问Mathematican=3;k=6;\[Epsilon][1]=-1,\[Epsilon][2]=0,\[Epsilon][3]=1；如何对Table[Sum[1./(x[j] - x[i]),{j,1,i - 1}] + Sum[1./(x[j] - x[i]),{j,i + 1,k}] + Sum[-rho[j]/(x[i] - 2 \[Epsilon][j]),{j,1,n}] == 0,{i,1,k}]进行FindRoot,每一组{xi}的

问Mathematican=3;k=6;\[Epsilon][1]=-1,\[Epsilon][2]=0,\[Epsilon][3]=1；如何对Table[Sum[1./(x[j] - x[i]),{j,1,i - 1}] + Sum[1./(x[j] - x[i]),{j,i + 1,k}] + Sum[-rho[j]/(x[i] - 2 \[Epsilon][j]),{j,1,n}] == 0,{i,1,k}]进行FindRoot,每一组{xi}的
问Mathematica
n=3;k=6;
\[Epsilon][1]=-1,\[Epsilon][2]=0,\[Epsilon][3]=1；
如何对Table[Sum[1./(x[j] - x[i]),{j,1,i - 1}] +
Sum[1./(x[j] - x[i]),{j,i + 1,k}] +
Sum[-rho[j]/(x[i] - 2 \[Epsilon][j]),{j,1,n}] == 0,{i,1,k}]
进行FindRoot,每一组{xi}的值位于区间(2\[Epsilon][1],2\[Epsilon][2]),(2\[Epsilon][2],2\[Epsilon][3])内.

问Mathematican=3;k=6;\[Epsilon][1]=-1,\[Epsilon][2]=0,\[Epsilon][3]=1；如何对Table[Sum[1./(x[j] - x[i]),{j,1,i - 1}] + Sum[1./(x[j] - x[i]),{j,i + 1,k}] + Sum[-rho[j]/(x[i] - 2 \[Epsilon][j]),{j,1,n}] == 0,{i,1,k}]进行FindRoot,每一组{xi}的
你好好看看Map（/@）和Function（#&）的帮助……
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仔细一看我发现……我又是根本不知道你要问啥……你的Table执行过后每个式子里有这么多 x[i] ,到底哪个在哪个区间啊?