一个代数方程根的级数解问题

kastin

我说这个式子很熟悉，感觉像是在哪见到过，应该是数学分析的某个辅导书上的(本科时候图书馆借了好多本，而且还做了不少笔记)。笔记本在家里，于是我在《拉马努金的笔记本》中找到一些线索：
$$\begin{align*}-\frac{\pi}{2}\tan\frac{\pi x}{2} &=\sum_{n=1}^{\infty}\frac{2x}{x^2-(2n-1)^2}\tag{1}\\
\pi\cot \pi x &=\frac{1}{x}+\sum_{n=1}^{\infty}\frac{2x}{x^2-n^2}=\frac{\Gamma\,'(1-x)}{\Gamma(1-x)}-\frac{\Gamma\,' (x)}{\Gamma(x)}\tag{2}\\
\pi\csc \pi x &=\frac{1}{x}+\sum_{n=1}^{\infty}(-1)^n\frac{2x}{x^2-n^2}\tag{3}\\
\frac{\pi}{4}\sec \frac{\pi x}{2}&=\sum_{n=1}^{\infty}(-1)^n\frac{2n-1}{x^2-(2n-1)^2}\tag{4}
\end{align*}$$上面将自变量用 `\mathrm ix` 替换，还能得到类似形式的双曲三角函数的级数形式。
上述公式似乎来源于这个式子，而这个式子的证明应该跟伯努利数或者Zeta函数有关$$\frac{1}{e^x-1}=\frac{1}{2}\coth\frac{x}{2}-\frac{1}{2}=\frac{1}{x}-\frac{1}{2}+\sum_{n=1}^{\infty}\frac{2x}{x^2+4\pi^2m^2}$$第一个问题对应的无穷级数并不收敛$$\frac{1}{2x}+\sum_{n=1}^{\infty}\frac{1}{x-n}$$看到`(2)`就能想到对数Gamma函数(PolyGamma)

1. f[x_,n_]=Sum[1/2 x + 1/(x - k), {k, 1, n}];
   
2. f[x,n]

复制代码

第一个函数得到 `-\psi (n-x+1)+\frac{1}{2 x}+\psi (1-x)`，这里的 `\psi(x)=\Gamma'(x)/\Gamma(x)` 为对数Gamma函数。
楼上也触及到PolyGamma了，其实可以直接展开的

1. InverseSeries[Series[f[u, n], {u, 0, 4}], a]

复制代码

如果还需要展开，可以接着对 `n` 进行无穷远处展开

1. Series[%,{n,Infinity,4}]

复制代码

数学星空

对于第\(k\)个根，\(x_k \in (k-1,k)\),我们可以来讨论一下

\(x_k, k=1...n\)的渐近表达式？

282842712474

数学星空 发表于 2016-7-7 19:40
对于第\(k\)个根，\(x_k \in (k-1,k)\),我们可以来讨论一下

\(x_k, k=1...n\)的渐近表达式？

如果要研究第$k$个根，就得在$x=k$附近来展开$\Gamma(x)$，这通常会导致异常复杂的结果，前面的最大根比较简单，是因为$\Gamma(x)$在无穷远处的展开式还算简单。

282842712474

kastin 发表于 2016-7-7 18:38
我说这个式子很熟悉，感觉像是在哪见到过，应该是数学分析的某个辅导书上的(本科时候图书馆借了好多本，而 ...

利用$(2)$，或许可以对结果进行改进。对于楼主的第二个方程，我们有
$$a=\frac{1}{x}+\sum_{k=1}^n \frac{2x}{x^2-n^2}=\pi \cot \pi x -\sum_{k=n+1}^{\infty} \frac{2x}{x^2-n^2}$$
后面这一项
$$h(x)=\sum_{k=n+1}^{\infty} \frac{2x}{x^2-n^2}$$
是一项没有任何奇点（在$[0,n]$内）的项，性态比较好，容易做估计。比如最简单地，用积分估计：
$$h(x)\approx 2x\int_{n+1}^{\infty} \frac{1}{x^2-k^2}dk=\ln\left(\frac{n+1+x}{n+1-x}\right)$$
于是可以考虑
$$a\approx \pi\cot \pi x-\ln\left(\frac{n+1+x}{n+1-x}\right)$$
可以考虑格式：
$$x_k=k-1+\frac{1}{\pi}\mathrm{arccot}\frac{1}{\pi}\left(a+\ln\left(\frac{n+1+x_k}{n+1-x_k}\right)\right)$$
对于比较小的$k$以及比较大的$n$，\( \ln \left( \frac{n+1+x_k}{n+1-x_k} \right) \)也比较小，这时候应该还算不错的近似。比如$x_1$，以\(\frac{1}{\pi}\mathrm{arccot}\frac{a}{\pi}\)为初值代入，得到近似
$$x_0\approx \left(1-\frac{2}{\left(a^2+\pi ^2\right) (n+1)}\right)\frac{1}{\pi}\mathrm{arccot}\frac{a}{\pi}$$

数学星空

如何求解下面方程的渐近级数解也不容易:

\(a=\pi \cot(\pi x)-\ln(\frac{n+1+x}{n+1-x})\)

根据楼上的结论是否有：

\[x_k=k-1+(1+\frac{a_1}{(a^2+\pi^2)(n+1)}+\frac{a_2}{(a^2+\pi^2)(n+1)^2}+\frac{a_3}{(a^2+\pi^2)(n+1)^3}+\dots)\frac{arccot(\frac{a}{\pi})}{\pi}\]

感觉渐近展开式中除第一项外，后面的也应该含有\(k\)吧？

xingyuntu

（2）左边是 pi*cos（pix）的级数展开式，故解为x=arccos（a/pi）/pi。

数学星空

首先我们利用渐近分析得到：

\(\frac{1}{x}+\sum_{j=1}^n \frac{2x}{x^2-j^2}=\pi\cot(x\pi)+\frac{2x}{n}-\frac{x}{n^2}+\frac{2x^3+x}{3n^3}-\frac{x^3}{n^4}+\frac{-x+10x^3+6x^5}{15n^5}-\frac{x^5}{n^6}+\frac{x-7x^3+21x^5+6x^7}{21n^7}-\frac{x^7}{n^8}+\dots\)

为了得到第k个根的一般表达式\(x_k\)的代数结构

我们利用:下面简记\(b=arccot(\frac{a}{\pi})\)

\(\pi\cot(x\pi)+\frac{2x}{n}=a\)

得到：\(x_k=k+\frac{b}{\pi}+\dots\)

\(\pi\cot(x\pi)+\frac{2x}{n}-\frac{x}{n^2}=a\)

得到：\(x_k=k+\frac{b}{\pi}+\frac{2(\pi k+b)}{\pi(\pi^2+a^2)n}+\dots\)

\(\pi\cot(x\pi)+\frac{2x}{n}-\frac{x}{n^2}+\frac{2x^3+x}{3n^3}=a\)

得到：\(x_k=k+\frac{b}{\pi}+\frac{2(\pi k+b)}{\pi(\pi^2+a^2)n}+\frac{(\pi k+b)(\pi^3+\pi(a^2-4ak-4)-4ab)}{(\pi(\pi^2+a^2)^2n^2}+\dots\)

因此我们可以设

\(x_k=k+\frac{b}{\pi}+(\pi k+b)(\frac{a_0}{\pi(\pi^2+a^2)n}+\frac{b_0+b_1 k}{(\pi(\pi^2+a^2)n)^2}+\frac{c_0+c_1 k+c_2 k^2}{(\pi(\pi^2+a^2)n)^3}+\frac{d_0+d_1 k+d_2 k^2+d_3 k^3}{(\pi(\pi^2+a^2)n)^4}+\frac{m_0+m_1 k+m_2 k^2+m_3 k^3+m_4 k^4}{(\pi(\pi^2+a^2)n)^5}+\frac{n_0+n_1 k+n_2 k^2+n_3 k^3+n_4 k^4+n_5 k^5}{(\pi(\pi^2+a^2)n)^6}+\dots)\)

代入可以得到

{a0 = 2, b0 = -Pi^3+(-a^2+4)*Pi+4*a*b, b1 = 4*Pi*a, c0 = -4*Pi^4-4*Pi^3*a*b+(-4*a^2-8*b^2*(1/3)+8)*Pi^2+(-4*a^3*b+24*a*b)*Pi+8*a^2*b^2, c1 = -4*Pi^4*a-16*Pi^3*b*(1/3)+(-4*a^3+24*a)*Pi^2+16*Pi*a^2*b, c2 = -(8/3)*Pi^4+8*a^2*Pi^2, d0 = (-b^2+1)*Pi^7+Pi^6*a*b+(-3*a^2*b^2+2*a^2+4*b^2-12)*Pi^5+(2*a^3*b-36*a*b)*Pi^4+(-3*a^4*b^2+a^4-8*a^2*b^2-12*a^2-(64/3)*b^2+16)*Pi^3+(a^5*b-36*a^3*b-16*a*b^3+96*a*b)*Pi^2+(-a^6*b^2-12*a^4*b^2+96*a^2*b^2)*Pi+16*a^3*b^3, d1 = -2*Pi^8*b+Pi^7*a+(-6*a^2*b+8*b)*Pi^6+(2*a^3-36*a)*Pi^5+(-6*a^4*b-16*a^2*b-(128/3)*b)*Pi^4+(a^5-36*a^3-48*a*b^2+96*a)*Pi^3+(-2*a^6*b-24*a^4*b+192*a^2*b)*Pi^2+48*Pi*a^3*b^2, d2 = -Pi^9+(-3*a^2+4)*Pi^7+(-3*a^4-8*a^2-64/3)*Pi^5-48*Pi^4*a*b+(-a^6-12*a^4+96*a^2)*Pi^3+48*Pi^2*a^3*b, d3 = -16*Pi^5*a+16*Pi^3*a^3, m0 = -(1/15)*Pi^12+(-4*a^2*(1/15)+2*b^2*(1/3))*Pi^10+(-(2/5)*a^4+(2/5)*b^4-10*b^2+(8/3)*a^2*b^2+6)*Pi^8+(-4*a*b^3+18*a*b)*Pi^7+(-(4/15)*a^6+12*a^2+(128/3)*b^2-22*a^2*b^2+4*a^4*b^2+(8/5)*a^2*b^4-32)*Pi^6+(-12*a^3*b^3+36*a^3*b+32*a*b^3-192*a*b)*Pi^5+(-(1/15)*a^8-32*a^2-(320/3)*b^2+(32/5)*b^4+6*a^4-14*a^4*b^2-(448/3)*a^2*b^2+(8/3)*a^6*b^2+(12/5)*a^4*b^4+32)*Pi^4+(18*a^5*b-192*a^3*b-(640/3)*a*b^3+320*a*b-12*a^5*b^3)*Pi^3+(-2*a^6*b^2-192*a^4*b^2-64*a^2*b^4+640*a^2*b^2+(2/3)*a^8*b^2+(8/5)*a^6*b^4)*Pi^2+(-4*a^7*b^3-32*a^5*b^3+320*a^3*b^3)*Pi+2*a^8*b^4*(1/5)+32*a^4*b^4, m1 = 4*Pi^11*b*(1/3)+((16/3)*a^2*b+(8/5)*b^3-20*b)*Pi^9+(-12*a*b^2+18*a)*Pi^8+(8*a^4*b+(32/5)*a^2*b^3-44*a^2*b+(256/3)*b)*Pi^7+(-36*a^3*b^2+36*a^3+96*a*b^2-192*a)*Pi^6+((16/3)*a^6*b-(896/3)*a^2*b+(48/5)*a^4*b^3-28*a^4*b+(128/5)*b^3-(640/3)*b)*Pi^5+(-36*a^5*b^2+18*a^5-192*a^3-640*a*b^2+320*a)*Pi^4+((4/3)*a^8*b-384*a^4*b+(32/5)*a^6*b^3-4*a^6*b-256*a^2*b^3+1280*a^2*b)*Pi^3+(-12*a^7*b^2-96*a^5*b^2+960*a^3*b^2)*Pi^2+((8/5)*a^8*b^3+128*a^4*b^3)*Pi, m2 = 2*Pi^12*(1/3)+(8*a^2*(1/3)+12*b^2*(1/5)-10)*Pi^10-12*Pi^9*a*b+(4*a^4-22*a^2+(48/5)*a^2*b^2+128/3)*Pi^8+(-36*a^3*b+96*a*b)*Pi^7+((8/3)*a^6-(448/3)*a^2+(192/5)*b^2-14*a^4+(72/5)*a^4*b^2-320/3)*Pi^6+(-36*a^5*b-640*a*b)*Pi^5+((2/3)*a^8+640*a^2-192*a^4-2*a^6-384*a^2*b^2+(48/5)*a^6*b^2)*Pi^4+(-12*a^7*b-96*a^5*b+960*a^3*b)*Pi^3+(192*a^4*b^2+(12/5)*a^8*b^2)*Pi^2, m3 = 8*Pi^11*b*(1/5)-4*Pi^10*a+32*Pi^9*a^2*b*(1/5)+(-12*a^3+32*a)*Pi^8+((48/5)*a^4*b+(128/5)*b)*Pi^7+(-12*a^5-(640/3)*a)*Pi^6+((32/5)*a^6*b-256*a^2*b)*Pi^5+(-4*a^7-32*a^5+320*a^3)*Pi^4+((8/5)*a^8*b+128*a^4*b)*Pi^3, m4 = 2*Pi^12*(1/5)+8*Pi^10*a^2*(1/5)+(12*a^4*(1/5)+32/5)*Pi^8+((8/5)*a^6-64*a^2)*Pi^6+((2/5)*a^8+32*a^4)*Pi^4, n0 = -4*Pi^13*(1/15)-4*Pi^12*a*b*(1/15)+(-b^4-(16/15)*a^2+(29/3)*b^2-1)*Pi^11+(-(16/15)*a^3*b+(14/3)*a*b^3-3*a*b)*Pi^10+(-3*a^2-(8/5)*a^4+(44/5)*b^4-72*b^2+24-5*a^2*b^4+(112/3)*a^2*b^2)*Pi^9+(-(8/5)*a^5*b+(56/3)*a^3*b^3+(8/5)*a*b^5-9*a^3*b-84*a*b^3+144*a*b)*Pi^8+(48*a^2-3*a^4+(800/3)*b^2-(16/15)*a^6-16*b^4-80-10*a^4*b^4+54*a^4*b^2+(96/5)*a^2*b^4-40*a^2*b^2)*Pi^7+(-800*a*b-(16/15)*a^7*b+28*a^5*b^3+(32/5)*a^3*b^5-9*a^5*b-204*a^3*b^3+288*a^3*b+(1600/3)*a*b^3)*Pi^6+(24*a^4-a^6-80*a^2-(1280/3)*b^2-(4/15)*a^8+(1472/15)*b^4-(4000/3)*a^2*b^2+64-10*a^6*b^4+(104/3)*a^6*b^2+(24/5)*a^4*b^4+136*a^4*b^2+144*a^2*b^4)*Pi^5+(-1600*a*b^3-800*a^3*b+960*a*b-(4/15)*a^9*b+(56/3)*a^7*b^3+(48/5)*a^5*b^5-3*a^7*b-156*a^5*b^3+144*a^5*b-(800/3)*a^3*b^3+64*a*b^5)*Pi^4+(-1600*a^4*b^2+3200*a^2*b^2-5*a^8*b^4+(25/3)*a^8*b^2-(64/5)*a^6*b^4+104*a^6*b^2+80*a^4*b^4-1280*a^2*b^4)*Pi^3+(3200*a^3*b^3+(14/3)*a^9*b^3+(32/5)*a^7*b^5-36*a^7*b^3-800*a^5*b^3-(640/3)*a^3*b^5)*Pi^2+(-a^10*b^4-(36/5)*a^8*b^4-80*a^6*b^4+960*a^4*b^4)*Pi+8*a^9*b^5*(1/5)+64*a^5*b^5, n1 = -4*Pi^13*a*(1/15)+((58/3)*b-4*b^3)*Pi^12+(-(16/15)*a^3-3*a+14*a*b^2)*Pi^11+(-144*b+(176/5)*b^3-20*a^2*b^3+(224/3)*a^2*b)*Pi^10+(-(8/5)*a^5+144*a-9*a^3+8*a*b^4-252*a*b^2+56*a^3*b^2)*Pi^9+((1600/3)*b-64*b^3+(384/5)*a^2*b^3-40*a^4*b^3+108*a^4*b-80*a^2*b)*Pi^8+(-(16/15)*a^7+288*a^3-9*a^5-800*a+32*a^3*b^4-612*a^3*b^2+84*a^5*b^2+1600*a*b^2)*Pi^7+(-(2560/3)*b+(5888/15)*b^3+(96/5)*a^4*b^3-40*a^6*b^3+576*a^2*b^3-(8000/3)*a^2*b+(208/3)*a^6*b+272*a^4*b)*Pi^6+(-(4/15)*a^9+144*a^5-3*a^7+960*a-800*a^3+320*a*b^4+48*a^5*b^4-468*a^5*b^2-4800*a*b^2+56*a^7*b^2-800*a^3*b^2)*Pi^5+(-(256/5)*a^6*b^3+320*a^4*b^3+6400*a^2*b-5120*a^2*b^3-20*a^8*b^3-3200*a^4*b+(50/3)*a^8*b+208*a^6*b)*Pi^4+(-(3200/3)*a^3*b^4-108*a^7*b^2+32*a^7*b^4+9600*a^3*b^2+14*a^9*b^2-2400*a^5*b^2)*Pi^3+(-(144/5)*a^8*b^3-4*a^10*b^3-320*a^6*b^3+3840*a^4*b^3)*Pi^2+(8*a^9*b^4+320*a^5*b^4)*Pi, n2 = (-6*b^2+29/3)*Pi^13+14*Pi^12*a*b+((112/3)*a^2+(264/5)*b^2-30*a^2*b^2-72)*Pi^11+(56*a^3*b+16*a*b^3-252*a*b)*Pi^10+(-40*a^2+54*a^4-96*b^2-60*a^4*b^2+800/3+(576/5)*a^2*b^2)*Pi^9+(84*a^5*b+64*a^3*b^3-612*a^3*b+1600*a*b)*Pi^8+(-(4000/3)*a^2+136*a^4+(2944/5)*b^2+(104/3)*a^6-60*a^6*b^2-1280/3+(144/5)*a^4*b^2+864*a^2*b^2)*Pi^7+(56*a^7*b+96*a^5*b^3-468*a^5*b-800*a^3*b+640*a*b^3-4800*a*b)*Pi^6+(-1600*a^4+104*a^6+3200*a^2+(25/3)*a^8-30*a^8*b^2-7680*a^2*b^2-(384/5)*a^6*b^2+480*a^4*b^2)*Pi^5+(9600*a^3*b+14*a^9*b+64*a^7*b^3-108*a^7*b-2400*a^5*b-(6400/3)*a^3*b^3)*Pi^4+(-6*a^10*b^2+5760*a^4*b^2-(216/5)*a^8*b^2-480*a^6*b^2)*Pi^3+(16*a^9*b^3+640*a^5*b^3)*Pi^2, n3 = -4*Pi^14*b+14*Pi^13*a*(1/3)+((176/5)*b-20*a^2*b)*Pi^12+((56/3)*a^3-84*a+16*a*b^2)*Pi^11+(-64*b+(384/5)*a^2*b-40*a^4*b)*Pi^10+(28*a^5+(1600/3)*a-204*a^3+64*a^3*b^2)*Pi^9+((5888/15)*b+(96/5)*a^4*b-40*a^6*b+576*a^2*b)*Pi^8+((56/3)*a^7-(800/3)*a^3-156*a^5-1600*a+96*a^5*b^2+640*a*b^2)*Pi^7+(-5120*a^2*b-(256/5)*a^6*b-20*a^8*b+320*a^4*b)*Pi^6+((14/3)*a^9-800*a^5-36*a^7+3200*a^3+64*a^7*b^2-(6400/3)*a^3*b^2)*Pi^5+(3840*a^4*b-(144/5)*a^8*b-4*a^10*b-320*a^6*b)*Pi^4+(16*a^9*b^2+640*a^5*b^2)*Pi^3, n4 = -Pi^15+(44/5-5*a^2)*Pi^13+8*Pi^12*a*b+((96/5)*a^2-16-10*a^4)*Pi^11+32*Pi^10*a^3*b+(144*a^2+(24/5)*a^4+1472/15-10*a^6)*Pi^9+(48*a^5*b+320*a*b)*Pi^8+(-1280*a^2+80*a^4-(64/5)*a^6-5*a^8)*Pi^7+(32*a^7*b-(3200/3)*a^3*b)*Pi^6+(960*a^4-80*a^6-(36/5)*a^8-a^10)*Pi^5+(8*a^9*b+320*a^5*b)*Pi^4, n5 = 8*Pi^13*a*(1/5)+32*Pi^11*a^3*(1/5)+((48/5)*a^5+64*a)*Pi^9+((32/5)*a^7-(640/3)*a^3)*Pi^7+((8/5)*a^9+64*a^5)*Pi^5}

例如：取\(n=10,a=1\)

我们得到第k个根近似公式为

{a0 = 2., b0 = -16.53098971, b1 = 12.56637062, c0 = -456.6267843, c1 = -337.5743483, c2 = -180.8007410, d0 = -7887.945822, d1 = -35043.29498, d2 = -39415.34927, d3 = -4400.214534, m0 = -56374.18536, m1 = -44467.07391, m2 = 28660.52806, m3 = 5.043098762*10^5, m4 = 5.462089913*10^5, n0 = 2.491507096*10^6, n1 = 1.491578265*10^7, n2 = 1.133089630*10^7, n3 = -1.291336512*10^7, n4 = 3.397063999*10^5, n5 = 8.117814619*10^6}

\(x_k=k+0.4019067381+(3.141592654k+1.262627256)(0.005703056713+0.00009671112355k-0.000007425991790k^2-2.231400907*10^{-7}k^3+1.178503973*10^{-7}k^4+5.119852668*10^{-9}k^5)\)

则有

\(x_1=1.427417044\) ,准确数值解：1.427653678
\(x_2=2.446178317\), 准确数值解：2.447241727
\(x_3=3.465295036\), 准确数值解：3.468263146
\(x_4=4.484773091\), 准确数值解：4.491215523
\(x_5=5.504794780\), 准确数值解：5.516810213
\(x_6=6.525804536\), 准确数值解：6.546154163
\(x_7=7.548606255\), 准确数值解：7.581177444
\(x_8=8.574472189\), 准确数值解：8.625897952
\(x_9=9.605263428\), 准确数值解：9.692115043