note-math

analytic-struct-product_(tag) Product space

Asymmetry: $𝑉 \neq 𝑊$ ==> No $𝑉 ⊙ 𝑊$

We have to use $(⊙^{𝑛} 𝑉) \otimes (⊙^{𝑚} 𝑊)$ and partial derivatives

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

The radius of convergence is at least $\min (𝑅_{𝐴}, 𝑅_{𝐵})$

(Related to Cauchy product. Try to find a better proof method)

$𝑐_{1} = 𝑎_{1} 𝑏_{0} + 𝑎_{0} 𝑏_{1}$

Restore $\frac{1}{𝑛!}$ in differentiation, $\sum 𝑎_{𝑛} 𝑣^{𝑛} \sim \sum \frac{1}{𝑛!} 𝑑^{𝑛} 𝑓 (𝑥) (𝑣^{𝑛})$

==> 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}

Collect the $𝑛 + 𝑚$ tensor $𝐴_{𝑛} (𝑣_{1} \dots 𝑣_{𝑛}) 𝐵_{𝑚} (𝑣_{𝑛 + 1} \dots 𝑣_{𝑛 + 𝑚})$

let $\forall 𝑖, 𝑣_{𝑖} = 𝑣$ to get the $𝑛 + 𝑚$ polynomial $𝐴_{𝑛} (𝑣^{𝑛}) 𝐵_{𝑚} (𝑣^{𝑚})$

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

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

==> 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 calculate directly

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

$𝑚 = 0$ : $1 = 𝐴_{0} 𝐵_{0}$

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

==> $𝐵_{𝑚} = - \frac{1}{𝐴_{0}} \sum_{𝑛 = 1}^{𝑚} 𝐴_{𝑛} 𝐵_{𝑚 - 𝑛}$ , 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, ${(\frac{1}{𝑥})}^{'} = - \frac{1}{𝑥^{2}}$

radius of convergence

try inductive proof $| 𝐵_{𝑚} | \leq 𝑅^{𝑚}$

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

To complete induction, 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 all possible sources of the compounded $𝑣^{𝑙}$

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

thus can only come from 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}}

$𝑐_{1} = 𝑎_{1} 𝑏_{1}$ . Written as a differential 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 $𝐵_{1}^{𝑖_{1}} \dots 𝐵_{𝑙}^{𝑖_{𝑙}} (𝑣^{1 \cdot 𝑖_{1} + \dots + 𝑙 \cdot 𝑖_{𝑙}}) = {(𝐵_{1} (𝑣^{⊙ 1}))}^{⊙ 𝑖_{1}} ⊙ \dots ⊙ {(𝐵_{𝑙} (𝑣^{⊙ 𝑙}))}^{⊙ 𝑖_{𝑙}}$

$𝐶_{1} (𝑣) = 𝐴_{1} (𝐵_{1} (𝑣))$ , written as a differential is chain-rule_(tag)

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

Generally written as the differential form

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

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

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

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

Get $\frac{𝑙!}{{(1!)}^{𝑖_{1}} \cdot 𝑖_{1!} \dots {(𝑙!)}^{𝑖_{𝑙}} \cdot 𝑖_{𝑙!}}$ (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 $𝑓^{- 1} (𝑦 + 𝑣) = \sum 𝐵_{𝑚} (𝑣^{𝑚})$

First-order differential calculation. $𝑓 \circ 𝑓^{- 1} = 𝑓^{- 1} \circ 𝑓 = 𝟙 : 𝑣 ⇝ 𝑣$ , use composite

$𝟙 (𝑣) = \sum 𝐶_{𝑙} (𝑣^{𝑙})$

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

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

Higher-order differential calculation. use induction for $𝐵_{1}, \dots, 𝐵_{𝑙 - 1}$

$𝐵_{𝑙}$ only comes from

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

$𝑖_{1} + \dots + 𝑖_{𝑙} = 1$

$(\binom{𝑙}{0, \dots, 𝑙}) = 1$

==> (omitting $(𝑣^{𝑙})$ )

\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}}

Because it may not converge, $𝑓 \sim \sum_{𝑛 \geq 1} 𝐴_{𝑛} (𝑣^{𝑛})$ cannot be directly used as a $𝕂^{𝑑} \to 𝕂^{𝑑}$ function

But it can be extended to $𝑓 : ⨁_{𝑛 = 1}^{\infty} Lin (⊙^{𝑛} 𝕂^{𝑑} \to 𝕂^{𝑑^{'}}) \to self$

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

The radius of convergence of the inverse function is non-zero (p.77 of ref-4)

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 $\in ℕ$ )

Construct a power series control with a non-zero radius of convergence for (almost) $ℝ_{\geq 0}$ such that $| 𝐴_{𝑘} | \leq 𝑎_{𝑘}, | 𝐵_{𝑘} | \leq 𝑏_{𝑘}$

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 $𝑘 \geq 2 ⟹ 𝑎_{𝑘} \in ℝ_{\geq 0}$

Its inverse is $\sum_{𝑗 \geq 1} 𝑏_{𝑗} 𝑣^{𝑗}$ with $𝑗 \geq 1 ⟹ 𝑏_{𝑗} \in ℝ_{\geq 0}$ . $ℝ_{\geq 0}$ to prove. radius of convergence is non-zero 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 the inverse power series $𝑏_{𝑘}$ of the power series $𝑎_{𝑘}$ has a non-zero radius of convergence

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

In order to find the inverse mapping $𝑓^{- 1} = 𝑔$ of $𝑓$ , solve the equation $𝛼 𝑔 (𝑣) + \frac{{(𝛽 𝑔 (𝑣))}^{2}}{1 - 𝛽 𝑔 (𝑣)} = 𝑣$

==> Quadratic equation of $𝑔 (𝑣)$ , there are two roots

use $𝑓 (0) = 0 ⟹ 𝑔 (0) = 0$ , select the correct root

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

use ${(1 + 𝑤)}^{\frac{1}{2}} \sim \sum_{𝑛 \in ℕ} (\binom{\frac{1}{2}}{𝑛}) 𝑤^{𝑛}$ radius of convergence $1$ ==> $𝑔 (𝑣) \sim \sum 𝑏_{𝑗} 𝑣^{𝑗}$ non-zero radius of convergence

use $| 𝐵_{𝑙} | \leq 𝑏_{𝑙}$ ==> $\sum 𝐵_{𝑗} 𝑣^{𝑗}$ non-zero radius of convergence

Although the exact radius of convergence cannot be given here, the method of proving the inverse function by the compression fixed point principle cannot give the exact maximal local reversible region for the pure differential method.

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

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

The calculation of differentials and differential functions does not require series in advance

$𝐶^{\infty}$ function with zero radius of convergence at finite points

$exp \frac{1}{1 - 𝑥^{2}}$ connected to $0$
Function that is $𝐶^{\infty}$ everywhere but has a radius of convergence of $0$ everywhere

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 $ℝ$

Continuous but nowhere differentiable

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 $𝑎 𝑏 > 1 + \frac{3}{2} 𝜋$

$𝑘$ -th order differentiable but not $𝑘 + 1$ -th order differentiable: use the integrals of each order of the Weierstrass function
$𝑘$ -th order differentiable but $𝑘$ -th order not continuously differentiable (although $𝑘$ -th order differentiable implies $𝑘 - 1$ -th order continuously differentiable): use $𝑥^{2} sin \frac{1}{𝑥}$ , $1$ -th order differentiable but not $1$ -th order continuously differentiable, use its integrals of each order
Continuous homeomorphism but not differentiable homeomorphism or analytic homeomorphism. $𝑥^{3}$
Diffeomorphism but not analytic diffeomorphism. Take the part of the smooth but everywhere non-analytic function where $𝑑 𝑓 \neq 0$ to get a local diffeomorphism. Local to global by using $\frac{𝑥}{1 - 𝑥}$ to get an analytic diffeomorphism from $(- 1, 1) \to ℝ$