建设部住房城乡建设厅网站我国大宗商品交易所
web/
2025/9/26 19:13:40/
文章来源:
建设部住房城乡建设厅网站,我国大宗商品交易所,网站建设文案详情,内蒙古建设工程造价信息网官网入口课程地址和说明
线性代数实现p4 本系列文章是我学习李沐老师深度学习系列课程的学习笔记#xff0c;可能会对李沐老师上课没讲到的进行补充。 本节是第二篇
矩阵计算
矩阵的导数运算
此处参考了视频#xff1a;矩阵的导数运算 为了方便看出区别#xff0c;我将所有的向量…课程地址和说明
线性代数实现p4 本系列文章是我学习李沐老师深度学习系列课程的学习笔记可能会对李沐老师上课没讲到的进行补充。 本节是第二篇
矩阵计算
矩阵的导数运算
此处参考了视频矩阵的导数运算 为了方便看出区别我将所有的向量都不按印刷体加粗而是按手写体在向量对应字母上加箭头的方式展现。
标量方程对向量的导数
在一元函数中求一个函数的极值点一般令导数为0该点切线斜率为0求得驻点最后通过极值点定义或推论判断其是否为极值点也就是如下过程 求多元函数极值的方法如下 这个图中给的自变量记成了 y y y实际上记成 x x x更顺眼 假设这个多元函数有 m m m个变量即 f ( x 1 , x 2 , . . . , x m ) f(x_{1},x_{2},...,x_{m}) f(x1,x2,...,xm)那么求其极值的偏导数方程组中的方程就有 m m m个这样写起来有一些麻烦于是我们将用一种简洁的方式表达它我们将所有这 m m m个变量写成一个列向量的形式即 x → [ x 1 x 2 ⋮ x m ] m × 1 \overrightarrow x\begin{bmatrix} x_{1}\\ x_{2}\\ \vdots \\ x_{m} \end{bmatrix}_{m\times 1} x x1x2⋮xm m×1此时我们将多元函数 f ( x 1 , x 2 , . . . , x m ) f(x_{1},x_{2},...,x_{m}) f(x1,x2,...,xm)转化为一个自变量是一个向量的方程即 f ( x → ) f(\overrightarrow x) f(x ) 【注意】此处 x → \overrightarrow x x 是一个由多个自变量汇总而成的 m m m维列向量 m × 1 m\times 1 m×1而 f ( x → ) f(\overrightarrow x) f(x )是函数值是一个标量所以对其求偏导数就是标量对向量求导。 此时我们可以定义标量方程对向量的偏导数形式有两种为 (1)分母布局Denominator Layout ∂ f ( x → ) ∂ x → [ ∂ f ( x → ) ∂ x 1 ∂ f ( x → ) ∂ x 2 ⋮ ∂ f ( x → ) ∂ x m ] m × 1 \frac{\partial {f(\overrightarrow x)}}{\partial\overrightarrow x} \begin{bmatrix} \frac{\partial {f(\overrightarrow x)}}{\partial{x_{1}}}\\ \frac{\partial {f(\overrightarrow x)}}{\partial{x_{2}}}\\ \vdots \\ \frac{\partial {f(\overrightarrow x)}}{\partial{x_{m}}} \end{bmatrix}_{m\times 1} ∂x ∂f(x ) ∂x1∂f(x )∂x2∂f(x )⋮∂xm∂f(x ) m×1 其中 ∂ f ( x → ) ∂ x → \frac{\partial {f(\overrightarrow x)}}{\partial\overrightarrow x} ∂x ∂f(x )为 m × 1 m\times 1 m×1的列向量。 (2)分子布局Numerator Layout ∂ f ( x → ) ∂ x → [ ∂ f ( x → ) ∂ x 1 , ∂ f ( x → ) ∂ x 2 , … , ∂ f ( x → ) ∂ x m ] 1 × m \frac{\partial {f(\overrightarrow x)}}{\partial\overrightarrow x} \begin{bmatrix} \frac{\partial {f(\overrightarrow x)}}{\partial{x_{1}}}, \frac{\partial {f(\overrightarrow x)}}{\partial{x_{2}}}, \dots, \frac{\partial {f(\overrightarrow x)}}{\partial{x_{m}}} \end{bmatrix}_{1\times m} ∂x ∂f(x )[∂x1∂f(x ),∂x2∂f(x ),…,∂xm∂f(x )]1×m 其中 ∂ f ( x → ) ∂ x → \frac{\partial {f(\overrightarrow x)}}{\partial\overrightarrow x} ∂x ∂f(x )为 1 × m 1\times m 1×m的行向量。 不同的资料采用的布局不一样分子布局与分母布局互为转置虽然在李沐老师的课程中标量对向量的导数采用了分子布局但是为了方便推导一些结论我们采用分母布局注意分母布局和分子布局的结论互为转置。 【例】已知 f ( x 1 , x 2 ) x 1 2 x 2 2 f(x_{1},x_{2})x_{1}^{2}x_{2}^{2} f(x1,x2)x12x22其中 x → [ x 1 x 2 ] \overrightarrow x\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix} x [x1x2]求 ∂ f ( x → ) ∂ x → \frac{\partial {f(\overrightarrow x)}}{\partial\overrightarrow x} ∂x ∂f(x ) 【答】 ∂ f ( x → ) ∂ x → [ ∂ f ( x → ) ∂ x 1 ∂ f ( x → ) ∂ x 2 ] [ 2 x 1 2 x 2 ] \frac{\partial {f(\overrightarrow x)}}{\partial\overrightarrow x} \begin{bmatrix} \frac{\partial {f(\overrightarrow x)}}{\partial{x_{1}}}\\ \frac{\partial {f(\overrightarrow x)}}{\partial{x_{2}}} \end{bmatrix}\begin{bmatrix} 2x_{1}\\ 2x_{2} \end{bmatrix} ∂x ∂f(x )[∂x1∂f(x )∂x2∂f(x )][2x12x2]
向量方程对向量的导数
设有如下函数它本身就是一个向量然后它的自变量也是向量由多个自变量组成的向量即 f → ( x → ) [ f 1 ( x → ) f 2 ( x → ) ⋮ f n ( x → ) ] n × 1 , x → [ x 1 x 2 ⋮ x m ] \overrightarrow{f}(\overrightarrow x)\begin{bmatrix} f_{1}(\overrightarrow x)\\ f_{2}(\overrightarrow x)\\ \vdots \\f_{n}(\overrightarrow x) \end{bmatrix}_{n\times 1},\overrightarrow x\begin{bmatrix} x_{1}\\ x_{2} \\ \vdots \\ x_{m} \end{bmatrix} f (x ) f1(x )f2(x )⋮fn(x ) n×1,x x1x2⋮xm 其中 f → ( x → ) \overrightarrow{f}(\overrightarrow x) f (x )是一个 n × 1 n\times 1 n×1的列向量 x → \overrightarrow x x 是一个 m × 1 m\times 1 m×1的列向量。 此时我们将其偏导数形式定义为 (1)分母布局 ∂ f → ( x → ) n × 1 ∂ x → m × 1 [ ∂ f ( x → ) ∂ x 1 ∂ f ( x → ) ∂ x 2 ⋮ ∂ f ( x → ) ∂ x m ] [ ∂ f 1 ( x → ) ∂ x 1 ∂ f 2 ( x → ) ∂ x 1 … ∂ f n ( x → ) ∂ x 1 ∂ f 1 ( x → ) ∂ x 2 ∂ f 2 ( x → ) ∂ x 2 … ∂ f n ( x → ) ∂ x 2 ⋮ ⋮ ⋱ ⋮ ∂ f 1 ( x → ) ∂ x m ∂ f 2 ( x → ) ∂ x m … ∂ f n ( x → ) ∂ x m ] m × n \frac{\partial {\overrightarrow{f}(\overrightarrow x)}_{n\times 1}}{\partial\overrightarrow x_{m\times 1}} \begin{bmatrix} \frac{\partial {{f}(\overrightarrow x)}}{\partial {x_{1}}}\\ \frac{\partial {{f}(\overrightarrow x)}}{\partial {x_{2}}}\\ \vdots \\ \frac{\partial {{f}(\overrightarrow x)}}{\partial {x_{m}}} \end{bmatrix}\begin{bmatrix} \frac{\partial {{f_{1}}(\overrightarrow x)}}{\partial {x_{1}}} \frac{\partial {{f_{2}}(\overrightarrow x)}}{\partial {x_{1}}} \dots \frac{\partial {{f_{n}}(\overrightarrow x)}}{\partial {x_{1}}} \\ \frac{\partial {{f_{1}}(\overrightarrow x)}}{\partial {x_{2}}} \frac{\partial {{f_{2}}(\overrightarrow x)}}{\partial {x_{2}}} \dots \frac{\partial {{f_{n}}(\overrightarrow x)}}{\partial {x_{2}}} \\ \vdots \vdots \ddots \vdots \\ \frac{\partial {{f_{1}}(\overrightarrow x)}}{\partial {x_{m}}} \frac{\partial {{f_{2}}(\overrightarrow x)}}{\partial {x_{m}}} \dots \frac{\partial {{f_{n}}(\overrightarrow x)}}{\partial {x_{m}}} \end{bmatrix}_{m\times n} ∂x m×1∂f (x )n×1 ∂x1∂f(x )∂x2∂f(x )⋮∂xm∂f(x ) ∂x1∂f1(x )∂x2∂f1(x )⋮∂xm∂f1(x )∂x1∂f2(x )∂x2∂f2(x )⋮∂xm∂f2(x )……⋱…∂x1∂fn(x )∂x2∂fn(x )⋮∂xm∂fn(x ) m×n (2)分子布局 ∂ f → ( x → ) n × 1 ∂ x → m × 1 [ ∂ f 1 ( x → ) ∂ x → ∂ f 2 ( x → ) ∂ x → … ∂ f n ( x → ) ∂ x → ] [ ∂ f 1 ( x → ) ∂ x 1 ∂ f 1 ( x → ) ∂ x 2 … ∂ f 1 ( x → ) ∂ x m ∂ f 2 ( x → ) ∂ x 1 ∂ f 2 ( x → ) ∂ x 2 … ∂ f 2 ( x → ) ∂ x m ⋮ ⋮ ⋱ ⋮ ∂ f n ( x → ) ∂ x 1 ∂ f n ( x → ) ∂ x 2 … ∂ f n ( x → ) ∂ x m ] n × m \frac{\partial {\overrightarrow{f}(\overrightarrow x)}_{n\times 1}}{\partial\overrightarrow x_{m\times 1}} \begin{bmatrix} \frac{\partial {{f_{1}}(\overrightarrow x)}}{\partial {\overrightarrow x}}\\ \frac{\partial {{f_{2}}(\overrightarrow x)}}{\partial {\overrightarrow x}}\\ \dots \\ \frac{\partial {{f_{n}}(\overrightarrow x)}}{\partial {\overrightarrow x}} \end{bmatrix}\begin{bmatrix} \frac{\partial {{f_{1}}(\overrightarrow x)}}{\partial {x_{1}}} \frac{\partial {{f_{1}}(\overrightarrow x)}}{\partial {x_{2}}} \dots \frac{\partial {{f_{1}}(\overrightarrow x)}}{\partial {x_{m}}} \\ \frac{\partial {{f_{2}}(\overrightarrow x)}}{\partial {x_{1}}} \frac{\partial {{f_{2}}(\overrightarrow x)}}{\partial {x_{2}}} \dots \frac{\partial {{f_{2}}(\overrightarrow x)}}{\partial {x_{m}}} \\ \vdots \vdots \ddots \vdots \\ \frac{\partial {{f_{n}}(\overrightarrow x)}}{\partial {x_{1}}} \frac{\partial {{f_{n}}(\overrightarrow x)}}{\partial {x_{2}}} \dots \frac{\partial {{f_{n}}(\overrightarrow x)}}{\partial {x_{m}}} \end{bmatrix}_{n\times m} ∂x m×1∂f (x )n×1 ∂x ∂f1(x )∂x ∂f2(x )…∂x ∂fn(x ) ∂x1∂f1(x )∂x1∂f2(x )⋮∂x1∂fn(x )∂x2∂f1(x )∂x2∂f2(x )⋮∂x2∂fn(x )……⋱…∂xm∂f1(x )∂xm∂f2(x )⋮∂xm∂fn(x ) n×m 【例】已知 f → ( x → ) [ f 1 ( x → ) f 2 ( x → ) ] [ x 1 2 x 2 2 x 3 x 3 2 2 x 1 ] 2 × 1 \overrightarrow{f}(\overrightarrow x)\begin{bmatrix} f_{1}( \overrightarrow {x})\\ f_{2}( \overrightarrow {x}) \end{bmatrix}\begin{bmatrix} x_{1}^{2}x_{2}^{2}x_{3} \\ x_{3}^{2}2x_{1} \end{bmatrix}_{2\times 1} f (x )[f1(x )f2(x )][x12x22x3x322x1]2×1 x → [ x 1 x 2 x 3 ] \overrightarrow {x}\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} x x1x2x3 求 ∂ f → ( x → ) ∂ x → \frac{\partial {\overrightarrow{f}(\overrightarrow x)}}{\partial\overrightarrow x} ∂x ∂f (x ) 【答】按分母布局 ∂ f → ( x → ) ∂ x → [ ∂ f ( x → ) ∂ x 1 ∂ f ( x → ) ∂ x 2 ∂ f ( x → ) ∂ x 3 ] [ ∂ f 1 ( x → ) ∂ x 1 ∂ f 2 ( x → ) ∂ x 1 ∂ f 1 ( x → ) ∂ x 2 ∂ f 2 ( x → ) ∂ x 2 ∂ f 1 ( x → ) ∂ x 3 ∂ f 2 ( x → ) ∂ x 3 ] [ 2 x 1 2 2 x 2 0 1 2 x 3 ] \frac{\partial {\overrightarrow{f}(\overrightarrow x)}}{\partial\overrightarrow x}\begin{bmatrix} \frac{\partial {{f}(\overrightarrow x)}}{\partial {x_{1}}}\\ \frac{\partial {{f}(\overrightarrow x)}}{\partial {x_{2}}}\\ \frac{\partial {{f}(\overrightarrow x)}}{\partial {x_{3}}} \end{bmatrix}\begin{bmatrix} \frac{\partial {{f_{1}}(\overrightarrow x)}}{\partial {x_{1}}} \frac{\partial {{f_{2}}(\overrightarrow x)}}{\partial {x_{1}}} \\ \frac{\partial {{f_{1}}(\overrightarrow x)}}{\partial {x_{2}}} \frac{\partial {{f_{2}}(\overrightarrow x)}}{\partial {x_{2}}} \\ \frac{\partial {{f_{1}}(\overrightarrow x)}}{\partial {x_{3}}} \frac{\partial {{f_{2}}(\overrightarrow x)}}{\partial {x_{3}}} \end{bmatrix}\begin{bmatrix} 2x_{1} 2 \\ 2x_{2} 0\\ 1 2x_{3} \end{bmatrix} ∂x ∂f (x ) ∂x1∂f(x )∂x2∂f(x )∂x3∂f(x ) ∂x1∂f1(x )∂x2∂f1(x )∂x3∂f1(x )∂x1∂f2(x )∂x2∂f2(x )∂x3∂f2(x ) 2x12x21202x3 按分子布局 ∂ f → ( x → ) ∂ x → [ ∂ f 1 ( x → ) ∂ x → ∂ f 2 ( x → ) ∂ x → ] [ ∂ f 1 ( x → ) ∂ x 1 ∂ f 1 ( x → ) ∂ x 2 ∂ f 1 ( x → ) ∂ x 3 ∂ f 2 ( x → ) ∂ x 1 ∂ f 2 ( x → ) ∂ x 2 ∂ f 2 ( x → ) ∂ x 3 ] [ 2 x 1 2 x 2 1 2 0 2 x 3 ] \frac{\partial {\overrightarrow{f}(\overrightarrow x)}}{\partial\overrightarrow x} \begin{bmatrix} \frac{\partial {{f_{1}}(\overrightarrow x)}}{\partial {\overrightarrow x}}\\ \frac{\partial {{f_{2}}(\overrightarrow x)}}{\partial {\overrightarrow x}} \end{bmatrix}\begin{bmatrix} \frac{\partial {{f_{1}}(\overrightarrow x)}}{\partial {x_{1}}} \frac{\partial {{f_{1}}(\overrightarrow x)}}{\partial {x_{2}}} \frac{\partial {{f_{1}}(\overrightarrow x)}}{\partial {x_{3}}}\\ \frac{\partial {{f_{2}}(\overrightarrow x)}}{\partial {x_{1}}} \frac{\partial {{f_{2}}(\overrightarrow x)}}{\partial {x_{2}}}\frac{\partial {{f_{2}}(\overrightarrow x)}}{\partial {x_{3}}} \\ \end{bmatrix}\begin{bmatrix} 2x_{1} 2x_{2} 1\\ 2 0 2x_{3} \end{bmatrix} ∂x ∂f (x )[∂x ∂f1(x )∂x ∂f2(x )][∂x1∂f1(x )∂x1∂f2(x )∂x2∂f1(x )∂x2∂f2(x )∂x3∂f1(x )∂x3∂f2(x )][2x122x2012x3]
本文来自互联网用户投稿,该文观点仅代表作者本人,不代表本站立场。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如若转载,请注明出处:http://www.mzph.cn/web/82340.shtml
如若内容造成侵权/违法违规/事实不符,请联系多彩编程网进行投诉反馈email:809451989@qq.com,一经查实,立即删除!