另类代数方程的根求解

数学星空

1.设\(x_n,n \geq 2\)是下面方程的唯一正数解：

      \(x^{-n}=\sum_{k=1}^{\infty}(x+k)^{-n}\)

   若设\(x_n=a_2+a_1n+\frac{a_3}{n}+\frac{a_4}{n^2}+\frac{a_5}{n^3}+\frac{a_6}{n^4}+\frac{a_7}{n^5}+\frac{a_8}{n^6}+...\)

   求\(a_1,a_2,a_3,a_4,a_5,a_6,a_7\)

本题来源于《Xionger 问题与解答级数等式渐近分析与麻蛋定理》，见楼下附件

2.设\(x_n,n \geq 2\)是下面方程的唯一正数解：

      \(x^{-n}=\sum_{k=1}^{\infty}(x+2k)^{-n}\)

   若设\(x_n=b_2+b_1n+\frac{b_3}{n}+\frac{b_4}{n^2}+\frac{b_5}{n^3}+\frac{b_6}{n^4}+\frac{b_7}{n^5}+\frac{b_8}{n^6}+...\)

   求\(b_1,b_2,b_3,b_4,b_5,b_6,b_7\)

3.设\(x_n,n \geq 2\)是下面方程的唯一正数解：

      \(x^{-n}=\sum_{k=1}^{\infty}(x+k^2)^{-n}\)

   若设\(x_n=c_2+c_1n+\frac{c_3}{n}+\frac{c_4}{n^2}+\frac{c_5}{n^3}+\frac{c_6}{n^4}+\frac{c_7}{n^5}+\frac{c_8}{n^6}+...\)

   求\(c_1,c_2,c_3,c_4,c_5,c_6,c_7\)

   若\(x_n\)的级数形式不对，可以表达成另外更简洁的形式讨论

数学星空

本题来源于附件

附件中已证明：

\(a_1=\frac{1}{\log2}\)

\(a_2=-\frac{3}{2}\)

\(a_3=-\frac{1}{2\log2}+\frac{3}{2}+\frac{23\log2}{24}+3(\log2)^2\)

mathe

https://en.wikipedia.org/wiki/Polygamma_function
\(\psi^{(n-1)}(x)=(-1)^n(n-1)!\sum_{k=0}^{+\infty}\frac1{(x+k)^n}\)
于是第一问相当于求$x$使得\(2(-1)^n(n-1)!x^{-n}=\psi^{(n-1)}(x)\)
再使用页面最后的Asymptotic expansion可以有比较好的方法
即\(\frac{2(n-1)!}{x_n} \sim \sum_{k=0}^{\infty}\frac{(k+n-2)!}{k!}\frac{B_k}{x_n^k}\)

kastin

第一个问题似乎在群里问过。\[2x^{-n}=\sum_{k=0}^{\oo}(x+k)^{-n}=\int_0^{\oo}(x+t)^{-n}\dif t+\frac{x^{1-n}}{n-1}+\frac{x^{-n}}{2}+\frac{nx^{-n-1}}{12}-\frac{n(n+1)(n+2)x^{-n-3}}{720}+\cdots+\frac{B_{2m}(n)_{2m-1}x^{-n-2m+1}}{(2m)!}+\cdots\]于是\[\frac{3}{2}=\frac{x}{n-1}+\frac{n}{12x}-\frac{n(n+1)(n+2)}{720x^3}+\cdots+\frac{B_{2m}(n)_{2m-1}}{(2m)!x^{2m-1}}+\cdots\]其中用到了升阶乘符号 `(a)_n=a(a+1)\cdots(a+n-1)`，又称为Pochhammer符号。

数学星空

通过级数展开的计算方法可以得到如下结果:

1.将\(x_n\)的级数形式代入方程并按\(n\)渐近展开可以得到

\(f=(1+\frac{k}{a_1n+a_2+\frac{a_3}{n}+\frac{a_4}{n^2}+\frac{a_5}{n^3}+\frac{a_6}{n^4}+\frac{a_7}{n^5}+\frac{a_8}{n^6}})^{-n}\)

\(=exp(\frac{-k}{a_1})(1+\frac{k(2a_2+k)}{2a_1^2n}+\frac{k(24a_1^2a_3-24a_1a_2^2-24a_1a_2k-8a_1k^2+12a_2^2k+12a_2k^2+3k^3)}{24a_1^4n^2}+\)

\(\frac{k(48a_1^4a_4-96a_1^3a_2a_3-48a_1^3a_3k+48a_1^2a_2^3+72a_1^2a_2^2k+48a_1^2a_2a_3k+48a_1^2a_2k^2+24a_1^2a_3k^2+12a_1^2k^3-48a_1a_2^3k-72a_1a_2^2k^2-40a_1a_2k^3-8a_1k^4+8a_2^3k^2+12a_2^2k^3+6a_2k^4+k^5)}{48a_1^6n^3}+\)

\(\frac{k(5760a_1^6a_5-11520a_1^5a_2a_4-5760a_1^5a_3^2-5760a_1^5a_4k+17280a_1^4a_2^2a_3+17280a_1^4a_2a_3k+5760a_1^4a_2a_4k+2880a_1^4a_3^2k+5760a_1^4a_3k^2+2880a_1^4a_4k^2-5760a_1^3a_2^4-11520a_1^3a_2^3k-17280a_1^3a_2^2a_3k-11520a_1^3a_2^2k^2-17280a_1^3a_2a_3k^2-5760a_1^3a_2k^3-4800a_1^3a_3k^3-1152a_1^3k^4+8640a_1^2a_2^4k+17280a_1^2a_2^3k^2+2880a_1^2a_2^2a_3k^2+14880a_1^2a_2^2k^3+2880a_1^2a_2a_3k^3+6240a_1^2a_2k^4+720a_1^2a_3k^4+1040a_1^2k^5-2880a_1a_2^4k^2-5760a_1a_2^3k^3-4560a_1a_2^2k^4-1680a_1a_2k^5-240a_1k^6+240a_2^4k^3+480a_2^3k^4+360a_2^2k^5+120a_2k^6+15k^7)}{5760a_1^8n^4}+\dots)\)

将上式关于\(k\)求和有

\(\sum_{k=1}^{\infty}f=s_0+\frac{s_1\frac{a_2}{a_1^2}+s_2\frac{1}{2a_1^2}}{n}+\frac{s_1\frac{24a_1^2a_3-24a_1a_2^2}{24a_1^4}+s_2\frac{-24a_1a_2+12a_2^2}{24a_1^4}+s_3\frac{-8a_1+12a_2}{24a_1^4}+s_4\frac{1}{8a_1^4}}{n^2}+\)

\(\frac{s_1\frac{48a_1^4a_4-96a_1^3a_2a_3+48a_1^2a_2^3}{48a_1^6}+s_2\frac{-48a_1^3a_3+72a_1^2a_2^2+48a_1^2a_2a_3-48a_1a_2^3}{48a_1^6}+s_3\frac{48a_1^2a_2+24a_1^2a_3-72a_1a_2^2+8a_2^3}{48a_1^6}+s_4\frac{12a_1^2-40a_1a_2+12a_2^2}{48a_1^6}+s_5\frac{-8a_1+6a_2}{48a_1^6}+s_6\frac{1}{48a_1^6}}{n^3}+\)

\(\frac{s_1\frac{5760a_1^6a_5-11520a_1^5a_2a_4-5760a_1^5a_3^2+17280a_1^4a_2^2a_3-5760a_1^3a_2^4}{5760a_1^8}+s_2\frac{-5760a_1^5a_4+17280a_1^4a_2a_3+5760a_1^4a_2a_4+2880a_1^4a_3^2-11520a_1^3a_2^3-17280a_1^3a_2^2a_3+8640a_1^2a_2^4}{5760a_1^8}+s_3\frac{5760a_1^4a_3+2880a_1^4a_4-11520a_1^3a_2^2-17280a_1^3a_2a_3+17280a_1^2a_2^3+2880a_1^2a_2^2a_3-2880a_1a_2^4}{5760a_1^8}+s_4\frac{-5760a_1^3a_2-4800a_1^3a_3+14880a_1^2a_2^2+2880a_1^2a_2a_3-5760a_1a_2^3+240a_2^4}{5760a_1^8}+s_5\frac{-1152a_1^3+6240a_1^2a_2+720a_1^2a_3-4560a_1a_2^2+480a_2^3}{5760a_1^8}+s_6\frac{1040a_1^2-1680a_1a_2+360a_2^2}{5760a_1^8}+s_7\frac{-240a_1+120a_2}{5760a_1^8}+s_8\frac{1}{384a_1^8}}{n^4}+\dots\)


其中

\(s_i=\sum_{k=1}^{\infty}k^iexp(\frac{-k}{a_1})\)

根据方程有

\(\sum_{k=1}^{\infty}f=1\)

对比左右关于n的系数知除第一项为1,其余系数均为0

\(s_0=1\) 得到

\(\sum_{k=1}^{\infty}exp(\frac{-k}{a_1})=1\) 解得\(a_1=\frac{1}{\ln2}\)

代入\(s_i\)表达式,依次求得\(s_1=2,s_2=6,s_3=26,s_4=150,s_5=1082,s_6=9366\)

由n的系数项全为0可以得到:

\(2a_2+3=0\)

\(24a_1^2a_3-24a_1a_2^2-72a_1a_2+36a_2^2-104a_1+156a_2+225=0\)

\(48a_1^4a_4-96a_1^3a_2a_3+48a_1^2a_2^3-144a_1^3a_3+216a_1^2a_2^2+144a_1^2a_2a_3-144a_1a_2^3+624a_1^2a_2+312a_1^2a_3-936a_1a_2^2+104a_2^3+900a_1^2-3000a_1a_2+900a_2^2-4328a_1+3246a_2+4683=0\)

\(1920a_1^6a_5-3840a_1^5a_2a_4-1920a_1^5a_3^2+5760a_1^4a_2^2a_3-1920a_1^3a_2^4-5760a_1^5a_4+17280a_1^4a_2a_3+5760a_1^4a_2a_4+2880a_1^4a_3^2-11520a_1^3a_2^3-17280a_1^3a_2^2a_3+8640a_1^2a_2^4+24960a_1^4a_3+12480a_1^4a_4-49920a_1^3a_2^2-74880a_1^3a_2a_3+74880a_1^2a_2^3+12480a_1^2a_2^2a_3-12480a_1a_2^4-144000a_1^3a_2-120000a_1^3a_3+372000a_1^2a_2^2+72000a_1^2a_2a_3-144000a_1a_2^3+6000a_2^4-207744a_1^3+1125280a_1^2a_2+129840a_1^2a_3-822320a_1a_2^2+86560a_2^3+1623440a_1^2-2622480a_1a_2+561960a_2^2-3783440a_1+1891720a_2+2729175=0\)

依次可得解得:

\(a_1=\frac{1}{a}=\frac{1}{\ln2}=1.442695041\)

\(a_2=\frac{-3}{2}\)

\(a_3=\frac{25a}{12}-3a^2=0.002697584\)

\(a_4=-25a^4+26a^3-6a^2=0.005045401895\)

\(a_5=\frac{21839a^3}{720}-\frac{925a^4}{4}+514a^5-\frac{1405a^6}{4}=0.00652697\)

\(a_6=\frac{-26861a^8}{4}+\frac{38401a^7}{3}-\frac{51673a^6}{6}+\frac{4725a^5}{2}-\frac{425a^4}{2}=0.00691115\)

\(a_7=\frac{-19139581a^{10}}{120}+\frac{4510615a^9}{12}-\frac{5401661a^8}{16}+\frac{426796a^7}{3}-\frac{3293911a^6}{120}+\frac{11234261a^5}{6048}=0.0063922\)


注:二楼附件中结果\(a_3\)是错误的


同样的计算方法可以得到

\(b_1=\frac{1}{b}=\frac{2}{\ln2}=2.885390082\)

\(b_2=-3\)

\(b_3=\frac{25b}{3}-24b^2=0.005395168\)

\(b_4=-800b^4+416b^3-48b^2=0.01009080811\)

\(b_5=\frac{21839b^3}{45}-7400b^4+32896b^5-44960b^6=0.0130539\)

\(b_6=-3438208b^8-\frac{3307072b^6}{3}+\frac{9830656b^7}{3}+151200b^5-6800b^4=0.0138221\)

\(b_7=\frac{-4899732736b^{10}}{15}+\frac{1154717440b^9}{3}-172853152b^8+\frac{109259776b^7}{3}-\frac{52702576b^6}{15}+\frac{22468522b^5}{189}=0.0127763\)



\(c_1=\frac{1}{c}=2.864788976\)

\(c_2=-2.148591733\)

\(c_3=5.642323308c-15.42933818c^2=0.089524310\)

\(c_4=-30.85867637c^2+246.2669228c^3-447.7245111c^4=0.067116029\)

\(c_5=17093.56323c^5-22234.96628c^6-4083.412841c^4+289.1531447c^3=0.03660989\)

\(c_6=-1521150c^8+1514736c^7-535615.3290c^6+77854.67872c^5-3755.910727c^4=-0.01396656\)

kastin

将假定的渐近展开式代入4楼结果，比较各阶系数
零阶：`3/2=x/(n-1)`，得 `a_2=-3/2`
一阶：`3/2=a_1+1/(12a_1)-1/(720a_1^3)+\cdots+B_{2m}/[(2m)!a_1^{2m-1}]+\cdots`，利用伯努利数母函数易得 `2f(z)=z/(\exp(z)-1)-z/(\exp(-z)-1)=z\coth \frac z 2=2\sum_{n\geqslant 0}\frac{B_{2m}}{(2m)!}z^{2m}` ，因而有 `a_1f(1/a_1)=3/2`，得到 `a_1=1/(2\,\mathrm{arccoth}\,3)=1/\ln 2`