多元复合函数二阶导数与向量微积分的思考
引入
对于形似\(z=f(u_1,u_2,...,u_n),\)其中\(u_i=g_i(x_i)\)的多元复合函数，对其二阶导数的考察经常会通过繁琐而重复的运算，且容易在连续运用链式法则时犯错。本文将提出该类题型的通解以及理论推导过程供参考。函数
例1：设\(z=f(x^2-y^2,e^{xy})\),其中\(f\)具备二阶连续偏导数，求 \(\frac{\partial ^2z}{ \partial x \partial y}\).spa
经过链式法则，咱们能够获得结果\(\frac{\partial ^2z}{ \partial x \partial y}=-4xyf^{''}_{11}+2(x^2-y^2)e^{xy}f^{''}_{12}+xye^{2xy}f{''}_{22}+e^{xy}(1+xy)f^{'}_2\)数学
对于式子中的\(f^{''}_{11}、f^{''}_{12}\)的出现，咱们能够联想到矩阵的下标，由此引起咱们对该式子简化形式甚至该类题型通解的思考。class
梯度矩阵
咱们定义[1]，对于一个函数\(f: ℝ^n\rightarrow ℝ ,\pmb{x} \rightarrow f(\pmb{x}),\pmb{x}\in ℝ^n\),即，\(\pmb{x}=[x_1,x_2,x_3,...,x_n]^T\),偏导数为：im
\[\frac{\partial f}{\partial x_1}= \lim_{h \rightarrow 0} f\frac{(x_1+h,x_2,...,x_n)-f(\pmb{x})}{h}\\.\\.\\.\\\frac{\partial f}{\partial x_n}= \lim_{h \rightarrow 0}\frac{f(x_1,x_2,...,x_n+h)-f(\pmb{x})}{h} \tag{2.1} \]
咱们写做行向量的形式，记做：总结
\[∇_{\pmb{x}}f=grad\ f=\left[\begin{matrix}\frac{\partial f(\pmb{x})}{\partial x_1} & \frac{\partial f(\pmb{x})}{\partial x_1} & ... & \frac{\partial f(\pmb{x})}{\partial x_n}\\\end{matrix} \right] \inℝ^n \tag{2.2} \]
例如，对于函数\(f(x,y)=(x+2y^3)^2\),咱们有：di
\[∇f=\left[\begin{matrix}2(x+2y^3) & 12(x+2y^3)y^2\end{matrix} \right] \inℝ^{1×2} \tag{2.3} \]
为了探求文章开始所提出问题通解形式的探讨，绕不开的一个重要步骤是对梯度矩阵\(∇f\)进行求导，咱们将在推导的过程当中单独进行分析。思考
多元复合函数的二阶导数与黑塞矩阵
设\(z=f(u_1,u_2,...,u_n),\)其中\(u_i=g_i(x_i)\)，求\(\frac{\partial ^2z}{ \partial x_i \partial x_j}\).display
\[\frac{\partial z}{ \partial x_i}=\frac{\partial z}{ \partial \pmb{u} }·\frac{\partial \pmb{u}}{ \partial x_i} =\left[\begin{matrix}\frac{\partial f}{\partial u_1} & \frac{\partial f}{\partial u_2} & ... & \frac{\partial f}{\partial u_n}\end{matrix} \right] \left[\begin{matrix}\frac{\partial u_1}{\partial x_i} \\ \frac{\partial u_2}{\partial x_i} \\ ... \\ \frac{\partial u_n}{\partial x_i}\end{matrix} \right] \tag{3.1} \]
为了简化形式，咱们令：math
\[\pmb{X_i}=\left[\begin{matrix}\frac{\partial u_1}{\partial x_i} & \frac{\partial u_2}{\partial x_i} & ... & \frac{\partial u_n}{\partial x_i}\end{matrix} \right]^T \tag{3.2} \]
那么:
\[\frac{\partial z}{ \partial x_i}=∇_{\pmb{u}}f·\pmb{X_i} \tag{3.3} \]
接下来，咱们须要求解
\[\frac{\partial {}}{ \partial x_i}(∇_{\pmb{u}}f·\pmb{X_i} \tag{3.4}) \]
\[\frac{\partial {}}{ \partial x_j}(∇_{\pmb{u}}f·\pmb{X_i} )=\frac{\partial {}}{ \partial x_j}∇_{\pmb{u}}f·\pmb{X_i} + ∇_{\pmb{u}}f·\frac{\partial {}}{ \partial x_j}\pmb{X_i} \tag{3.5} \]
\(\frac{\partial {}}{ \partial x_j}\pmb{X_i}\)的答案容易获得的，咱们着重于讨论\(\frac{\partial {}}{ \partial x_j}∇_{\pmb{u}}f·\pmb{X_i}\),尤为是\(\frac{\partial {}}{ \partial x_j}∇_{\pmb{u}}f\)的结果。
经分析：
\[\frac{\partial {}}{ \partial x_j}∇_{\pmb{u}}f=\frac{\partial {}}{ \partial \pmb{u}^T}·\frac{\partial {\pmb{u}^T}}{ \partial x_j}·∇_{\pmb{u}}f=\frac{\partial {\pmb{u}^T}}{ \partial x_j}·\frac{\partial {}}{ \partial \pmb{u}^T}·∇_{\pmb{u}}f \tag{3.6} \]
问题被简化转化为解决向量\((∇_{\pmb{u}}f)\)对向量\((\pmb{u}^T)\)求导的问题。
咱们对这个运算进行进一步分析，这个运算的实质是梯度矩阵中的元素逐个对\(u_i\)分别求导，结果显然是一个\(2×2\)的方阵，而这个矩阵在数学上被定义为 黑塞矩阵（Hessian Matrix），记做\(H(f)\),它的具体形式是：
\[A= \left[\begin{matrix} \frac{\partial^2 f}{\partial x_1\partial x_1} & \frac{\partial^2 f}{\partial x_1\partial x_2} & \cdots & \frac{\partial^2 f}{\partial x_1\partial x_n}\\ \frac{\partial^2 f}{\partial x_2\partial x_1} & \frac{\partial^2 f}{\partial x_2\partial x_2} & \cdots & \frac{\partial^2 f}{\partial x_2\partial x_n} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial^2 f}{\partial x_n\partial x_1} & \frac{\partial^2 f}{\partial x_n\partial x_2} & \cdots & \frac{\partial^2 f}{\partial x_1\partial x_n} \end{matrix} \right]\tag{3.7} \]
其规律是显而易见的。
因而，引入\(H(f)\)后，咱们能够继续化简：
\[\frac{\partial {\pmb{u}^T}}{ \partial x_j}·\frac{\partial {}}{ \partial \pmb{u}^T}·∇_{\pmb{u}}f=\frac{\partial {\pmb{u}^T}}{ \partial x_j}·\left[\begin{matrix} \frac{\partial^2 f}{\partial u_1\partial u_1} & \frac{\partial^2 f}{\partial u_1\partial u_2} & \cdots & \frac{\partial^2 f}{\partial x_1\partial u_n}\\ \frac{\partial^2 f}{\partial x_2\partial x_1} & \frac{\partial^2 f}{\partial u_2\partial u_2} & \cdots & \frac{\partial^2 f}{\partial u_2\partial u_2} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial^2 f}{\partial u_n\partial u_1} & \frac{\partial^2 f}{\partial u_n\partial u_2} & \cdots & \frac{\partial^2 f}{\partial u_1\partial u_n} \end{matrix} \right]=\pmb{X_j}^T·H_{\pmb{u}}(f)\tag{3.7} \]
因此
\[\frac{\partial ^2z}{ \partial x_i \partial x_j}=\pmb{X_j}^T·H_{\pmb{u}}(f)·\pmb{X_i}+∇_{\pmb{u}}f·\frac{\partial {}}{ \partial x_j}\pmb{X_i}=\pmb{X_j}^T·H_{\pmb{u}}(f)·\pmb{X_i}+∇_{\pmb{u}}f·\pmb{X_{ij}}\tag{3.8} \]
其中
\[\pmb{X_{ij}}=\left[\begin{matrix}\frac{\partial^2 u_1}{\partial x_i\partial x_j} & \frac{\partial^2 u_2}{\partial x_i\partial x_j} & ... & \frac{\partial^2 u_n}{\partial x_i\partial x_j}\end{matrix} \right]^T\tag{3.9} \]
固然在实际计算过程当中，因为\(\pmb{X_i}\)的值已经被计算，因此直接计算\(\frac{\partial {}}{ \partial x_j}\pmb{X_i}\)或许更为便捷。
总结
设\(z=f(u_1,u_2,...,u_n),\)其中\(u_i=g_i(x_i)\)，求\(\frac{\partial ^2z}{ \partial x_i \partial x_j}\).
\[\frac{\partial z}{ \partial x_i}=∇_{\pmb{u}}f·\pmb{X_i} \\\frac{\partial ^2z}{ \partial x_i \partial x_j}=\pmb{X_j}^T·H_{\pmb{u}}(f)·\pmb{X_i}+∇_{\pmb{u}}f·\pmb{X_{ij}}\tag{end} \]
参考
- [1] 《MATHEMATICS FOR MACHINE LEARNING》（Marc Peter Deisenroth,A. Aldo Faisal ,Cheng Soon Ong）