12楼: Originally posted by xzczd at 2016-01-08 14:38:05
看看楼上的某些发言就觉得Mathematica推广工作任重而道远……楼主你这个问题是可以用Mathematica推导的。不过在给出解法前得先和你讨论清楚一件事：你给出的
w_j的表达式是不是错了。我手算和用软件推导（当然编程 ...

13楼: Originally posted by xzczd at 2016-01-08 14:44:20
刚不放心所以又带具体的数算了一遍，如果各个式子的定义真的如你所写，那么分母各项的指数毫无疑问就是2j而不是j+1。...

13楼: Originally posted by xzczd at 2016-01-08 14:44:20
刚不放心所以又带具体的数算了一遍，如果各个式子的定义真的如你所写，那么分母各项的指数毫无疑问就是2j而不是j+1。...

14楼: Originally posted by loujing at 2016-01-08 17:03:58
您好，感谢您的回复。w_j最终的闭解，是原书上给出的。所以我想利用软件验证一下。

我在网上搜了一下，李航老师在他新浪博客上给出了勘误表，http://blog.sina.com.cn/s/blog_7ad48fee01017dpi.html，如截图。
...

Sum[1/2 a[b], {b, c, e}]

Sum[ff a[b], {b, c, e}]

Clear[m, n]

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

(*以下规则对应前面提及的第一点*)
rule[w_] := Sum[expr_, {i_, downup__}] /; ! FreeQ[w, i] :> expr
(*以下三条规则对应第二点*)
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}];
(*把相应的变换规则代入：*)
D[l /. rule[w@j], w[j]] //. {ruleExpand, ruleDistribute, ruleExtract}
Solve[% == 0, w[j]]

rule2[w_] := sum[expr_, {i_, downup__}] /; ! FreeQ[w, i] :> expr

(*d[a_ +b_,x_]:=d[a,x]+d[b,x]*)
d[a_ b_, x_] /; FreeQ[a, x] := a d[b, x]
d[a_^b_, x_] /; FreeQ[b, x] := b a^(b - 1) d[a, x]
d[sum[expr_, i_], x_] := If[FreeQ[i, x], sum[d[expr, x], i], d[expr, x]]
d[expr_, i_] := D[expr, i]

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

f2[i_, m_] := sum[w[j]*x[i]^j, {j, 0, m}]
l2 = 1/2 sum[(f2[i, m] - y[i])^2, {i, n}];

d[l2 /. rule2[w@j], w[j]]

Solve[% == 0, w[j]] /. sum -> Sum

12楼: Originally posted by xzczd at 2016-01-08 14:38:05
看看楼上的某些发言就觉得Mathematica推广工作任重而道远……楼主你这个问题是可以用Mathematica推导的。不过在给出解法前得先和你讨论清楚一件事：你给出的
w_j的表达式是不是错了。我手算和用软件推导（当然编程 ...

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

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

...

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' :> ...