Sure! The equation for the error function, E, is:

E(t,a)=12|t&#x2212;a|2=12&#x2211;k=1K(tk&#x2212;ak)2" role="presentation" style="text-align: center; position: relative;">E(t,a)=12|t−a|2=12∑k=1K(tk−ak)2E(t,a)=12|t−a|2=12∑k=1K(tk−ak)2E(t, a) = \frac{1}{2} |t - a|^2 = \frac{1}{2} \sum\_{k=1}^K (t\_k - a\_k)^2

Here, t" role="presentation" style="position: relative;">ttt is the target vector, a" role="presentation" style="position: relative;">aaa is the actual output of the MLP, and K" role="presentation" style="position: relative;">KKK is the number of output neurons in the MLP. The norm of the difference between t" role="presentation" style="position: relative;">ttt and a" role="presentation" style="position: relative;">aaa is denoted by the double vertical bars, |&#x22C5;|" role="presentation" style="position: relative;">|⋅||⋅|| \cdot |, which represent the Euclidean norm.

To find the partial derivative of E" role="presentation" style="position: relative;">EEE with respect to the k" role="presentation" style="position: relative;">kkkth element of a" role="presentation" style="position: relative;">aaa, we can use the chain rule:

&#x2202;E&#x2202;ak=&#x2202;&#x2202;ak(12&#x2211;j=1K(tj&#x2212;aj)2)" role="presentation" style="text-align: center; position: relative;">∂E∂ak=∂∂ak(12∑j=1K(tj−aj)2)∂E∂ak=∂∂ak(12∑j=1K(tj−aj)2)\frac{\partial E}{\partial a\_k} = \frac{\partial}{\partial a\_k} \left( \frac{1}{2} \sum\_{j=1}^K (t\_j - a\_j)^2 \right)

=&#x2212;&#x2211;j=1K(tj&#x2212;aj)&#x2202;aj&#x2202;ak" role="presentation" style="text-align: center; position: relative;">=−∑j=1K(tj−aj)∂aj∂ak=−∑j=1K(tj−aj)∂aj∂ak= -\sum\_{j=1}^K (t\_j - a\_j) \frac{\partial a\_j}{\partial a\_k}

=&#x2212;(tk&#x2212;ak)&#x2202;ak&#x2202;ak&#x2212;&#x2211;j&#x2260;k(tj&#x2212;aj)&#x2202;aj&#x2202;ak" role="presentation" style="text-align: center; position: relative;">=−(tk−ak)∂ak∂ak−∑j≠k(tj−aj)∂aj∂ak=−(tk−ak)∂ak∂ak−∑j≠k(tj−aj)∂aj∂ak= -(t\_k - a\_k) \frac{\partial a\_k}{\partial a\_k} - \sum\_{j \neq k} (t\_j - a\_j) \frac{\partial a\_j}{\partial a\_k}

=&#x2212;(tk&#x2212;ak)&#x22C5;1&#x2212;&#x2211;j&#x2260;k(tj&#x2212;aj)&#x22C5;0" role="presentation" style="text-align: center; position: relative;">=−(tk−ak)⋅1−∑j≠k(tj−aj)⋅0=−(tk−ak)⋅1−∑j≠k(tj−aj)⋅0= -(t\_k - a\_k) \cdot 1 - \sum\_{j \neq k} (t\_j - a\_j) \cdot 0

=ak&#x2212;tk" role="presentation" style="text-align: center; position: relative;">=ak−tk=ak−tk= a\_k - t\_k

So the partial derivative of E" role="presentation" style="position: relative;">EEE with respect to the k" role="presentation" style="position: relative;">kkkth element of a" role="presentation" style="position: relative;">aaa is simply ak&#x2212;tk" role="presentation" style="position: relative;">ak−tkak−tka\_k - t\_k.
물론이죠! 오류 함수 E의 방정식은 다음과 같습니다:E(t,a)=12|t&#x2212;a|2=12&#x2211;k=1K(tk&#x2212;ak)2" role="presentation" style="text-align: 가운데; 위치: relative;">E(t,a)=12|t−a|2=12∑k=1K(tk−ak)2E(t,a)=12|t−a|2=12∑k=1K(tk−ak)2E(t, a) = \frac{1}{2} |t - a|^2 = \frac{1}{2} \sum\_{k=1}^K (t\_k - a\_k)^2여기서 t" role="presentation" style="위치: 상대적;">ttt는 목표 벡터, a" role="presentation" style="위치: 상대적;">aaa는 MLP의 실제 출력, K" role="presentation" style="위치: 상대적;">KKK는 MLP의 출력 뉴런의 수입니다. t" role="presentation" style="위치: 상대적;">ttt와 a" role="presentation" style="위치: 상대적;">aaa 사이의 차이의 규범은 유클리드 규범을 나타내는 이중 수직 막대인 |&#x22C5;|" role="presentation" style="위치: 상대적;">|⋅||⋅|| \cdot |으로 표시됩니다.k" role="presentation" style="위치: 상대적;">에 대한 E" role="presentation" style="위치: 상대적;">의 부분 도함수를 찾으려면 a" role="presentation" style="위치: 상대적;">aaa에 대해 연쇄 규칙을 사용할 수 있습니다:&#x2202;E&#x2202;ak=&#x2202;&#x2202;ak(12&#x2211;j=1K(tj&#x2212;aj)2)" role="presentation" style="text-align: center; position: 상대;">∂E∂ak=∂∂ak(12∑j=1K(tj-aj)2)∂E∂ak=∂∂ak(12∑j=1K(tj-aj)2)\frac{\partial E}{\partial a\_k} = \frac{\partial}{\partial a\_k} \왼쪽( \frac{1}{2} \sum\_{j=1}^K (t\_j - a\_j)^2 \right)=&#x2212;&#x2211;j=1K(tj&#x2212;aj)&#x2202;aj&#x2202;ak" role="presentation" style="text-align: 가운데; 위치: 상대;">=-∑j=1K(tj-aj)∂aj∂ak=-∑j=1K(tj-aj)∂aj∂ak= -\sum\_{j=1}^K (t\_j - a\_j) \frac{\partial a\_j}{\partial a\_k}=&#x2212;(tk&#x2212;ak)&#x2202;ak&#x2202;ak&#x2212;&#x2211;j&#x2260;k(tj&#x2212;aj)&#x2202;aj&#x2202;ak" role="presentation" style="text-align: center; position: 상대;">=-(tk-ak)∂ak∂ak-∑j≠k(tj-aj)∂aj∂ak=-(tk-ak)∂ak∂ak-∑j≠k(tj-aj)∂aj∂ak= -(t\_k - a\_k) \frac{\partial a\_k}{\partial a\_k} - 합계\_{j \neq k} (t\_j - a\_j) \frac{\partial a\_j}{\partial a\_k}=&#x2212;(tk&#x2212;ak)&#x22C5;1&#x2212;&#x2211;j&#x2260;k(tj&#x2212;aj)&#x22C5;0" role="presentation" style="text-align: center; position: 상대;">=-(tk-ak)⋅1-∑j≠k(tj-aj)⋅0=-(tk-ak)⋅1-∑j≠k(tj-aj)⋅0= -(t\_k - a\_k) \cdot 1 - \sum\_{j \neq k} (t\_j - a\_j) \cdot 0=ak&#x2212;tk" role="presentation" style="text-align: 가운데; 위치: 상대적;">=ak-tk=ak-tk= a\_k - t\_k따라서 k" role="presentation" style="위치: 상대적;">에 대한 E" role="presentation" style="위치: 상대적;">의 부분 도함수는 a" role="presentation" style="위치: 상대적;">aaa에 대한 ak&#x2212;tk" role="presentation" style="위치: 상대적;">ak-tkak-tka\_k - t\_k가 됩니다.