已知记忆单元的计算公式：
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s_{c_j}(t) = s_{c_j}(t-1) + g(net_{c_j}(t)) y^{in_j}(t)
scj(t)=scj(t−1)+g(netcj(t))yinj(t) 我们使用截断求导规则来计算误差在时间步
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(30)
\begin{aligned} \frac{\partial s_{c_j}(t-k)}{\partial s_{c_j}(t-k-1)} &= 1 + \frac{\partial g(net_{c_j}(t-k))y^{in_j}(t-k)}{\partial s_{c_j}(t-k-1)}\\ &=1+ \frac{\partial y^{in_j}(t-k)}{\partial s_{c_j}(t-k-1)}g(net_{c_j}(t-k)) + \frac{\partial g(net_{c_j}(t-k))}{\partial s_{c_j}(t-k-1)}y^{in_j}(t-k)\\ &=1 + \sum_u[\frac{\partial y^{in_j}(t-k)}{\partial y^u(t-k-1)}\frac{\partial y^u(t-k-1)}{\partial s_{c_j}(t-k-1)}]g(net_{c_j}(t-k)) \\ &\quad + y^{in_j}(t-k)g'(net_{c_j}(t-k))\sum_u [\frac{\partial net_{c_j}(t-k)}{\partial y^u(t-k-1)}\frac{\partial y^u(t-k-1)}{\partial s_{c_j}(t-k-1)}]\\ &\approx_{tr} 1.\tag{30} \end{aligned}
∂scj(t−k−1)∂scj(t−k)=1+∂scj(t−k−1)∂g(netcj(t−k))yinj(t−k)=1+∂scj(t−k−1)∂yinj(t−k)g(netcj(t−k))+∂scj(t−k−1)∂g(netcj(t−k))yinj(t−k)=1+u∑[∂yu(t−k−1)∂yinj(t−k)∂scj(t−k−1)∂yu(t−k−1)]g(netcj(t−k))<
