$x^2+\epsilon x-1=0$的渐近解

hbghlyj

$\abs\epsilon< 1, x^2+\epsilon x-1=0$的渐近解为
$$x_+\sim1-\frac{1}{2} \epsilon + \frac{1}{8} \epsilon^{2}+O\left(\epsilon^{4}\right)$$
与
$$x_-\sim-1-\frac{1}{2} \epsilon - \frac{1}{8} \epsilon^{2}+O\left(\epsilon^{4}\right)$$

Czhang271828

直截了当的方法是
\begin{align*}
x_\pm &=\dfrac{-\epsilon\pm \sqrt {\epsilon^2+4}}{2}\\
&=\pm \sqrt{1+\epsilon^2/4}-\epsilon/2 \\
&=\pm (1+\frac{\epsilon^2}{8}-\frac{\epsilon^4}{128}+\mathcal O(\epsilon ^6))-\dfrac\epsilon 2.
\end{align*}
但求根公式未必能解析地写出. 一般地, 就是给定在 $(x_0,f(x_0))$ 处可逆的光滑函数 $f$, 并且已知 $f^{-1}$ 的表达式, 试求 $f(x_0)$ 附近的 Taylor 展开. 用 $f$ 各阶导数表示 $f^{-1}$ 各阶导数的算法估计有很多, 此处不议.

hbghlyj

Czhang271828 发表于 2023-2-28 07:21
用 $f$ 各阶导数表示 $f^{-1}$ 各阶导数的算法估计有很多, 此处不议.

In mathematical analysis, the Lagrange inversion theorem, also known as the Lagrange–Bürmann formula, gives the Taylor series expansion of the inverse function of an analytic function.

例子. 如何求Lambert W函数的级数?

hbghlyj

Czhang271828 发表于 2023-2-28 07:21
一般地, 就是给定在 $(x_0,f(x_0))$ 处可逆的光滑函数 $f$, 并且已知 $f^{-1}$ 的表达式

我用同样的方法来解$x\tan x=ϵ\;(0<ϵ\ll1)$, 即$\tan x=\fracϵx$. 注意$k≠0$与$k=0$时leading order不同:
Putting $x∼x_0+ϵx_1+O(ϵ^2)$ into the equation, expanding $\tan x$ at $x_0$,$$\left(\tan x_0+(1+\tan^2 x_0)ϵx_1+O(ϵ^2)\right)\left(x_0+ϵx_1+O(ϵ^2)\right)=ϵ$$Equating the constant terms\[x_0\tan x_0=0⇒x_0=kπ,k∈\Bbb Z\]Putting into the equation, we get$$\left(ϵx_1+O(ϵ^2)\right)\left(kπ+ϵx_1+O(ϵ^2)\right)=ϵ$$For $k≠0$, equating the coefficients of $ϵ$\[x_1=\frac1{kπ}\]we obtain the asymptotic expansion$$x\sim kπ+\fracϵ{kπ}+O(ϵ^2)$$

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For $k=0$, equating the coefficient of $ϵ^1$ we get $0=1$, contradiction😯
When $x\to0$, $\tan x=x+O(x^3)$, so $x(x+O(x^3))=ϵ$, we must have $x=O(ϵ^{1/2})$. Let$$x∼ϵ^{1/2}(x_0+ϵx_1+O(ϵ^2))$$Plugging into the equation, if expanding $\tan x$ at $x_0$,\[\left(\tan(ϵ^{1/2}x_0)+(1+\tan^2(ϵ^{1/2}x_0))ϵ^{3/2}x_1+O(ϵ^{5/2})\right)ϵ^{1/2}\left(x_0+ϵx_1+O(ϵ^2)\right)=ϵ\]which is cumbersome😵‍💫To get rid of $\tan$ let us expand at $0$ instead😀\[\left(ϵ^{1/2}x_0+ϵ^{3/2}x_1+O(ϵ^{5/2})+\frac13\left(ϵ^{1/2}x_0+O(ϵ^{3/2})\right)^3+O(ϵ^{5/2})\right)ϵ^{1/2}\left(x_0+ϵx_1+O(ϵ^2)\right)=ϵ\]
Equating the coefficient of $ϵ^1$ we get$$x_0^2=1⇒x_0=±1$$
Equating the coefficient of $ϵ^2$ we get$$2x_0x_1+\frac{x_0^4}3=0⇒x_1=-\frac16x_0$$
we obtain the asymptotic expansions $±ϵ^{1/2}(1-\fracϵ6+O(ϵ^2))$

hbghlyj

1. InverseSeries[Series[x Tan[x], {x, Pi, 1}]]

复制代码

\[\pi +\frac{x}{\pi }+O\left(x^2\right)\]

1. InverseSeries[Series[x Tan[x], {x, 0, 4}]]

复制代码

\[\sqrt{x}-\frac{x^{3/2}}{6}+O\left(x^2\right)\]

hbghlyj

以下是InverseSeries的输出:

级数	反演级数
$x^2+O\left(x^3\right)$	$\sqrt{x}+O\left(x^1\right)$
$x^2+\frac{x^4}{3}+O\left(x^5\right)$	$\sqrt{x}-\frac{x^{3/2}}{6}+O\left(x^2\right)$
$x^2+\frac{x^4}{3}+\frac{2 x^6}{15}+O\left(x^7\right)$	$\sqrt{x}-\frac{x^{3/2}}{6}+\frac{11 x^{5/2}}{360}+O\left(x^3\right)$
$x^2+\frac{x^4}{3}+\frac{2 x^6}{15}+\frac{17 x^8}{315}+O\left(x^9\right)$	$\sqrt{x}-\frac{x^{3/2}}{6}+\frac{11 x^{5/2}}{360}-\frac{17 x^{7/2}}{5040}+O\left(x^4\right)$

级数的最高阶是$2,4,6,8$
反演级数的最高阶是$\frac12,\frac32,\frac52,\frac72$这能否直接看出来

hbghlyj

$$x^2+\frac{x^4}{3}+\frac{2 x^6}{15}+\frac{17 x^8}{315}+O\left(x^9\right)$$
的InverseSeries结果为$$\sqrt{x}-\frac{x^{3/2}}{6}+\frac{11 x^{5/2}}{360}-\frac{17 x^{7/2}}{5040}+O\left(x^4\right)$$再做一次InverseSeries结果为$$x^2+\frac{x^4}{3}+\frac{2 x^6}{15}+O\left(x^7\right)$$与原级数相比, 最高阶数减少了$1$, 最高阶项的数据丢失了!
再做一次InverseSeries结果为
\[\sqrt{x}-\frac{x^{3/2}}{6}+\frac{11 x^{5/2}}{360}+O\left(x^3\right)\]再做一次InverseSeries结果为
\[x^2+\frac{x^4}{3}+O\left(x^5\right)\]
最高阶数又减少了$1$!