20楼: Originally posted by loujing at 2016-01-08 18:15:38
追问一下，在得到这里的联立方程组后，符号推导的工作是不是就可以算结束了？

如果接下来，我给出具体的一组(x_i,y_i)，如何再利用这里的推导，计算出具体的w_j？...

(*之前的代码我就不复制了*)
n = 3; m = 2;
points = RandomReal[1, {n, 2}];

{x[i_], y[i_]} := {points[[i, 1]], points[[i, 2]]};

var = w /@ Range[0, m];
Solve[Table[eqn, {j, 0, m}], var][[1]]
(var /. %).x^Range[0, m]
Show[Plot[%, {x, Min[points[[All, 1]]], Max[points[[All, 1]]]}], ListPlot[points],
PlotRange -> All]
Clear[x, y]

{xlist, ylist} = points\[Transpose];
With[{xm = #^Range[0, m] & /@ xlist}, Solve[ylist.xm == var.xm\[Transpose].xm, var]]

f[i_, m_, x_] := Sum[w@j*x[[i]]^j, {j, 0, m}]
l[m_, n_, x_, y_] := 1/2 Sum[(f[i, m, x] - y[[i]])^2, {i, n}];
n = 4; m = 2;
var = w /@ Range[0, m];
{xlist, ylist} = Transpose@RandomReal[1, {n, 2}];
NSolve[D[l[m, n, xlist, ylist], {var}], var]
Show[ListPlot[Transpose[{xlist, ylist}]],
Plot[Evaluate[var.x^Range[0, m] /. %], {x, 0, 1}], PlotRange -> All]

21楼: Originally posted by xzczd at 2016-01-08 21:36:58
说结束也算结束了，说没结束也没结束，4楼的推导在这之后又多了一步，是把之前的结果写成了矩阵乘法的形式，这个其实是矩阵点乘规则的直接应用，，但是如果你想把这步也交给软件推导的话，我可以告诉你这个规则又是 ...

eqn = (D[l /. rulej1[w@j], w[j]] /. {rulej2, rulej3} //. {ruleExpand, ruleDistribute,
      ruleExtract} // Expand) == 0

18楼: Originally posted by xzczd at 2016-01-08 17:59:09
……刚发现我为了凑出和你的顶楼相似的结果犯了个错误。盯着四楼的图看了半天（真伤眼），现在我基本确定，正确的推导是：

f := Sum
l = 1/2 Sum;

rulej1 := Sum /; ! FreeQ :> sum
rulej2 = sum' :> ...