作业帮问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
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![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](/uploads/image/z/12882326-14-6.jpg?t=%E9%97%AEMathematican%3D3%3Bk%3D6%3B%5C%5BEpsilon%5D%5B1%5D%3D-1%2C%5C%5BEpsilon%5D%5B2%5D%3D0%2C%5C%5BEpsilon%5D%5B3%5D%3D1%EF%BC%9B%E5%A6%82%E4%BD%95%E5%AF%B9Table%5BSum%5B1.%2F%28x%5Bj%5D+-+x%5Bi%5D%29%2C%7Bj%2C1%2Ci+-+1%7D%5D+%2B+Sum%5B1.%2F%28x%5Bj%5D+-+x%5Bi%5D%29%2C%7Bj%2Ci+%2B+1%2Ck%7D%5D+%2B+Sum%5B-rho%5Bj%5D%2F%28x%5Bi%5D+-+2+%5C%5BEpsilon%5D%5Bj%5D%29%2C%7Bj%2C1%2Cn%7D%5D+%3D%3D+0%2C%7Bi%2C1%2Ck%7D%5D%E8%BF%9B%E8%A1%8CFindRoot%2C%E6%AF%8F%E4%B8%80%E7%BB%84%7Bxi%7D%E7%9A%84)
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?