记\(g_{(1)}(x)=g(x)=\frac{(1+x)^3}{x^2}, g_{(k+1)}(x)=g(g_{(k)}(x))\)我们知道\(h(g(x))=h(x)+3\)
那么我们知道对于所有\(n>0\),方程\(g_{(n)}(x)=x\)的解都必然是\(h(x)\)的非解析点.
试做了一下\(g_{(5)}(x)=x\)的零点图如下
wayne 发表于 2016-1-3 21:04
对于$b$数列，$x_1=t, g(x) = x(1+1/x)^3$，则不停迭代之，展开式为:
$g^{(n)}(t) =(3+t+1/t^2+3/t)+3n+ ...
没别的意思,纯粹玩玩,先记录之,
接29#, 设$g(n,t) = g^{(n)}(t) $, $g(0,x) = 3+x+1/x^2+3/x$, 设$\phi(x,y) = \sum_{n=0}^{\infty}g(n,x)*y^n$
则$0<y<1,\phi(x,y) = (3+x+1/x^2+3/x)*1/{1-y}+(3+3x-9x^3)*y/(1-y)^2 +\frac{28 y^2-19 y}{(y-1)^3}x^4 -\frac{18 (5 y^2-2 y)}{(y-1)^3} x^5 + \frac{3 (99 y^3-103 y^2+22 y)}{(y-1)^4} x^6 +...$
对于\(a_{n+1}=a_n+\frac{1}{a_n^2}\)
1.令\(b_n=a_n^3\),得到
\(b_{n+1}=b_n+3+\frac{3}{b_n}+\frac{1}{b_n^2}\)
2.由\(m'-\frac{\ln(m')}{3}=m+1-\frac{\ln(m)}{3}\)渐近展开：
\(m'=J(m)=m+1+\frac{1}{3m}-\frac{1}{18m^2}-\frac{1}{54m^3}+\frac{7}{324m^4}-\frac{31}{4860m^5}-\frac{107}{29160m^6}+\frac{2833}{612360m^7}-\)
\(\frac{641}{459270m^8}-\frac{3001}{2755620m^9}+\frac{87961}{66134880m^{10}}-\frac{817951}{2182451040m^{11}}-\frac{24880117}{65473531200m^{12}}+\)
\(\frac{1118093573}{2553467716800m^{13}}-\frac{67495517}{612832252032m^{14}}-\frac{603350659}{4178401718400m^{15}}+\frac{150300871447}{965210796950400m^{16}}+\cdots\)
3.由\(m(x+3+\frac{3}{x}+\frac{1}{x^2})=J(m(x))\),设\(m(x)=\frac{x}{3}+\sum_{k=1}^\infty \frac{u_k}{x^k}\),渐近展开：
\(m(x)=\frac{x}{3}+\frac{5}{18x}+\frac{1}{2x^2}+\frac{239}{324x^3}+\frac{5227}{6480x^4}+\frac{7297}{32400x^5}-\frac{382271}{226800x^6}-\frac{12524549}{3175200x^7}+\frac{709670963}{342921600x^8}+\)
\(\frac{28761656717}{1028764800x^9}-\frac{6137702104697}{226328256000x^{10}}-\frac{516199738730261}{829870272000x^{11}}-\frac{1429968790523743403}{4369266982080000x^{12}}+\)
\(\frac{2050189135375425012401}{113600941534080000x^{13}}+\frac{35895037045177279350019}{795206590738560000x^{14}}-\frac{6198573918031962345943781}{9542479088862720000x^{15}}-\frac{53515807461516509490031169}{16031364869289369600x^{16}}+\cdots\)
4.极限值\(a_0\)由下式求出：
\(m(a_n)-\frac{\ln(m(a_n))}{3}=n+\frac{a_0}{3}\)
5.关于\(b_n\)渐近展开式计算：\(y=\frac{1}{n},X=\ln(n)+a_0\)
\(F_1=f(y,X)=\frac{3}{y}+X+(b_{00}+b_{01}X)y+(b_{10}+b_{11}X+b_{12}X^2)y^2+(b_{20}+b_{21}X+b_{22}X^2+b_{23}X^3)y^3+(b_{30}+b_{31}X+b_{32}X^2+b_{33}X^3+b_{34}X^4)y^4+(b_{40}+b_{41}X+b_{42}X^2+b_{43}X^3+b_{44}X^4+b_{45}X^5)y^5+\)
\((b_{50}+b_{51}X+b_{52}X^2+b_{53}X^3+b_{54}X^4+b_{55}X^5+b_{56}X^6)y^6+(b_{60}+b_{61}X+b_{62}X^2+b_{63}X^3+b_{64}X^4+b_{65}X^5+b_{66}X^6+b_{67}X^7)y^7+(b_{70}+b_{71}X+b_{72}X^2+b_{73}X^3+b_{74}X^4+b_{75}X^5+b_{76}X^6+b_{77}X^7+b_{78}X^8)y^8+\cdots\)
\(F_2=f(\frac{y}{1+y},X+\ln(1+y))\)
由\(F_2=F_1+3+\frac{3}{F_1}+\frac{1}{F_1^2}\)求解得到：
\(b_{00}=\frac{-5}{18}, b_{01} =\frac{1}{3}, b_{10} =\frac{-1}{6}, b_{11} =\frac{11}{54}, b_{12}=\frac{-1}{18}, b_{20}=\frac{-157}{1458}, b_{21}=\frac{29}{162}, b_{22}=\frac{-7}{81}, b_{23}=\frac{1}{81}, b_{30}=\frac{-13327}{174960}, b_{31}=\frac{122}{729}, b_{32}=\frac{-115}{972},b_{33}=\frac{8}{243}, b_{34}=\frac{-1}{324}, b_{40}=\frac{-444191}{7873200}, b_{41}=\frac{20647}{131220}\)
\(b_{42}=\frac{-1321}{8748}, b_{43}=\frac{139}{2187}, b_{44}=\frac{-35}{2916}, b_{45}=\frac{1}{1215},b_{50}=\frac{-14533499}{330674400}, b_51=\frac{138391}{944784}, b_{52}=\frac{-28573}{157464}, b_{53}=\frac{8273}{78732}, b_{54}=\frac{-200}{6561}, b_{55}=\frac{187}{43740}, b_{56}=\frac{-1}{4374}, b_{60}=\frac{-777322409}{20832487200}, b_{61}=\frac{33909461}{248005800}\)
\(b_{62}=\frac{-65179}{314928}, b_{63}=\frac{2047}{13122}, b_{64}=\frac{-1097}{17496}, b_{65}=\frac{1787}{131220}, b_{66}=\frac{-197}{131220}, b_{67}=\frac{1}{15309}\)
类似于mathe提供的求解思路，我们来讨论稍一般的情形
对于\(a_{n+1}=a_n+\frac{1}{a_n^{k-1}}\)
1.令\(a_n^k=b_n\),得到
\(b_{n+1}=b_n+k+\frac{k(k-1)}{2b_n}+\frac{k(k-1)(k-2)}{6b_n^2}+\frac{k(k-1)(k-2)(k-3)}{24b_n^3}+\frac{k(k-1)(k-2)(k-3)(k-4)}{120b_n^4}+\frac{k(k-1)(k-2)(k-3)(k-4)(k-5)}{720b_n^5}+\cdots\)
2.由\(m'-\frac{\ln(m')}{k}=m+1-\frac{\ln(m)}{k}\)渐近展开：
\(m'=J(m)=m+1+\frac{1}{km}-\frac{k-2}{2k^2m^2}+\frac{-9k+6+2k^2}{6(k^3m^3)}-\frac{3k^3-22k^2+36k-12}{12k^4m^4}+\frac{12k^4-125k^3+350k^2-300k+60}{60k^5m^5}-\)
\(\frac{-120+900k-1700k^2+1125k^3-274k^4+20k^5}{120k^6m^6}+\frac{-8820k+24500k^2-25725k^3+11368k^4+840-2058k^5+120k^6}{840k^7m^7}-\frac{-2520+35280k+315k^7-6534k^6-135240k^2+205800k^3-142149k^4+45962k^5}{2520k^8m^8}+\)
\(\frac{-45360k+229320k^2-476280k^3+471429k^4+2520-235494k^5+59062k^6-6849k^7+280k^8}{2520k^9m^9}-\frac{-1461600k^2+3969000k^3-5314932k^4+3770550k^5+226800k-1447360k^6+293175k^7-28516k^8+1008k^9-10080}{10080k^{10}m^{10}}+\cdots\)
3.由\(m(x+k+\frac{k(k-1)}{2x}+\frac{k(k-1)(k-2)}{6x^2}+\frac{k(k-1)(k-2)(k-3)}{24x^3}+\frac{k(k-1)(k-2)(k-3)(k-4)}{120x^4}+\frac{k(k-1)(k-2)(k-3)(k-4)(k-5)}{720x^5}+\cdots)=J(m(x))\)
设\(m(x)=\frac{x}{k}+\sum_{k=1}^\infty \frac{u_k}{x^k}\),渐近展开：
注：上面\(m(x)\)形式解不正确，需要重新研究其形式才能求解渐近展开式
4.极限值\(a_0\)由下式求出：
\(m(a_n)-\frac{\ln(m(a_n))}{k}=n+\frac{a_0}{k}\)
5.关于\(b_n\)渐近展开式计算：\(y=\frac{1}{n},X=\ln(n)+a_0\)
\(F_1=f(y,X)=\frac{k}{y}+X+(b_{00}+b_{01}X)y+(b_{10}+b_{11}X+b_{12}X^2)y^2+(b_{20}+b_{21}X+b_{22}X^2+b_{23}X^3)y^3+(b_{30}+b_{31}X+b_{32}X^2+b_{33}X^3+b_{34}X^4)y^4+(b_{40}+b_{41}X+b_{42}X^2+b_{43}X^3+b_{44}X^4+b_{45}X^5)y^5+\)
\((b_{50}+b_{51}X+b_{52}X^2+b_{53}X^3+b_{54}X^4+b_{55}X^5+b_{56}X^6)y^6+(b_{60}+b_{61}X+b_{62}X^2+b_{63}X^3+b_{64}X^4+b_{65}X^5+b_{66}X^6+b_{67}X^7)y^7+(b_{70}+b_{71}X+b_{72}X^2+b_{73}X^3+b_{74}X^4+b_{75}X^5+b_{76}X^6+b_{77}X^7+b_{78}X^8)y^8+\cdots\)
\(F_2=f(\frac{y}{1+y},X+\ln(1+y))\)
由\(F_2=x+k+\frac{k(k-1)}{2F_1}+\frac{k(k-1)(k-2)}{6F_1^2}+\frac{k(k-1)(k-2)(k-3)}{24F_1^3}+\frac{k(k-1)(k-2)(k-3)(k-4)}{120F_1^4}+\frac{k(k-1)(k-2)(k-3)(k-4)(k-5)}{720F_1^5}+\cdots\)求解得到：
注：上面\(F_1\)形式解不正确，需要重新研究其形式才能求解渐近展开式
数学星空 发表于 2016-1-6 18:28
对于\(a_{n+1}=a_n+\frac{1}{a_n^2}\)
1.令\(b_n=a_n^3\),得到
mathe,能否给一个k=4时的分析结果？
猜测：
对于\(k=4\)
\
\
做个小结：
考察满足递推关系$a_{n+1}=\alpha a_n+\frac{P_k(n)}{a_n^m}$的初值为$x$的数列的极限行为。这里的$P_k(n)=\sum_{i=0}^{k}\alpha_i n^i$，不妨设$\alpha_k=1$。所有的参数都是非负正实数，特别是$m$也不限于整数。现在根据$\alpha$取值范围分情况讨论。
1. $\alpha>1$
此时$a_n ~h(x)\alpha^{n-1}+\frac{Q_k(n)}{h(x)^m}\alpha^{-mn}+\cdots$。多项式$Q_k(n)$满足$Q_k(n+1)-\alpha^{m+1}Q_k(n)=\alpha^{2m}P_k(n)$。第二项已经很小了，后面的没必要计算下去。
若$P_k(n)=1$，则$h(x)$满足$h(\alpha x+\frac{1}{x^m})=\alpha h(x)$，由这个等式可以求得$h(x)~x(1+\frac{\alpha^m}{(\alpha^{m+1}-1)x^{m+1}}-\sum_{i=2}^{\infty}\frac{m}{(i-1)!}\frac{R(i,\alpha^{m+1})}{\prod_{j=1}^{i}(\alpha^{(m+1)j}-1)}(-\frac{\alpha^m}{x^{m+1}})^i)$。$R(i,t)$是的$t$的$\frac{i(i-1)}{2}-1$次多项式，系数除了$1$次、$\frac{i(i-1)}{2}-2$次为零之外都是$m$的$i-2$次正整系数多项式，而且$\frac{i(i-1)}{2}-1$次系数为$\frac{(m+i-2)!}{m!}$，常数项为$\prod_{j=1}^{i-2}(im+j)$，二次项系数为$(i-2)(m+1)\prod_{j=2}^{i-2}(im+j)$。
前几个$R(i,t)$是$R(2,t)=1$，$R(3,t)=3m+1+(m+1)t^2$。
2. $0<\alpha<1$
作代换$a_n=(\frac{\alpha_k n^k}{1-\alpha})^{\frac{1}{m+1}}b_n$可以得到$b_{n+1}=\alpha b_n+\frac{1-\alpha}{b_n^m}+O(\frac{1}{n})$。普适极限方程$S_{n+1}=\alpha S_n+\frac{1-\alpha}{S_n^m}$有不动点$S_n=1$，随着参数的变化数列的趋势会出现分岔、混沌现象。当$0<m<\frac{1+\alpha}{1-\alpha}$时$S_n \rightarrow 1$，$m>\frac{1+\alpha}{1-\alpha}$时不动点不稳定，开始变成两周期以至更长周期的循环，而且很快就变成混沌的。
3. $\alpha=1$
作代换$b_n=a_n^{m+1}$可以得到$b_{n+1}=b_n+m+1+\sum_{i=2}^{\infty}C_{m+1}^{i}(\frac{P_k(n)}{b_n})^{i-1}$，所以$b_n~Q_{k+1}(n)+\frac{m}{2}\ln n+h(P_k,x)+O(\frac{\ln n}{n})$，多项式$Q_{k+1}(n)$满足$Q_{k+1}(n+1)-Q_{k+1}(n)=P_{k}(n)$。
如果$P_{k}(n)=1$可以进一步得到$b_n~(m+1)n+\frac{m}{2}\ln(n+\frac{\frac{m}{2}\ln n+h(x)}{m+1})+h(x)-\frac{m(2m+1)}{12((m+1)n+\frac{m}{2}\ln n+h(x))}+\frac{\frac{m^2}{4} \ln(1+\frac{\frac{m}{2}\ln n+h(x)}{2(m+1)n})}{(m+1)x+\frac{m}{2}\ln n+h(x)}-\frac{7m^3+4m^2}{48(m+1)^2 n^2}+\sum_{i=3}^{\infty}\frac{R_{i-2}(\frac{m}{2}\ln n+h(x))}{n^i}$
这时候$h(x)$满足$h(\frac{(1+x)^{m+1}}{x^m})=h(x)+m+1$，$h(x)~x-\frac{m}{2}\ln\frac{x}{m+1}-(m+1)+\frac{m(2m+1)}{12x}+\frac{m^2(3m+2)}{48x^2}+\cdots$。后面的$-k$次项系数含$m((k+1)m+k)$因子，如果$k$是偶数还有一个额外的$m$因子。
受楼上Buffalo 网友给出的精彩小结启发，终于得到如下结论：
对于\(k=4\),我们可以得到：
对于\(a_{n+1}=a_n+\frac{1}{a_n^3}\)
1.令\(a_n^4=b_n\),得到
\(b_{n+1}=b_n+4+\frac{6}{b_n}+\frac{4}{b_n^2}+\frac{1}{b_n^3}\)
2.由\(m'-\frac{3\ln(m')}{8}=m+1-\frac{3\ln(m)}{8}\)渐近展开：
\(m'=J(m)=m+1+\frac{3}{8m}-\frac{3}{64m^2}-\frac{17}{512m^3}+\frac{105}{4096m^4}-\frac{297}{163840m^5}-\frac{11111}{1310720m^6}+\frac{76737}{14680064m^7}+\)
\(\frac{492243}{587202560m^8}-\frac{40474549}{14092861440m^9}+\frac{9274267}{7516192768m^{10}}+\ldots\)
3.由\(m(x+4+\frac{6}{x}+\frac{4}{x^2}+\frac{1}{x^3})=J(m(x))\),设\(m(x)=\frac{x}{4}+\sum_{k=1}^\infty \frac{u_k}{x^k}\),渐近展开：
\(m(x)=\frac{x}{4}+\frac{7}{16x}+\frac{75}{64x^2}+\frac{1001}{384x^3}+\frac{11637}{2560x^4}+\frac{292843}{76800x^5}-\frac{8813443}{716800x^6}-\frac{1962519689}{30105600x^7}+\frac{709670963}{342921600x^8}+\)
\(\frac{28761656717}{1028764800x^9}-\frac{6137702104697}{226328256000x^{10}}+\ldots\)
4.极限值\(a_0\)由下式求出：
\(m(a_n)-\frac{3\ln(a_n)}{8}=n+\frac{a_0}{4}\)
5.关于\(b_n\)渐近展开式计算：\(y=\frac{1}{n},X=\frac{3\ln(n)}{2}+a_0\)
\(F_1=f(y,X)=\frac{4}{y}+X+(b_{00}+b_{01}X)y+(b_{10}+b_{11}X+b_{12}X^2)y^2+(b_{20}+b_{21}X+b_{22}X^2+b_{23}X^3)y^3+(b_{30}+b_{31}X+b_{32}X^2+b_{33}X^3+b_{34}X^4)y^4+(b_{40}+b_{41}X+b_{42}X^2+b_{43}X^3+b_{44}X^4+b_{45}X^5)y^5+\)
\((b_{50}+b_{51}X+b_{52}X^2+b_{53}X^3+b_{54}X^4+b_{55}X^5+b_{56}X^6)y^6+(b_{60}+b_{61}X+b_{62}X^2+b_{63}X^3+b_{64}X^4+b_{65}X^5+b_{66}X^6+b_{67}X^7)y^7+(b_{70}+b_{71}X+b_{72}X^2+b_{73}X^3+b_{74}X^4+b_{75}X^5+b_{76}X^6+b_{77}X^7+b_{78}X^8)y^8\)
\(F_2=f(\frac{y}{1+y},X+\frac{3\ln(1+y)}{2})\)
由\(F_2=F_1+4+\frac{6}{F_1}+\frac{4}{F_1^2}+\frac{1}{F_1^3}\)求解得到：
\(b_{00}=\frac{-7}{16}, b_{01} =\frac{3}{8}, b_{10} =\frac{-75}{256}, b_{11} =\frac{1}{4}, b_{12}=\frac{-3}{64}, b_{20}=\frac{-1295}{6144}, b_{21}=\frac{123}{512}, b_{22}=\frac{-41}{512}, b_{23}=\frac{1}{128}, b_{30}=\frac{-27387}{163840}, b_{31}=\frac{2033}{8192}, b_{32}=\frac{-123}{1024},b_{33}=\frac{47}{2048}, b_{34}=\frac{-3}{2048}, b_{40}=\frac{-2743793}{1960800}, b_{41}=\frac{85369}{327680}\)
\(b_{42}=\frac{-2771}{16384}, b_{43}=\frac{797}{16384}, b_{44}=\frac{-103}{16384}, b_{45}=\frac{3}{10240},b_{50}=\frac{-89215957}{734003200}, b_{51}=\frac{855727}{3145728}, b_{52}=\frac{-118521}{524288}, b_{53}=\frac{34883}{393216}, b_{54}=\frac{-4603}{262144}, b_{55}=\frac{551}{327680}, b_{56}=\frac{-1}{16384}, b_{60}=\frac{-4615268183}{41104179200}, b_{61}=\frac{13043753}{45875200}\)
\(b_{62}=\frac{-605645}{2097152}, b_{63}=\frac{38351}{262144}, b_{64}=\frac{-83575}{2097152}, b_{65}=\frac{7731}{1310720}, b_{66}=\frac{-581}{1310720}, b_{67}=\frac{3}{229376}\)
类似于mathe提供的求解思路，我们来讨论稍一般的情形
对于\(a_{n+1}=a_n+\frac{1}{a_n^{k-1}}\)
1.令\(a_n^k=b_n\),得到
\(b_{n+1}=b_n+k+\frac{k(k-1)}{2b_n}+\frac{k(k-1)(k-2)}{6b_n^2}+\frac{k(k-1)(k-2)(k-3)}{24b_n^3}+\frac{k(k-1)(k-2)(k-3)(k-4)}{120b_n^4}+\frac{k(k-1)(k-2)(k-3)(k-4)(k-5)}{720b_n^5}+\ldots\)
2.由\(m'-\frac{(k-1)\ln(m')}{2k}=m+1-\frac{(k-1)\ln(m)}{2k}\)渐近展开：
\(m'=J(m)=m+1+\frac{k-1}{2km}-\frac{k-1}{4k^2m^2}-\frac{-5k^2+2k^3+3}{24k^3m^3}+\frac{6k+k^4+4k^3-14k^2+3}{48k^4m^4}+\frac{-73k^4+50k^3+13k^5+100k^2-75k-15}{480k^5m^5}-\frac{325k^3+50k^2-248k^4+17k^6+6k^5-135k-15}{960k^6m^6}+\)
\(\frac{1225k^2-6125k^3+2744k^5-1441k^6+1911k^4+1470k+105+111k^7}{13440k^7m^7}+\frac{-32340k^3-17052k^4-13584k^6-2652k^7+43316k^5+14700k^2+6300k+997k^8+315}{80640k^8m^8}+\frac{102312k^4-112896k^5+42108k^7-10721k^8-12092k^6+35280k^3-35280k^2+109k^9-8505k-315}{161280k^9m^9}+\ldots\)
3.由\(m(x+k+\frac{k(k-1)}{2x}+\frac{k(k-1)(k-2)}{6x^2}+\frac{k(k-1)(k-2)(k-3)}{24x^3}+\frac{k(k-1)(k-2)(k-3)(k-4)}{120x^4}+\frac{k(k-1)(k-2)(k-3)(k-4)(k-5)}{720x^5}+\ldots)=J(m(x))\),设\(m(x)=\frac{x}{k}+\sum_{k=1}^\infty \frac{u_k}{x^k}\),渐近展开：
\(m(x)=\frac{x}{k}+\frac{1+2k^2-3k}{12kx}+\frac{7k^3+13k-17k^2-3}{48kx^2}+\frac{-442k-1420k^3+437k^4+1490k^2-67+2k^5}{4320kx^3}+\frac{-6573k+42315k^3-17661k^2+7014k^5-30072k^4+4901+76k^6}{12096kx^4}+\frac{-117880k^2+301049k^4-151018k^3+25666k^6-160888k^5+119816k+738k^7-17483}{1209600kx^5}-\)
\(\frac{3163257k^2+411621k^4-2145479k^3-35273k^6+207669k^5-2315013k-27849k^7+738931+2136k^8}{4838400kx^6}-\frac{-357561276k^3-98841666k^5+225077818k^4-20753016k^7+46893306k^6+320183186k^2+4159831k^8-134911902k-37644k^9+15791307+56k^{10}}{203212800kx^7}+\cdot\)
4.极限值\(a_0\)由下式求出：
\(m(a_n)-\frac{(k-1)\ln(a_n)}{2k}=n+\frac{a_0}{k}\)
5.关于\(b_n\)渐近展开式计算：\(y=\frac{1}{n},X=\frac{(k-1)\ln(n)}{2}+a_0\)
\(F_1=f(y,X)=\frac{k}{y}+X+(b_{00}+b_{01}X)y+(b_{10}+b_{11}X+b_{12}X^2)y^2+(b_{20}+b_{21}X+b_{22}X^2+b_{23}X^3)y^3+(b_{30}+b_{31}X+b_{32}X^2+b_{33}X^3+b_{34}X^4)y^4+(b_{40}+b_{41}X+b_{42}X^2+b_{43}X^3+b_{44}X^4+b_{45}X^5)y^5+\)
\((b_{50}+b_{51}X+b_{52}X^2+b_{53}X^3+b_{54}X^4+b_{55}X^5+b_{56}X^6)y^6+(b_{60}+b_{61}X+b_{62}X^2+b_{63}X^3+b_{64}X^4+b_{65}X^5+b_{66}X^6+b_{67}X^7)y^7+(b_{70}+b_{71}X+b_{72}X^2+b_{73}X^3+b_{74}X^4+b_{75}X^5+b_{76}X^6+b_{77}X^7+b_{78}X^8)y^8\)
\(F_2=f(\frac{y}{1+y},X+\frac{(k-1)\ln(1+y)}{2})\)
由\(F_2=F_1+k+\frac{k(k-1)}{2F_1}+\frac{k(k-1)(k-2)}{6F_1^2}+\frac{k(k-1)(k-2)(k-3)}{24F_1^3}+\frac{k(k-1)(k-2)(k-3)(k-4)}{120F_1^4}+\frac{k(k-1)(k-2)(k-3)(k-4)(k-5)}{720F_1^5}+\ldots\)求解得到：
\(b_{00}=\frac{-(2k-1)(k-1)}{12k}, b_{01} =\frac{k-1}{2k}, b_{10} =\frac{-(7k-3)(k-1)^2}{48k^2}, b_{11} =\frac{(5k-4)(k-1)}{12k^2}, b_{12}=\frac{-(k-1)}{4k^2}, b_{20}=\frac{-(k-1)(2k^4+559k^3-1221k^2+659k+37)}{4320k^3}, b_{21}=\frac{(12k-7)(k-1)^2}{24k^3}, b_{22}=\frac{-(13k-11)(k-1)}{24k^3}, b_{23}=\frac{k-1}{6k^3}, b_{30}=\frac{-(k-1)(76k^5+15910k^4-48812k^3+46423k^2-10298k-3011)}{120960k^4}, b_{31}=\frac{(k-1)(2k^4+919k^3-2151k^2+1439k-173)}{1440k^4}, b_{32}=\frac{-(49k-32)(k-1)^2}{48k^4},b_{33}=\frac{(15k-13)(k-1)}{24k^4}, b_{34}=\frac{-(k-1)}{8k^4}\)
\(b_{40}=\frac{-(k-1)(3334k^6+511322k^5-2033962k^4+2779595k^3-1359379k^2+30191k+38659)}{3628800k^5}, b_{41}=\frac{(k-1)(194k^5+51077k^4-162094k^3+168236k^2-54448k-2389)}{60480k^5} ,b_{42}=\frac{-(k-1)(4k^4+2573k^3-6252k^2+4573k-826)}{1440k^5}, b_{43}=\frac{(241k-167)(k-1)^2}{144k^5},b_{44}=\frac{-(33k-29)(k-1)}{48k^5}, b_{45}=\frac{k-1}{10k^5},b_{50}=\frac{-(k-1)(18908k^7+2272111k^6-11151258k^5+19937499k^4-14363838k^3+637575k^2+4729396k-2469193)}{14515200k^6}, b_{51}=\frac{(k-1)(4498k^6+816620k^5-3312988k^4+4761575k^3-2695483k^2+342545k+52993)}{725760k^6}\)
\(b_{52}=\frac{-(k-1)(1138k^5+363283k^4-1181120k^3+1295830k^2-498998k+22747)}{120960k^6} , b_{53}=\frac{(k-1)(4k^4+3296k^3-8199k^2+6298k-1327)}{864k^6}, b_{54}=\frac{-(1403k-1009)(k-1)^2}{576k^6}, b_{55}=\frac{(177k-157)(k-1)}{240k^6}, b_{56}=\frac{-(k-1)}{12k^6}\)
http://spaces.ac.cn/archives/3696/
一个是隐式解，并可以从中解得显式解和极限值$a$。
@282842712474
回顾这个问题，看看博客中的方法是否还能有效？
kastin 发表于 2016-6-18 15:55
@282842712474
回顾这个问题，看看博客中的方法是否还能有效？
那个问题两边取对数后，让$\ln a_n=b_n$
$$b_{n+1}=b_n+\frac{b_n^2}{2}-\frac{b_n^4}{12}+\dots$$
就跟本题差不多了
282842712474 发表于 2016-6-19 11:33
那个问题两边取对数后，让$\ln a_n=b_n$
$$b_{n+1}=b_n+\frac{b_n^2}{2}-\frac{b_n^4}{12}+\dots$$
...
能证明那个不等式吗？