note-math

analytic-struct-product_(tag) 积空间

不对称性: $𝑉 \neq 𝑊$ ==> 没有 $𝑉 ⊙ 𝑊$

只好使用 $(⊙^{𝑛} 𝑉) \otimes (⊙^{𝑚} 𝑊)$ 和偏微分

something like

𝑑^{𝑙} 𝑓 (𝑥, 𝑦) : 𝑣, 𝑤 ⇝ \sum_{𝑛 + 𝑚 = 𝑙} (\binom{𝑙}{𝑛, 𝑚}) \frac{\partial^{𝑛 + 𝑚} 𝑓}{\partial 𝑣^{𝑛} \partial 𝑤^{𝑚}} (𝑥, 𝑦) (𝑣^{𝑛} \otimes 𝑤^{𝑚})

mulitplication-analytic_(tag)

$𝕂 \to 𝕂$

(\sum_{𝑛 \in ℕ} 𝑎_{𝑛} 𝑣^{𝑛}) (\sum_{𝑚 \in ℕ} 𝑏_{𝑚} 𝑣^{𝑚}) = \sum_{𝑙 \in ℕ} 𝑐_{𝑙} 𝑣^{𝑙}

with $𝑐_{𝑙} = \sum_{𝑛 + 𝑚 = 𝑙} 𝑎_{𝑛} 𝑏_{𝑚}$

收敛半径至少是 $\min (𝑅_{𝐴}, 𝑅_{𝐵})$

(联系于 Cauchy product. 尝试寻找更好的证明方法)

$𝑐_{1} = 𝑎_{1} 𝑏_{0} + 𝑎_{0} 𝑏_{1}$

恢复微分中的 $\frac{1}{𝑛!}$ , $\sum 𝑎_{𝑛} 𝑣^{𝑛} \sim \sum \frac{1}{𝑛!} 𝑑^{𝑛} 𝑓 (𝑥) (𝑣^{𝑛})$

==> Leibniz-law-1d_(tag)

$𝑑 (𝑓 𝑔) (𝑥) (𝑣) = (𝑑 𝑓 (𝑥) (𝑣)) 𝑔 (𝑥) + 𝑓 (𝑥) (𝑑 𝑔 (𝑥) (𝑣))$ , or

$\frac{\partial (𝑓 𝑔)}{\partial 𝑣} = \frac{\partial 𝑓}{\partial 𝑣} 𝑔 + 𝑓 \frac{\partial 𝑔}{\partial 𝑣}$ , or

${(𝑓 𝑔)}^{'} = 𝑓^{'} 𝑔 + 𝑓 𝑔^{'}$

$𝕂^{𝑑} \to 𝕂$

(\sum_{𝑛 \in ℕ} 𝑎_{𝑛} 𝑣^{𝑛}) (\sum_{𝑚 \in ℕ} 𝑏_{𝑚} 𝑣^{𝑚}) = \sum_{𝑙 \in ℕ} 𝑐_{𝑙} 𝑣^{𝑙}

收集 $𝑛 + 𝑚$ 重张量 $𝐴_{𝑛} (𝑣_{1} \dots 𝑣_{𝑛}) 𝐵_{𝑚} (𝑣_{𝑛 + 1} \dots 𝑣_{𝑛 + 𝑚})$

let $\forall 𝑖, 𝑣_{𝑖} = 𝑣$ 得到 $𝑛 + 𝑚$ 多项式 $𝐴_{𝑛} (𝑣^{𝑛}) 𝐵_{𝑚} (𝑣^{𝑚})$

==> $𝐶_{𝑙} (𝑣^{𝑙}) = \sum_{𝑛 + 𝑚 = 𝑙} 𝐴_{𝑛} (𝑣^{𝑛}) 𝐵_{𝑚} (𝑣^{𝑚})$

$𝐶_{1} 𝑣 = 𝐴_{1} (𝑣) 𝐵_{0} + 𝐴_{0} 𝐵_{1} (𝑣)$

==> Leibniz-law_(tag) $\frac{\partial (𝑓 𝑔)}{\partial 𝑣} = \frac{\partial 𝑓}{\partial 𝑣} 𝑔 + 𝑓 \frac{\partial 𝑔}{\partial 𝑣}$

mulitplication-inverse-analytic_(tag)

let $𝑓 : 𝕂^{𝑑} \to 𝕂$ , $𝑓 = \sum 𝐴_{𝑛} (𝑣^{𝑛})$ , $𝐴_{0} \neq 0$

use $𝑓 = 1 - 𝑔$ and $\frac{1}{1 - 𝑔} = 1 + 𝑔 + 𝑔^{2} + \dots$

or 直接计算

let $\frac{1}{𝑓} = \sum 𝐵_{𝑚} 𝑣^{𝑚}$ , $\frac{1}{𝑓} \cdot 𝑓 = 1$ , use 乘法

$𝑚 = 0$ : $1 = 𝐴_{0} 𝐵_{0}$

$𝑚 \geq 1$ : $0 = 𝐶_{𝑙} = \sum_{𝑛 + 𝑚 = 𝑙} 𝐴_{𝑛} 𝐵_{𝑚}$

==> $𝐵_{𝑚} = - \frac{1}{𝐴_{0}} \sum_{𝑛 = 1}^{𝑚} 𝐴_{𝑛} 𝐵_{𝑚 - 𝑛}$ , use induction $𝐵_{0}, \dots, 𝐵_{𝑚 - 1}$

differential-of-multiplication-inverse_(tag) use Leibniz law

$0 = \partial_{𝑣} (\frac{1}{𝑓} \cdot 𝑓) = \partial_{𝑣} (\frac{1}{𝑓}) 𝑓 + \frac{1}{𝑓} \partial_{𝑣} 𝑓 ⟹ \partial_{𝑣} (\frac{1}{𝑓}) = - \frac{1}{𝑓^{2}} \partial_{𝑣} 𝑓$ , or

${(\frac{1}{𝑓})}^{'} = - \frac{𝑓^{'}}{𝑓^{2}}$

in particular, ${(\frac{1}{𝑥})}^{'} = - \frac{1}{𝑥^{2}}$

收敛半径

try 归纳证明 $| 𝐵_{𝑚} | \leq 𝑅^{𝑚}$

$\begin{matrix} | 𝐴_{0} | | 𝐵_{𝑚} | & \leq & \sum_{𝑛 = 1}^{𝑚} | 𝐴_{𝑛} | | 𝐵_{𝑚 - 𝑛} | \\ \leq & \sum_{𝑛 = 1}^{𝑚} | 𝐴_{𝑛} | 𝑅^{𝑚 - 𝑛} (by induction 𝐵_{1}, \dots, 𝐵_{𝑚 - 1}) \\ = & 𝑅^{𝑚} \sum_{𝑛 = 1}^{𝑚} | 𝐴_{𝑛} | {(\frac{1}{𝑅})}^{𝑛} \\ \leq & 𝑅^{𝑚} \sum_{𝑛 = 1}^{\infty} | 𝐴_{𝑛} | {(\frac{1}{𝑅})}^{𝑛} \end{matrix}$

为完成归纳, use $𝑅$ with $\frac{1}{| 𝐴_{0} |} \sum_{𝑛 = 1}^{\infty} | 𝐴_{𝑛} | {(\frac{1}{𝑅})}^{𝑛} \leq 1$

compose-op-analytic_(tag)

$𝕂 \to 𝕂$

let $𝑓 (𝑦 + 𝑤) = \sum 𝑎_{𝑛} 𝑣^{𝑛}$ , $𝑔 (𝑥 + 𝑣) = \sum 𝑏_{𝑚} 𝑤^{𝑚}$

with $𝑎_{0} = 𝑓 (𝑦) = 𝑓 (𝑔 (𝑥)) = 𝑓 (𝑏_{0})$

$\begin{matrix} 𝑓 (𝑔 (𝑥 + 𝑣)) \circ & = & 𝑓 (𝑏_{0} + \sum_{𝑚 = 1}^{\infty} 𝑏_{𝑚} 𝑣^{𝑚}) \\ = & 𝑎_{0} + \sum_{𝑛 = 1}^{\infty} 𝑎_{𝑛} {(\sum_{𝑚 = 1}^{\infty} 𝑏_{𝑚} 𝑣^{𝑚})}^{𝑛} \\ = & 𝑐_{0} + \sum_{𝑙 = 1}^{\infty} 𝑐_{𝑙} 𝑣^{𝑙} \end{matrix}$

where 复合后的 $𝑣^{𝑙}$ 的所有可能来源

$𝑣^{𝑙} = {(𝑣^{1})}^{𝑖_{1}} \dots {(𝑣^{𝑙})}^{𝑖_{𝑙}} = 𝑣^{1 \cdot 𝑖_{1} + \dots + 𝑙 \cdot 𝑖_{𝑙}}$ with $𝑙 = 1 \cdot 𝑖_{1} + \dots + 𝑙 \cdot 𝑖_{𝑙}$

从而只能来自 for $𝑘 = 1, \dots, 𝑙$

${(𝑏_{1} 𝑣 + \dots + 𝑏_{𝑙} 𝑣^{𝑙})}^{𝑘} = \sum_{𝑖_{1} + \dots + 𝑖_{𝑙} = 𝑘} (\binom{𝑘}{𝑖_{1}, \dots, 𝑖_{𝑙}}) 𝑏_{1}^{𝑖_{1}} \dots 𝑏_{𝑙}^{𝑖_{𝑙}}$ (cf. #link(<multi-combination>)[])

==>

𝑐_{𝑙} 𝑣^{𝑙} = \sum_{𝑘 = 1, \dots, 𝑙} \sum_{\begin{matrix} 𝑖_{1}, \dots, 𝑖_{𝑙} \in ℕ \\ 𝑖_{1} + \dots + 𝑖_{𝑙} = 𝑘 \\ 1 \cdot 𝑖_{1} + \dots + 𝑙 \cdot 𝑖_{𝑙} = 𝑙 \end{matrix}} (\binom{𝑘}{𝑖_{1}, \dots, 𝑖_{𝑙}}) 𝑎_{𝑘} 𝑏_{1}^{𝑖_{1}} \dots 𝑏_{𝑙}^{𝑖_{𝑙}} 𝑣^{1 \cdot 𝑖_{1} + \dots + 𝑙 \cdot 𝑖_{𝑙}}

$𝑐_{1} = 𝑎_{1} 𝑏_{1}$ . 写为微分 chain-rule-1d_(tag)

$𝑑 (𝑓 \circ 𝑔) (𝑥) (𝑣) = 𝑑 𝑓 (𝑔 (𝑥)) (𝑑 𝑔 (𝑥) (𝑣))$ , or

${(𝑓 \circ 𝑔)}^{'} (𝑥) = 𝑓^{'} (𝑔 (𝑥)) 𝑔^{'} (𝑥)$

$𝕂^{𝑑} \to 𝕂^{𝑑^{'}}$

𝐶_{𝑙} (𝑣^{𝑙}) = \sum_{𝑘 = 1, \dots, 𝑙} \sum_{\begin{matrix} 𝑖_{1}, \dots, 𝑖_{𝑙} \in ℕ \\ 𝑖_{1} + \dots + 𝑖_{𝑙} = 𝑘 \\ 1 \cdot 𝑖_{1} + \dots + 𝑙 \cdot 𝑖_{𝑙} = 𝑙 \end{matrix}} (\binom{𝑘}{𝑖_{1}, \dots, 𝑖_{𝑙}}) 𝐴_{𝑘} (𝐵_{1}^{𝑖_{1}} \dots 𝐵_{𝑙}^{𝑖_{𝑙}} (𝑣^{1 \cdot 𝑖_{1} + \dots + 𝑙 \cdot 𝑖_{𝑙}}))

where $𝐵_{1}^{𝑖_{1}} \dots 𝐵_{𝑙}^{𝑖_{𝑙}} (𝑣^{1 \cdot 𝑖_{1} + \dots + 𝑙 \cdot 𝑖_{𝑙}}) = {(𝐵_{1} (𝑣^{⊙ 1}))}^{⊙ 𝑖_{1}} ⊙ \dots ⊙ {(𝐵_{𝑙} (𝑣^{⊙ 𝑙}))}^{⊙ 𝑖_{𝑙}}$

$𝐶_{1} (𝑣) = 𝐴_{1} (𝐵_{1} (𝑣))$ , 写为微分就是 chain-rule_(tag)

$𝑑 (𝑓 \circ 𝑔) (𝑥) (𝑣) = 𝑑 𝑓 (𝑔 (𝑥)) (𝑑 𝑔 (𝑥) (𝑣))$

一般的写为微分的形式

$𝐶_{𝑙} \sim \frac{1}{𝑙!} 𝑑^{𝑙} (𝑓 \circ 𝑔)$
$𝐴_{𝑘} \sim \frac{1}{𝑘!} 𝑑^{𝑘} 𝑓$
$𝐵_{𝑗} \sim \frac{1}{𝑗!} 𝑑^{𝑗} 𝑔$

in $𝑑^{𝑙} (𝑓 \circ 𝑔) (𝑥) (𝑣^{𝑙}) = formula of 𝑑^{𝑘} 𝑓, 𝑑^{𝑗} 𝑔$

将 $𝑙!, \frac{1}{𝑘!}, {(\frac{1}{𝑗!})}^{𝑖_{𝑗}}$ 提取

置于 $(\binom{𝑘}{𝑖_{1}, \dots, 𝑖_{𝑙}}) = \frac{𝑘!}{𝑖_{1!} \dots 𝑖_{𝑙!}}$

得到 $\frac{𝑙!}{{(1!)}^{𝑖_{1}} \cdot 𝑖_{1!} \dots {(𝑙!)}^{𝑖_{𝑙}} \cdot 𝑖_{𝑙!}}$ (this is not $\frac{𝑙!}{(1 \cdot 𝑖_{1})! \dots (𝑙 \cdot 𝑖_{𝑙})!} = (\binom{𝑙}{1 \cdot 𝑖_{1}, \dots, 𝑙 \cdot 𝑖_{𝑙}})$ )

𝑑^{𝑙} (𝑓 \circ 𝑔) = \sum_{𝑘 = 1, \dots, 𝑙} \sum_{\begin{matrix} 𝑖_{1}, \dots, 𝑖_{𝑙} \in ℕ \\ 𝑖_{1} + \dots + 𝑖_{𝑙} = 𝑘 \\ 1 \cdot 𝑖_{1} + \dots + 𝑙 \cdot 𝑖_{𝑙} = 𝑙 \end{matrix}} \frac{𝑙!}{{(1!)}^{𝑖_{1}} \cdot 𝑖_{1!} \dots {(𝑙!)}^{𝑖_{𝑙}} \cdot 𝑖_{𝑙!}} 𝑑^{𝑘} 𝑓 {(𝑑^{1} 𝑔)}^{𝑖_{1}} \dots {(𝑑^{𝑙} 𝑔)}^{𝑖_{𝑙}}

inverse-analytic_(tag)

let $𝑓 \sim \sum 𝐴_{𝑛} (𝑣^{𝑛})$ , $𝕂^{𝑑} \to 𝕂^{𝑑}$ , $𝐴_{1} \in GL (𝑑, 𝕂)$

let $𝑓^{- 1} (𝑦 + 𝑣) = \sum 𝐵_{𝑚} (𝑣^{𝑚})$

一阶微分计算. $𝑓 \circ 𝑓^{- 1} = 𝑓^{- 1} \circ 𝑓 = 𝟙 : 𝑣 ⇝ 𝑣$ , use 复合

$𝟙 (𝑣) = \sum 𝐶_{𝑙} (𝑣^{𝑙})$

$\begin{matrix} 𝐶_{0} = 0 \\ 𝐶_{1} = 𝟙 \\ 𝐶_{𝑙} = 0, \forall 𝑙 \geq 2 \end{matrix}$

$𝟙 (𝑣) = 𝐶_{1} (𝑣) = 𝐴_{1} (𝐵_{1} (𝑣)) ⟹ 𝐵_{1} = 𝐴_{1}^{- 1}$ by $𝐴_{1} \in GL (𝑑, 𝕂)$

高阶微分计算. use 归纳 for $𝐵_{1}, \dots, 𝐵_{𝑙 - 1}$

$𝐵_{𝑙}$ 只来自

$𝑖_{1} = \dots = 𝑖_{𝑙 - 1} = 0$ and $𝑖_{𝑙} = 1$ ==> $1 \cdot 𝑖_{1} + \dots + 𝑙 \cdot 𝑖_{𝑙} = 𝑙$

$𝑖_{1} + \dots + 𝑖_{𝑙} = 1$

$(\binom{𝑙}{0, \dots, 𝑙}) = 1$

==> (省略 $(𝑣^{𝑙})$ )

\begin{matrix} 0 & = & 𝐶_{𝑙} \\ = & 𝐴_{1} 𝐵_{𝑙} \\ + \sum_{𝑘 = 2, \dots, 𝑙} \sum_{\begin{matrix} 𝑖_{1}, \dots, 𝑖_{𝑙} \in ℕ \\ 𝑖_{1} + \dots + 𝑖_{𝑙} = 𝑘 \\ 1 \cdot 𝑖_{1} + \dots + 𝑙 \cdot 𝑖_{𝑙} = 𝑙 \end{matrix}} (\binom{𝑘}{𝑖_{1}, \dots, 𝑖_{𝑙}}) 𝐴_{𝑘} 𝐵_{1}^{𝑖_{1}} \dots 𝐵_{𝑙 - 1}^{𝑖_{𝑙 - 1}} \end{matrix}

$𝐴_{1} \in GL$ ==>

𝐵_{𝑙} = - 𝐴_{1}^{- 1} \sum_{𝑘 = 2, \dots, 𝑙} \sum_{\begin{matrix} 𝑖_{1}, \dots, 𝑖_{𝑙} \in ℕ \\ 𝑖_{1} + \dots + 𝑖_{𝑙} = 𝑘 \\ 1 \cdot 𝑖_{1} + \dots + 𝑙 \cdot 𝑖_{𝑙} = 𝑙 \end{matrix}} (\binom{𝑘}{𝑖_{1}, \dots, 𝑖_{𝑙}}) 𝐴_{𝑘} 𝐵_{1}^{𝑖_{1}} \dots 𝐵_{𝑙 - 1}^{𝑖_{𝑙 - 1}}

由于可能不收敛, $𝑓 \sim \sum_{𝑛 \geq 1} 𝐴_{𝑛} (𝑣^{𝑛})$ 无法直接作为 $𝕂^{𝑑} \to 𝕂^{𝑑}$ 函数

但是可以扩充到 $𝑓 : ⨁_{𝑛 = 1}^{\infty} Lin (⊙^{𝑛} 𝕂^{𝑑} \to 𝕂^{𝑑^{'}}) \to self$

使得 $𝑓 \circ 𝑓^{- 1} = 𝑓^{- 1} \circ 𝑓 = 𝟙 of ⨁_{𝑛 = 1}^{\infty} Lin (⊙^{𝑛} 𝕂^{𝑑} \to 𝕂^{𝑑^{'}})$

逆函数的收敛半径非零 (p.77 of ref-4)

Reference

𝐵_{𝑙} = - 𝐴_{1}^{- 1} \sum_{𝑘 = 2, \dots, 𝑙} \sum_{\begin{matrix} 𝑖_{1}, \dots, 𝑖_{𝑙} \in ℕ \\ 𝑖_{1} + \dots + 𝑖_{𝑙} = 𝑘 \\ 1 \cdot 𝑖_{1} + \dots + 𝑙 \cdot 𝑖_{𝑙} = 𝑙 \end{matrix}} (\binom{𝑘}{𝑖_{1}, \dots, 𝑖_{𝑙}}) 𝐴_{𝑘} 𝐵_{1}^{𝑖_{1}} \dots 𝐵_{𝑙 - 1}^{𝑖_{𝑙 - 1}}

==>

\begin{matrix} | 𝐵_{𝑙} | & \leq & \frac{1}{| 𝐴_{1} |} \sum_{𝑘 = 2, \dots, 𝑙} \sum_{\begin{matrix} 𝑖_{1}, \dots, 𝑖_{𝑙} \in ℕ \\ 𝑖_{1} + \dots + 𝑖_{𝑙} = 𝑘 \\ 1 \cdot 𝑖_{1} + \dots + 𝑙 \cdot 𝑖_{𝑙} = 𝑙 \end{matrix}} (\binom{𝑘}{𝑖_{1}, \dots, 𝑖_{𝑙}}) | 𝐴_{𝑘} | | 𝐵_{1} |^{𝑖_{1}} \dots | 𝐵_{𝑙 - 1} |^{𝑖_{𝑙 - 1}} \end{matrix}

use $(\binom{𝑘}{𝑖_{1}, \dots, 𝑖_{𝑙}}) \in ℝ_{\geq 0}$ (indeed $\in ℕ$ )

构造 (几乎) $ℝ_{\geq 0}$ 的非零收敛半径的幂级数控制 $| 𝐴_{𝑘} | \leq 𝑎_{𝑘}, | 𝐵_{𝑘} | \leq 𝑏_{𝑘}$

if by induction, for $𝐴_{2}, \dots, 𝐴_{𝑙}$ , $𝐵_{1}, \dots, 𝐵_{𝑙 - 1}$ , $| 𝐴_{𝑘} | \leq 𝑎_{𝑘}$ , $| 𝐵_{𝑗} | \leq 𝑏_{𝑗}$

where $\sum_{𝑘 \geq 1} 𝑎_{𝑘} 𝑣^{𝑘}$ with $𝑘 \geq 2 ⟹ 𝑎_{𝑘} \in ℝ_{\geq 0}$

其逆是 $\sum_{𝑗 \geq 1} 𝑏_{𝑗} 𝑣^{𝑗}$ with $𝑗 \geq 1 ⟹ 𝑏_{𝑗} \in ℝ_{\geq 0}$ . $ℝ_{\geq 0}$ to prove. 收敛半径非零 to prove

use case of $𝕂 \to 𝕂$

𝑏_{𝑙} = - \frac{1}{𝑎_{1}} \sum_{𝑘 = 2, \dots, 𝑙} \sum_{\begin{matrix} 𝑖_{1}, \dots, 𝑖_{𝑙} \in ℕ \\ 𝑖_{1} + \dots + 𝑖_{𝑙} = 𝑘 \\ 1 \cdot 𝑖_{1} + \dots + 𝑙 \cdot 𝑖_{𝑙} = 𝑙 \end{matrix}} (\binom{𝑘}{𝑖_{1}, \dots, 𝑖_{𝑙}}) 𝑎_{𝑘} 𝑏_{1}^{𝑖_{1}} \dots 𝑏_{𝑙 - 1}^{𝑖_{𝑙 - 1}}

to get $𝑏_{𝑗} \geq 0$ , use $𝑎_{1} < 0$

==>

\begin{matrix} | 𝐵_{𝑙} | & \leq & \frac{1}{| 𝐴_{1} |} \sum_{𝑘 = 2, \dots, 𝑙} \sum_{\begin{matrix} 𝑖_{1}, \dots, 𝑖_{𝑙} \in ℕ \\ 𝑖_{1} + \dots + 𝑖_{𝑙} = 𝑘 \\ 1 \cdot 𝑖_{1} + \dots + 𝑙 \cdot 𝑖_{𝑙} = 𝑙 \end{matrix}} (\binom{𝑘}{𝑖_{1}, \dots, 𝑖_{𝑙}}) | 𝐴_{𝑘} | | 𝐵_{1} |^{𝑖_{1}} \dots | 𝐵_{𝑙 - 1} |^{𝑖_{𝑙 - 1}} \\ \leq & \frac{- 𝑎_{1}}{| 𝐴_{1} |} 𝑏_{𝑙} \end{matrix}

$| 𝐵_{1} | = \frac{1}{| 𝐴_{1} |}$ , $𝑏_{1} = \frac{1}{- 𝑎_{1}}$

to get $| 𝐵_{𝑙} | \leq 𝑏_{𝑙}$ , $| 𝐵_{1} | \leq 𝑏_{1}$ , use $𝑎_{1} = - | 𝐴_{1} | ≕ 𝛼$

to get $𝑘 \geq 2 ⟹ | 𝐴_{𝑘} | \leq 𝑎_{𝑘}$ , use $𝑎_{𝑘} = {(\sup_{𝑘 \geq 2} {| 𝐴_{𝑘} |^{\frac{1}{𝑘}}})}^{𝑘} ≕ 𝛽^{𝑘}$

now prove 幂级数 $𝑎_{𝑘}$ 的逆幂级数 $𝑏_{𝑘}$ 收敛半径非零

\begin{matrix} \sum_{𝑛 \geq 1} 𝑎_{𝑘} 𝑣^{𝑘} & = & 𝛼 𝑣 + \sum_{𝑛 \geq 2} 𝛽^{𝑘} 𝑣^{𝑘} \\ \sim & 𝛼 𝑣 + \frac{1}{1 - 𝛽 𝑣} - 1 - 𝛽 𝑣 \\ = & 𝛼 𝑣 + \frac{{(𝛽 𝑣)}^{2}}{1 - 𝛽 𝑣} \end{matrix}

let $𝑓 (𝑣) = 𝛼 𝑣 + \frac{{(𝛽 𝑣)}^{2}}{1 - 𝛽 𝑣} \sim \sum 𝑎_{𝑘} 𝑣^{𝑘}$ , $𝑓^{- 1} (𝑣) = 𝑔 (𝑣) \sim \sum 𝑏_{𝑗} 𝑣^{𝑗}$

为了求 $𝑓$ 的逆映射 $𝑓^{- 1} = 𝑔$ , 解方程 $𝛼 𝑔 (𝑣) + \frac{{(𝛽 𝑔 (𝑣))}^{2}}{1 - 𝛽 𝑔 (𝑣)} = 𝑣$

==> $𝑔 (𝑣)$ 的二次方程, 有两个根

use $𝑓 (0) = 0 ⟹ 𝑔 (0) = 0$ , 选取正确的根

𝑔 (𝑣) = \frac{- (𝛼 + 𝛽 𝑣) - {({(𝛼 + 𝛽 𝑣)}^{2} + 4 𝛽 (𝛽 - 𝛼) 𝑣)}^{\frac{1}{2}}}{2 𝛽 (𝛽 - 𝛼)}

use ${(1 + 𝑤)}^{\frac{1}{2}} \sim \sum_{𝑛 \in ℕ} (\binom{\frac{1}{2}}{𝑛}) 𝑤^{𝑛}$ 收敛半径 $1$ ==> $𝑔 (𝑣) \sim \sum 𝑏_{𝑗} 𝑣^{𝑗}$ 非零收敛半径

use $| 𝐵_{𝑙} | \leq 𝑏_{𝑙}$ ==> $\sum 𝐵_{𝑗} 𝑣^{𝑗}$ 非零收敛半径

虽然这里无法给出确切的收敛半径, 但是对于纯微分方法, 压缩不动点原理证明逆函数的方法也不能给出确切的极大局部可逆区域

differential-of-inverse_(tag)

$\begin{matrix} 𝟙 of power series space = 𝑓 \circ 𝑓^{- 1} \\ ⟹ 𝟙 of GL = 𝑑 𝑓 (𝑓^{- 1} (𝑥)) 𝑑 𝑓^{- 1} (𝑥) \\ ⟹ 𝑑 𝑓^{- 1} (𝑥) = {(𝑑 𝑓 (𝑓^{- 1} (𝑥)))}^{- 1} \end{matrix}$

or ${(𝑓^{- 1})}^{'} (𝑥) = \frac{1}{𝑓^{'} (𝑓^{- 1} (𝑥))}$

implicit-function_(tag)

use #link(<analytic-struct-product>)[]

$𝐹 (𝑥, 𝑦) = 0$ and $\frac{\partial 𝐹}{\partial 𝑦} (𝑥, 𝑦) \in GL$

==> $𝐹 (𝑥, 𝑓 (𝑥)) = 0$ , $𝑑 𝑓 (𝑥) = - {(\frac{\partial 𝐹}{\partial 𝑦})}^{- 1} \frac{\partial 𝐹}{\partial 𝑥} (𝑥, 𝑓 (𝑥))$

微分和微分函数的计算不需要预先有级数

有限点处收敛半径为零的 $𝐶^{\infty}$ 函数

$exp \frac{1}{1 - 𝑥^{2}}$ 接上 $0$
处处 $𝐶^{\infty}$ 但处处收敛半径 $0$ 的函数

wiki: Non-analytic_smooth_function

𝐹 (𝑥) ≔ \sum_{𝑘 \in ℕ} 𝑒^{- \sqrt{2^{𝑘}}} cos (2^{𝑘} 𝑥)

Since the series $\sum_{𝑘 \in ℕ} 𝑒^{- \sqrt{2^{𝑘}}} {(2^{𝑘})}^{𝑛}$ converges for forall $𝑛 \in ℕ$ , this function is easily seen to be of class $𝐶^{\infty}$ , by a standard inductive application of the Weierstrass M-test to demonstrate uniform convergence of each series of derivatives.

We now show that $𝐹 (𝑥)$ is not analytic at any dyadic rational multiple of $𝜋$ , that is, at any $𝑥 ≔ 𝜋 \cdot 𝑝 \cdot 2^{- 𝑞}$ with $𝑝 \in ℤ$ and $𝑞 \in ℕ$ .

Since the sum of the first q terms is analytic, we need only consider $𝐹_{> 𝑞} (𝑥)$ , the sum of the terms with $𝑘 > 𝑞$ .

For forall orders of derivation $𝑛 = 2^{𝑚}$ with $𝑚 \in 𝑁$ , $𝑚 \geq 2$ and $𝑚 > \frac{𝑞}{2}$ we have

\begin{matrix} 𝐹_{> 𝑞}^{(𝑛)} & ≔ & \sum_{\begin{matrix} 𝑘 \in ℕ \\ 𝑘 > 𝑞 \end{matrix}} 𝑒^{- \sqrt{2^{𝑘}}} {(2^{𝑘})}^{𝑛} cos (2^{𝑘} 𝑥) \\ = & \sum_{\begin{matrix} 𝑘 \in ℕ \\ 𝑘 > 𝑞 \end{matrix}} 𝑒^{- \sqrt{2^{𝑘}}} {(2^{𝑘})}^{𝑛} \\ \geq & 𝑒^{- 𝑛} 𝑛^{2 𝑛} (as 𝑛 \to \infty) \end{matrix}

where we used the fact that $cos (2^{𝑘} 𝑥) = 1$ for forall $2^{𝑘} > 2^{𝑞}$ , and we bounded the first sum from below by the term with $2^{𝑘} = 2^{2 𝑚} = 𝑛^{2}$ .

As a consequence, at any such $𝑥 \in ℝ$ ,

\underset{𝑛 \to \infty}{lim sup} {(\frac{| 𝐹_{> 𝑞}^{(𝑛)} |}{𝑛!})}^{\frac{1}{𝑛}} = \infty

Since the set of analyticity of a function is an open set, and since dyadic rationals are dense, we conclude that $𝐹_{> 𝑞}$ , and hence $𝐹$ , is nowhere analytic in $ℝ$

连续但处处不可微

wiki: Weierstrass_function

𝑓 (𝑥) = \sum_{𝑛 = 0}^{\infty} 𝑎^{𝑛} cos (𝑏^{𝑛} 𝜋 𝑥)

where $0 < 𝑎 < 1$ , $𝑏$ is positive odd integer, and $𝑎 𝑏 > 1 + \frac{3}{2} 𝜋$

$𝑘$ 阶可微但不 $𝑘 + 1$ 阶可微: 使用 Weierstrass 函数的各阶积分
$𝑘$ 阶可微但 $𝑘$ 阶不连续可微 (虽然 $𝑘$ 阶可微蕴含 $𝑘 - 1$ 阶连续可微): 使用 $𝑥^{2} sin \frac{1}{𝑥}$ , $1$ 阶可微但不 $1$ 阶连续可微, 使用其各阶积分
连续同胚但不微分同胚不解析同胚. $𝑥^{3}$
微分同胚但不解析同胚. 光滑但处处不解析函数中取 $𝑑 𝑓 \neq 0$ 的部分得到局部微分同胚. 局部到全局 by 用 $\frac{𝑥}{1 - 𝑥}$ 得到 $(- 1, 1) \to ℝ$ 的解析同胚