \left(\frac{(2+\log (2))^2}{16 \log ^2(2) (2 \log (2)-1)^2}-\frac{-3-\log ^2(2)-3 \log (2)}{6 \log ^2(2) (2 \log (2)-1)}-\frac{\left(-\frac{1}{\log (2) (2 \log (2)-1)}\\-\frac{1}{2 (2 \log (2)-1)}\right) (2+\log (2))}{2 \log (2) (2 \log (2)-1)}\right)}{4 \log (2) (2 \log (2)-1)}+\frac{(2+\log (2)) \left(-3-\log ^2(2)-3 \log (2)\right)}{16 \log ^3(2) (2 \log (2)-1)^2}-\frac{12+3 \log ^3(2)+11 \log ^2(2)+18 \log (2)}{96 \log ^3(2) (2 \log (2)-1)}\right)}{\left(1-\frac{1}{2 \log (2)}\right)^4}+O\left(x^5\right)\end{multline*}当$x=0$只剩第一项1, 即$f_+^{-1}(0)=1$.
而在0.7 (在$x_1$附近)反演的级数$f_-^{-1}$满足$f_-^{-1}(0)=x_1.$

1. Normal@InverseSeries[Series[Log[x + 1, 2] + x - 2, {x, 0.7, 5}]] /. x -> 0

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输出0.715006
$f^{-1}_+$在$x=1$的级数可以像Lambert W函数那样求出收敛半径吗