深度学习中的反向传播数学计算过程

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假设：
X = (x1, x2, x3)
Y=2*X = (2*x1, 2*x2, 2*x3) = (y1, y2, y3)
实际上有：
y1=f(x1,x2,x3)=2*x1
y2=f(x1,x2,x3)=2*x2
y3=f(x1,x2,x3)=2*x3
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1 计算关于X关于的雅可比矩阵

J = ( ∂ y 1 ∂ x 1 ∂ y 1 ∂ x 2 ∂ y 1 ∂ x 3 ∂ y 2 ∂ x 1 ∂ y 2 ∂ x 2 ∂ y 2 ∂ x 3 ∂ y 3 ∂ x 1 ∂ y 3 ∂ x 2 ∂ y 3 ∂ x 3 ) \begin{equation*} J= \begin{pmatrix} \dfrac{\partial y_1}{\partial x_1}&\dfrac{\partial y_1}{\partial x_2}&\dfrac{\partial y_1}{\partial x_3} \\[2.5ex] \dfrac{\partial y_2}{\partial x_1}&\dfrac{\partial y_2}{\partial x_2}&\dfrac{\partial y_2}{\partial x_3} \\[2.5ex] \dfrac{\partial y_3}{\partial x_1}&\dfrac{\partial y_3}{\partial x_2}&\dfrac{\partial y_3}{\partial x_3} \\[1.5ex] \end{pmatrix} \end{equation*} J= ​∂x1​∂y1​​∂x1​∂y2​​∂x1​∂y3​​​∂x2​∂y1​​∂x2​∂y2​​∂x2​∂y3​​​∂x3​∂y1​​∂x3​∂y2​​∂x3​∂y3​​​ ​​

2 计算各分量的偏导和 / v投影各方向上的累加和

$$\begin{gathered} \frac{dY}{d{x}_1}=\frac{\partial y_1}{\partial x_1}+\frac{\partial y_1}{\partial x_1}+\frac{\partial y_1}{\partial x_1} \\ \frac{dY}{d{x}_2}=\frac{\partial y_1}{\partial x_2}+\frac{\partial y_2}{\partial x_2}+\frac{\partial y_3}{\partial x_2} \\ \frac{dY}{d{x}_3}=\frac{\partial y_1}{\partial x_3}+\frac{\partial y_2}{\partial x_3}+\frac{\partial y_3}{\partial x_3} \end{gathered}$$

3 确定最终分量的梯度计算表达式

$$\frac{dY}{dX}=\left(\begin{array}{ccc}\frac{dY}{d{x}_1}& \frac{dY}{d{x}_2}& \frac{dY}{d{x}_3}\end{array}\right)$$

4 y.backward(v) 根据函数中有无参数v进行计算

若是v=（m，n，q），则偏导数计算过程中，偏导数前应该乘上分量对应投影值
比如，若v=(1,2,3),则在表示在偏导计算过程中，对应分量x1，x2，x3应该乘上对应的投影值
$$以\frac{dY}{d{x}_1}=1\ast \frac{\partial y_1}{\partial x_1}+2\ast \frac{\partial y_1}{\partial x_1}+3\ast \frac{\partial y_1}{\partial x_1}为例$$
更广泛的：
$$\frac{d{Y}_j}{d{x}_i}=m\ast \frac{\partial y_j}{\partial x_i}+n\ast \frac{\partial y_j}{\partial x_i}+q\ast \frac{\partial y_j}{\partial x_i}$$

所以当v=(1,1,1)时，有无投影没区别

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