我的目标是求解abcd2的值，对于整个问题我的思路是将函数U拟合成D-10乘x的-10次到D10乘x的10次的和，其中D-10到D10都是实部+虚部，然后将φψ函数也表示为洛朗级数，然后比对每个x的n次方前面的系数来求解问题。我的实部部分求解的没有问题，但是虚部部分算出来就明显不对，希望得到帮助

Y2 = Join[ixs2, ics2, {A0i, A0i1}]
K2 = Table[0, {i, 1, 4 k3 + 2}, {j, 1, 4 k3 + 2}];
K2[[1, 1]] = 2;
K2[[1, 2]] = -2;
K2[[1, k3 + 1]] = -1;
K2[[1, k3 + 2]] = -2;
K2[[1, k3 + 3]] = 3;
K2[[1, 3 k3 + 1]] = -2;
K2[[1, 3 k3 + 2]] = 2;
For[i = 2, i <= 8, i++,
K2[[i, i]] = 2;
K2[[i, i + 1]] = -2;
K2[[i, k3 + i - 1]] = i - 1;
K2[[i, k3 + i]] = -i;
K2[[i, k3 + i + 1]] = -(i + 1);
K2[[i, k3 + i + 2]] = i + 2;
K2[[i, 3 k3 + i]] = -2;
K2[[i, 3 k3 + i + 1]] = 2;]
K2[[k3 - 1, k3 - 1]] = 2;
K2[[k3 - 1, k3]] = -2;
K2[[k3 - 1, 2 k3 - 2]] = 8;
K2[[k3 - 1, 2 k3 - 1]] = -9;
K2[[k3 - 1, 2 k3]] = -10;
K2[[k3 - 1, 4 k3 - 1]] = -2;
K2[[k3 - 1, 4 k3]] = 2;
K2[[k3, k3]] = 2;
K2[[k3, 2 k3 - 1]] = 9;
K2[[k3, 2 k3]] = -10;
K2[[k3, 4 k3]] = -2;
K2[[k3 + 1, 1]] = 1;
K2[[k3 + 1, 2]] = -2;
K2[[k3 + 1, k3 + 1]] = 3;
K2[[k3 + 1, 2 k3 + 1]] = -2;
K2[[k3 + 1, 4 k3 + 1]] = -2;
K2[[k3 + 1, 4 k3 + 2]] = 2;
K2[[k3 + 2, 1]] = 1;
K2[[k3 + 2, 2]] = 2;
K2[[k3 + 2, 3]] = -3;
K2[[k3 + 2, k3 + 1]] = -2;
K2[[k3 + 2, k3 + 2]] = 2;
K2[[k3 + 2, 2 k3 + 1]] = 2;
K2[[k3 + 2, 2 k3 + 2]] = -2;
For[i = 3, i <= 9, i++,
K2[[k3 + i, i - 2]] = -( i - 2);
K2[[k3 + i, i - 1]] = i - 1;
K2[[k3 + i, i]] = i;
K2[[k3 + i, i + 1]] = -(i + 1);
K2[[k3 + i, k3 + i - 1]] = -2;
K2[[k3 + i, k3 + i]] = 2;
K2[[k3 + i, 2 k3 + i - 1]] = 2;
K2[[k3 + i, 2 k3 + i]] = -2]
K2[[2 k3, k3 - 2]] = -8;
K2[[2 k3, k3 - 1]] = 9;
K2[[2 k3, k3]] = 10;
K2[[2 k3, 2 k3 - 1]] = -2;
K2[[2 k3, 2 k3]] = 2;
K2[[2 k3, 3 k3 - 1]] = 2;
K2[[2 k3, 3 k3]] = -2;
K2[[2 k3 + 1, 1]] = 2 \[Alpha]^-1;
K2[[2 k3 + 1, 2]] = -2 \[Alpha]^-1;
K2[[2 k3 + 1, k3 + 1]] = \[Alpha]^3 - 2 \[Alpha];
K2[[2 k3 + 1, k3 + 2]] = 2 \[Alpha] - 4 \[Alpha]^3;
K2[[2 k3 + 1, k3 + 3]] = 3 \[Alpha]^3;
K2[[2 k3 + 1, 3 k3 + 1]] = -2 \[Alpha];
K2[[2 k3 + 1, 3 k3 + 2]] = 2 \[Alpha]^3;
For[i = 2, i <= 8, i++,
K2[[2 k3 + i, i]] = 2 \[Alpha]^-i;
K2[[2 k3 + i, i + 1]] = -2 \[Alpha]^-i;
K2[[2 k3 + i, k3 + i - 1]] = (i - 1) \[Alpha]^i;
K2[[2 k3 + i, k3 + i]] = i (\[Alpha]^2 - 2) \[Alpha]^i;
K2[[2 k3 + i, k3 + i + 1]] = -(i + 1) (2 \[Alpha]^2 - 1) \[Alpha]^i;
K2[[2 k3 + i, k3 + i + 2]] = (i + 2) \[Alpha]^(i + 2);
K2[[2 k3 + i, 3 k3 + i]] = -2 \[Alpha]^i;
K2[[2 k3 + i, 3 k3 + i + 1]] = 2 \[Alpha]^(i + 2);]
K2[[3 k3 - 1, k3 - 1]] = 2 \[Alpha]^-9;
K2[[3 k3 - 1, k3]] = -2 \[Alpha]^-9;
K2[[3 k3 - 1, 2 k3 - 2]] = 8 \[Alpha]^9;
K2[[3 k3 - 1, 2 k3 - 1]] = 9 (\[Alpha]^2 - 2) \[Alpha]^9;
K2[[3 k3 - 1, 2 k3]] = -10 (2 \[Alpha]^2 - 1) \[Alpha]^9;
K2[[3 k3 - 1, 4 k3 - 1]] = -2 \[Alpha]^9;
K2[[3 k3 - 1, 4 k3]] = 2 \[Alpha]^11;
K2[[3 k3, k3]] = 2 \[Alpha]^-10;
K2[[3 k3, 2 k3 - 1]] = 9 \[Alpha]^10;
K2[[3 k3, 2 k3]] = 10 (\[Alpha]^2 - 2) \[Alpha]^10;
K2[[3 k3, 4 k3]] = -2 \[Alpha]^10;
K2[[3 k3 + 1, 1]] = -\[Alpha] + 2 \[Alpha]^-1;
K2[[3 k3 + 1, 2]] = -2 \[Alpha]^-1;
K2[[3 k3 + 1, k3 + 1]] = 3 \[Alpha];
K2[[3 k3 + 1, 2 k3 + 1]] = -2 \[Alpha]^-1;
K2[[3 k3 + 1, 4 k3 + 1]] = -2 \[Alpha];
K2[[3 k3 + 1, 4 k3 + 2]] = 2 \[Alpha];
K2[[3 k3 + 2, 1]] = 2 - \[Alpha]^-2;
K2[[3 k3 + 2, 2]] = -2 + 4 \[Alpha]^-2;
K2[[3 k3 + 2, 3]] = -3 \[Alpha]^-2;
K2[[3 k3 + 2, k3 + 1]] = -2 \[Alpha]^2;
K2[[3 k3 + 2, k3 + 2]] = 2 \[Alpha]^2;
K2[[3 k3 + 2, 2 k3 + 1]] = 2;
K2[[3 k3 + 2, 2 k3 + 2]] = -2 \[Alpha]^-2;
For[i = 3, i <= 9, i++,
K2[[3 k3 + i, i - 2]] = -(i - 2) \[Alpha]^(-i + 2);
K2[[3 k3 + i, i - 1]] = -(i - 1) \[Alpha]^-i +
2 (i - 1) \[Alpha]^(-i + 2);
K2[[3 k3 + i, i]] = -i \[Alpha]^(-i + 2) + 2 i \[Alpha]^-i;
K2[[3 k3 + i, i + 1]] = -(i + 1) \[Alpha]^-i;
K2[[3 k3 + i, k3 + i - 1]] = -2 \[Alpha]^i;
K2[[3 k3 + i, k3 + i]] = 2 \[Alpha]^i;
K2[[3 k3 + i, 2 k3 + i - 1]] = 2 \[Alpha]^(-i + 2);
K2[[3 k3 + i, 2 k3 + i]] = -2 \[Alpha]^-i]
K2[[4 k3, k3 - 2]] = -8 \[Alpha]^-8;
K2[[4 k3, k3 - 1]] = -9 \[Alpha]^-10 + 18 \[Alpha]^-8;
K2[[4 k3, k3]] = -10 \[Alpha]^-8 + 20 \[Alpha]^-10;
K2[[4 k3, 2 k3 - 1]] = -2 \[Alpha]^10;
K2[[4 k3, 2 k3]] = 2 \[Alpha]^10;
K2[[4 k3, 3 k3 - 1]] = 2 \[Alpha]^-8;
K2[[4 k3, 3 k3]] = -2 \[Alpha]^-10;
K2[[4 k3 + 1, 1]] = -3;
K2[[4 k3 + 1, k3 + 1]] = -1;
K2[[4 k3 + 1, k3 + 2]] = 2;
K2[[4 k3 + 1, 3 k3 + 1]] = 2;
K2[[4 k3 + 1, 4 k3 + 1]] = 2;
K2[[4 k3 + 1, 4 k3 + 2]] = -2;
K2[[4 k3 + 2, 1]] = -3;
K2[[4 k3 + 2, k3 + 1]] = -(2 \[Alpha]^2 - 1);
K2[[4 k3 + 2, k3 + 2]] = 2 \[Alpha]^2;
K2[[4 k3 + 2, 3 k3 + 1]] = 2 \[Alpha]^2;
K2[[4 k3 + 2, 4 k3 + 1]] = 2;
K2[[4 k3 + 2, 4 k3 + 2]] = -2;
abcd2 = LinearSolve[K2, Y2]
Dot[K2, abcd2]

这是我求解的思路，然后我实部的部分求解已经算出来了，矩阵系数应该没有问题了，希望有大佬能看看为什么？

关于矩阵的系数我也贴在这里
U11 = Sum[(2*Subscript[a, k, 1] +
I*2*Subscript[a, k, 2])*\[Zeta]^-k, {k, 0, 10}] +
Sum[(-2*Subscript[a, k, 1] +
I*-2*Subscript[a, k, 2])*\[Zeta]^(-k + 1), {k, 0, 11}] +
Sum[(2*Subscript[b, k, 1] + I*2*Subscript[b, k, 2])*\[Zeta]^k, {k,
1, 10}] +
Sum[(-2*Subscript[b, k, 1] + I*-2*Subscript[b, k, 2])*\[Zeta]^(
k + 1), {k, 1, 9}] +
Sum[(-k*Subscript[a, k, 1] + I*k*Subscript[a, k, 2])*\[Zeta]^(
k + 1), {k, 1, 9}] +
Sum[(k*Subscript[a, k, 1] + I*-k*Subscript[a, k, 2])*\[Zeta]^(
k + 2), {k, 1, 8}] +
Sum[(-k*Subscript[a, k, 1] + I*k*Subscript[a, k, 2])*\[Zeta]^k, {k,
1, 10}] +
Sum[(k*Subscript[a, k, 1] + I*-k*Subscript[a, k, 2])*\[Zeta]^(
k - 1), {k, 1, 11}] +
Sum[(k*Subscript[b, k, 1] + I*-k*Subscript[b, k, 2])*\[Zeta]^(-k +
1), {k, 1, 11}] +
Sum[(-k*Subscript[b, k, 1] + I*k*Subscript[b, k, 2])*\[Zeta]^(-k +
2), {k, 1, 12}] +
Sum[(k*Subscript[b, k, 1] +
I*-k*Subscript[b, k, 2])*\[Zeta]^-k, {k, 1, 10}] +
Sum[(-k*Subscript[b, k, 1] + I*k*Subscript[b, k, 2])*\[Zeta]^(-k -
1), {k, 1, 9}] +
Sum[(2*Subscript[c, k, 1] + I*-2*Subscript[c, k, 2])*\[Zeta]^k, {k,
0, 10}] +
Sum[(-2*Subscript[c, k, 1] + I*2*Subscript[c, k, 2])*\[Zeta]^(
k + 1), {k, 0, 9}] +
Sum[(2*Subscript[d, k, 1] +
I*-2*Subscript[d, k, 2])*\[Zeta]^-k, {k, 1, 10}] +
Sum[(-2*Subscript[d, k, 1] + I*2*Subscript[d, k, 2])*\[Zeta]^(-k +
1), {k, 1, 11}];
Collect[U11, \[Zeta], FullSimplify]
U1\[Alpha] =
Sum[(2*Subscript[a, k, 1] +
I*2*Subscript[a, k, 2])*\[Alpha]^-k*\[Zeta]^-k, {k, 0, 10}] +
Sum[(-2*Subscript[a, k, 1] +
I*-2*Subscript[a, k, 2])*\[Alpha]^(-k + 1)*\[Zeta]^(-k +
1), {k, 0, 11}] +
Sum[(2*Subscript[b, k, 1] + I*2*Subscript[b, k, 2])*\[Alpha]^
k*\[Zeta]^k, {k, 1, 10}] +
Sum[(-2*Subscript[b, k, 1] + I*-2*Subscript[b, k, 2])*\[Alpha]^(
k + 1)*\[Zeta]^(k + 1), {k, 1, 9}] +
Sum[(2 \[Alpha]^2 - 1)*(-k*Subscript[a, k, 1] +
I*k*Subscript[a, k, 2])*\[Alpha]^(-k - 1)*\[Zeta]^(k + 1), {k,
1, 9}] +
Sum[(k*Subscript[a, k, 1] +
I*-k*Subscript[a, k, 2])*\[Alpha]^-k*\[Zeta]^(k + 2), {k, 1,
8}] + Sum[(2 \[Alpha] - \[Alpha]^3)*(-k*Subscript[a, k, 1] +
I*k*Subscript[a, k, 2])*\[Alpha]^(-k - 1)*\[Zeta]^k, {k, 1,
10}] + Sum[(k*Subscript[a, k, 1] +
I*-k*Subscript[a, k, 2])*\[Alpha]^(-k + 1)*\[Zeta]^(k - 1), {k,
1, 11}] +
Sum[(2 \[Alpha]^2 - 1)*(k*Subscript[b, k, 1] +
I*-k*Subscript[b, k, 2])*\[Alpha]^(k - 1)*\[Zeta]^(-k + 1), {k,
1, 11}] +
Sum[(-k*Subscript[b, k, 1] + I*k*Subscript[b, k, 2])*\[Alpha]^
k*\[Zeta]^(-k + 2), {k, 1, 12}] +
Sum[(2 \[Alpha] - \[Alpha]^3)*(k*Subscript[b, k, 1] +
I*-k*Subscript[b, k, 2])*\[Alpha]^(k - 1)*\[Zeta]^-k, {k, 1,
10}] + Sum[(-k*Subscript[b, k, 1] +
I*k*Subscript[b, k, 2])*\[Alpha]^(k + 1)*\[Zeta]^(-k - 1), {k,
1, 9}] +
Sum[(2*Subscript[c, k, 1] +
I*-2*Subscript[c, k, 2])*\[Alpha]^-k*\[Zeta]^k, {k, 0, 10}] +
Sum[(-2*Subscript[c, k, 1] +
I*2*Subscript[c, k, 2])*\[Alpha]^(-k + 1)*\[Zeta]^(k + 1), {k,
0, 9}] +
Sum[(2*Subscript[d, k, 1] + I*-2*Subscript[d, k, 2])*\[Alpha]^
k*\[Zeta]^-k, {k, 1, 10}] +
Sum[(-2*Subscript[d, k, 1] + I*2*Subscript[d, k, 2])*\[Alpha]^(
k + 1)*\[Zeta]^(-k + 1), {k, 1, 11}];
Collect[U1\[Alpha], \[Zeta], FullSimplify]

2楼一直被删提交申诉就一直说操作失败也恢复不了，请问有什么办法吗？

你的原问题是什么？你的代码太长长长了

……你能不能用一个比较小的例子重现这个现象呢？比如说，使用洛朗级数展开一个形式非常简单的函数，再把级数展开的项数取的少一点，是不是依旧会观察到这个错误呢？