一种任意收敛阶数的根求解方法
密码学与安全
2026-02-02 v1 计算机与社会
摘要
设 a ∈ R + \ { 0 } a\in \mathbb{R}^{+}\backslash\left\{0\right\} a ∈ R + \ { 0 } M ∈ N M\in\mathbb{N} M ∈ N t M − a = 0 t^M-a=0 t M − a = 0 1 − t M a = 0 1-\frac{t^M}{a}=0 1 − a t M = 0 a M \sqrt[M]{a} M a a M \sqrt[M]{a} M a F ( x ) = ∏ l = 1 P ( 1 + 1 l ⋅ M ) ∫ 0 x ( 1 − t M a ) P d t = ∑ k = 0 P ( − 1 ) k a k ⋅ ( P k ) ⋅ x k ⋅ M + 1 k ⋅ M + 1 F\left(x\right) =\prod_{l=1}^{P}\left(1+\frac{1}{l\cdot M}\right) \int\limits_{0}^{x}\!\left(1-\frac{{t}^{M}}{a}\right)^{P}{\rm d}t =\sum\limits_{k=0}^{P}\frac{\left(-1\right)^{\,k}}{a^{\,k}}\cdot\binom{P}{k}\cdot\frac{x^{\,k\,\cdot M+1}}{k\cdot M+1} F ( x ) = l = 1 ∏ P ( 1 + l ⋅ M 1 ) 0 ∫ x ( 1 − a t M ) P d t = k = 0 ∑ P a k ( − 1 ) k ⋅ ( k P ) ⋅ k ⋅ M + 1 x k ⋅ M + 1 ( P ⋅ M + 1 ) \left(P\cdot M+1\right) ( P ⋅ M + 1 ) ( P + 1 ) \left(P+1\right) ( P + 1 ) a M \sqrt[M]{a} M a a ∈ R + \ { 0 } a\in \mathbb{R}^{+}\backslash\left\{0\right\} a ∈ R + \ { 0 } M ∈ N M\in\mathbb{N} M ∈ N t M − a = 0 t^M-a=0 t M − a = 0 1 − t M a = 0 1-\frac{t^M}{a}=0 1 − a t M = 0 a M \sqrt[M]{a} M a a M \sqrt[M]{a} M a F ( x ) = ∏ l = 1 P ( 1 + 1 l ⋅ M ) ∫ 0 x ( 1 − t M a ) P d t = ∑ k = 0 P ( − 1 ) k a k ⋅ ( P k ) ⋅ x k ⋅ M + 1 k ⋅ M + 1 F\left(x\right) =\prod_{l=1}^{P}\left(1+\frac{1}{l\cdot M}\right) \int\limits_{0}^{x}\!\left(1-\frac{{t}^{M}}{a}\right)^{P}{\rm d}t =\sum\limits_{k=0}^{P}\frac{\left(-1\right)^{\,k}}{a^{\,k}}\cdot\binom{P}{k}\cdot\frac{x^{\,k\cdot M+1}}{k\cdot M+1} F ( x ) = l = 1 ∏ P ( 1 + l ⋅ M 1 ) 0 ∫ x ( 1 − a t M ) P d t = k = 0 ∑ P a k ( − 1 ) k ⋅ ( k P ) ⋅ k ⋅ M + 1 x k ⋅ M + 1 ( P ⋅ M + 1 ) \left(P\cdot M+1\right) ( P ⋅ M + 1 ) ( P + 1 ) \left(P+1\right) ( P + 1 )
引用
@article{arxiv.2601.22185,
title = {MemeChain: A Multimodal Cross-Chain Dataset for Meme Coin Forensics and Risk Analysis},
author = {Alberto Maria Mongardini and Alessandro Mei},
journal= {arXiv preprint arXiv:2601.22185},
year = {2026}
}