例子1：
解：
首先有如下结论：
E k = 1 2 ∑ k = 1 n ( z k − f k ( x k ) ) 2 \color{green} E_k = \frac{1}{2}\sum_{k=1}^{n} ( z_k - f_k(x_k))^2 Ek=21k=1∑n(zk−fk(xk))2
z k = ∑ i = 1 , j = 1 3 , 2 y i v i j z_k = \sum_{i=1,j =1}^{3,2}y_i v_{ij} zk=i=1,j=1∑3,2yivij
∂ E k ∂ z 1 = z k − f k \frac{\partial E_k}{\partial z_{1}} = z_k - f_k ∂z1∂Ek=zk−fk
错误： E k = S ( ∑ i = 1 n z i ) = S ( ∑ i = 1 n ∑ j = 1 n v i j y i j ) \color{red} 错误：E_k = S(\sum_{i=1}^{n} z_i) = S(\sum_{i=1}^{n} \sum_{j=1}^{n}v_{ij}y_{ij}) 错误：Ek=S(i=1∑nzi)=S(i=1∑nj=1∑nvijyij)
并且可手动推导验证如下公式：
s ′ = s ( 1 − s ) s' = s(1-s) s′=s(1−s)
E k E_k Ek对 v 11 v_{11} v11的偏导数只跟 z 1 z_1 z1有关，跟 z 2 z_2 z2无关。同时， z 1 z_1 z1对 v 11 v_{11} v11的偏导数只跟 y 1 y_1 y1和 v 11 v_{11} v11有关，跟 y 2 y_2 y2, y 3 y_3 y3， v 21 v_{21} v21, v 12 v_{12} v12, v 21 v_{21} v21等都无关。
∂ E k ∂ v 11 = ∂ E k ∂ z 1 ∂ z 1 ∂ v 11 = ( z 1 − f 1 ) s ( z 1 ) ′ = ( z 1 − f 1 ) s ( z 1 ) ( 1 − s ( z 1 ) ) y 1 = ( z 1 − f 1 ) z 1 ( 1 − z 1 ) y 1 \frac{\partial E_k}{\partial v_{11}}=\frac{\partial E_k}{\partial z_{1}} \frac{\partial z_1}{\partial v_{11}} =( z_1 - f_1)s(z_1)' =\\ ( z_1 - f_1)s(z_1)(1-s(z_1)) y_1 = ( z_1 - f_1)z_1(1-z_1) y_1 ∂v11∂Ek=∂z1∂Ek∂v11∂z1=(z1−f1)s(z1)′=(z1−f1)s(z1)(1−s(z1))y1=(z1−f1)z1(1−z1)y1
∂ E k ∂ v 21 = ∂ E k ∂ z 2 ∂ z 2 ∂ v 21 = ( z 2 − f 2 ) s ( z 2 ) ′ = ( z 2 − f 2 ) s ( z 2 ) ( 1 − s ( z 2 ) ) y 1 = ( z 2 − f 2 ) z 2 ( 1 − z 2 ) y 1 \frac{\partial E_k}{\partial v_{21}}=\frac{\partial E_k}{\partial z_{2}} \frac{\partial z_2}{\partial v_{21}} =( z_2 - f_2)s(z_2)' =\\ ( z_2 - f_2)s(z_2)(1-s(z_2)) y_1 = ( z_2 - f_2)z_2(1-z_2) y_1 ∂v21∂Ek=∂z2∂Ek∂v21∂z2=(z2−f2)s(z2)′=(z2−f2)s(z2)(1−s(z2))y1=(z2−f2)z2(1−z2)y1
例子2：
解答：
例子3：
解答：
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