f[Subscript[w, j]] = 1/2 \!\(\*UnderoverscriptBox[\(\[Sum]\), \(i = 1\), \(n\)]\*SuperscriptBox[\((\*UnderoverscriptBox[\(\[Sum]\), \(j = 0\), \(m\)]\*SubscriptBox[\(w\), \(j\)] \*SubsuperscriptBox[\(x\), \(i\), \(j\)] - \*SubscriptBox[\(y\), \(i\)])\), \(2\)]\)

L := proc (w__j) options operator, arrow; (1/2)*(sum((sum(w__j*x__i^j, j = 0 .. m)-y__i)^2, i = 1 .. n)) end proc

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

2楼: Originally posted by 赵梦92 at 2016-01-08 12:31:47
大前提错了吧。我的书上和你的不一样呢