求一个数列的近似公式

KeyTo9_Fans

$A_9=-0.5928207421224596$ For $i=10$ to $\infty$: $A_i=A_{i-1}+1/((0.4187*i-3.1557)^3.03)$ 如何求$A_n$的近似公式？（要求精确到$o(1/n^4)$）

wayne

1# KeyTo9_Fans 就是对后面的那个式子求有限项的和吧

Buffalo

Euler-Maclaurin 求和公式

wayne

3# Buffalo 正是！

KeyTo9_Fans

根据$3#$的提示，在 http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula 里面找到一个公式： 剩下的问题是： $1.$ 只取$\sum_{n=a}^b f(n)=\int_a^b f(x)dx+(f(a)+f(b))/2$，精确程度是多少？ $2.$ 已知$f(n)=1/((0.4187*n-3.1557)^3.03)$，如何求$\int_a^b f(x)dx$？

数学星空

好像计算是发散的:记 $a=0.4187 ,b=-3.1557,m=3.03$ $A_n=A_9+sum_{k=10}^{n}1/(a*k+b)^m$ $sum_{k=10}^{n}1/(a*k+b)^m=int_{10}^{n}f(k)dk+1/2*(f(n)+f(10))+sum_{k=1}^{infty}(B_(2*k)/((2k)!))*(f^(2k-1)(n)-f^(2k-1)(10))+R$ $int_{10}^{n}1/(a*k+b)^mdk=-0.849961/(0.4187*n-3.1557)^2.03+0.7984126565$ $1/2*(f(n)+f(10))=0.4554213649+1/(2*(0.4187*n-3.1557)^3.03)$ $sum_{k=1}^{infty}(B_(2*k)/((2k)!))*(f^(2k-1)(n)-f^(2k-1)(10)) $ (计算$k=1..30$) $=-0.1057217500/(0.4187*n-3.1557)^4.03+0.006261696964/(0.4187*n-3.1557)^6.03 -0.001107952941/(0.4187*n-3.1557)^8.03+ $ $0.0003521036610/(0.4187*n-3.1557)^10.03 -0.0001724478253/(0.4187*n-3.1557)^12.03+ 0.0001199469163/(0.4187*n-3.1557)^14.03 -$ $0.0001122979070/(0.4187*n-3.1557)^16.03+0.0001361275585/(0.4187*n-3.1557)^18.03 -0.0002074062199/(0.4187*n-3.1557)^20.03+$ $0.0003879599905/(0.4187*n-3.1557)^22.03 -0.0008740600072/(0.4187*n-3.1557)^24.03+ 0.002334543536/(0.4187*n-3.1557)^26.03-$ $0.007294043631/(0.4187*n-3.1557)^28.03+ 0.02635631385/(0.4187*n-3.1557)^30.03 -0.1090606200/(0.4187*n-3.1557)^32.03+$ $0.5123652409/(0.4187*n-3.1557)^34.03-2.712238839/(0.4187*n-3.1557)^36.03+16.06911985/(0.4187*n-3.1557)^38.03 -$ $105.9163855/(0.4187*n-3.1557)^40.03+ 772.4963636/(0.4187*n-3.1557)^42.03 -6204.022510/(0.4187*n-3.1557)^44.03+$ $54622.32505/(0.4187*n-3.1557)^46.03-(5.250882752*10^5)/(0.4187*n-3.1557)^48.03+(5.491019501*10^6)/(0.4187*n-3.1557)^50.03-$ $(6.225227714*10^7)/(0.4187*n-3.1557)^52.03+( 7.627406287*10^8)/(0.4187*n-3.1557)^54.03 -(1.007065100*10^10)/(0.4187*n-3.1557)^56.03+$ $(1.428985181*10^11)/(0.4187*n-3.1557)^58.03-(2.173698292*10^12)/(0.4187*n-3.1557)^60.03+(3.536365639*10^13)/(0.4187*n-3.1557)^62.03$ $-(-00.9337330300+ 0.005199725162 -0.0008650468363+0.0002584751599-0.0001190243884+0.00007783900724-0.00006851881246+$ $0.00007809337373-0.0001118716056+ 0.0001967501169-0.0004167727091+0.001046622255-000.3074583621+ 0.01044558067-$ $0.04063926490+0 .1795095281 -0.8934407178+ 4.976910890 -30.84330376+211.5067090 -1597.096042+13220.80243 -1.194950509*10^5+$ $1.174899038*10^6 -1.252370326*10^7+1.442727704*10^8 -1.790997589*10^9+2.389434582*10^10 -3.417406559*10^11+ 5.227385287*10^12)+....$

数学星空

由余项公式: $(p=30)$ $|R|<=(2*zeta(2p))/(2*pi)^(2p)*int_{10}^{n}|f^(2p)(x)|dx$ $=(7.567804464*10^47)/(0.4187*n-3.155700000)^62.03-1.118657790*10^47$ 显然估计很不理想!

数学星空

对于$sum_{k=10}^{oo}1/(0.4187*k-3.1557)^3.03=1.649447$

lsrong314

In[3]:= -0.5928207421224596 + Sum[(0.4187*i - 3.1557)^-3.03, {i, 10, Infinity}] Out[3]= 1.05663

lsrong314

其实你应该说明这些系数怎么来的，如果只是某个计算题，最后的答案用计算机算就可以了