如果我们先扩展矩阵元素,再帮助整合一下,即使需要花费一点点的精力,也能够完成任务。
在装有Windows和Mathematica 8.0.4的四核笔记本电脑上,对于DIM=200,以下代码运行大约需要13分钟;对于DIM=50,代码运行时间为6秒。
$starttime = AbsoluteTime[]; Quiet[LaunchKernels[]];
DIM = 200;
Print["$Version = ", $Version, " ||| ", "Number of Kernels : ", Length[Kernels[]]];
f[r_] := f[r] = Sum[(((-1)^n*(-(2*n) + 2*r - 7)!!)*x^(-(2*n) + r - 1))/(2^n*n!*(-(2*n) + r - 1)!), {n, 0, r/2}];
Nw = Transpose[Table[f[j], {i, 1}, {j, 5, DIM, 1}]];
nw2 = Nw . Transpose[Nw];
Print["Seconds for expanding Nw.Transpose[Nm] ", Round[First[AbsoluteTiming[nw3 = ParallelMap[Expand, nw2]; ]]]];
Print["do the integral once: ", Integrate[x^n, {x, -1, 1}, Assumptions -> n > -1]];
Print["the integration can be written as a simple rule: ", intrule = (pol_Plus)?(PolynomialQ[
(Select[pol, !FreeQ[
Print["Seconds for integrating Nw.Transpose[Nw] : ", Round[First[AbsoluteTiming[X1 = ParallelTable[row /. intrule, {row, nw3}]; ]]]];
Print["expanding: ", Round[First[AbsoluteTiming[preY1 = ParallelMap[Expand, D[Nw, {x, 2}] . Transpose[D[Nw, {x, 2}]]]; ]]]];
Print["Seconds for integrating : ", Round[First[AbsoluteTiming[Y1 = ParallelTable[py /. intrule, {py, preY1}]; ]]]];
Print["X1 = ", (Shallow[
Print["Y1 = ", (Shallow[
Print["seq Y1 : ", Simplify[FindSequenceFunction[Diagonal[Y1], n]]];
Print["seq X1 0 : ",Simplify[FindSequenceFunction[Diagonal[X1, 0], n]]];
Print["seq X1 2: ",Simplify[FindSequenceFunction[Diagonal[X1, 2], n]]];
Print["seq X1 4: ",Simplify[FindSequenceFunction[Diagonal[X1, 4], n]]];
Print["overall time needed in seconds: ", Round[AbsoluteTime[] - $starttime]];