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

17楼: Originally posted by loujing at 2016-01-08 17:28:16
原书截图

...

f[i_, m_] := Sum[w[j]*x[i]^j, {j, 0, m}]
l = 1/2 Sum[(f[i, m] - y[i])^2, {i, n}];

rulej1[w_] := Sum[expr_, {i_, downup__}] /; ! FreeQ[w, i] :> sum[{i, downup}][expr]
rulej2 = sum[_]'[_] :> 1;
rulej3 = sum[i_][expr_] :> Sum[expr, i];

ruleExpand = Sum[(a_ + b_) c_, i_] :> Sum[(a + b) c // Expand, i];
ruleDistribute = Sum[expr1_ + expr2_, i_] :> Sum[expr1, i] + Sum[expr2, i];
ruleExtract = Sum[a_ b_., {i_, downup__}] /; FreeQ[a, i] :> a Sum[b, {i, downup}];

(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