鲁鹏 北京理工大学，宇航学院 2019.05.09

最近经常会遇到常数和向量对向量求导的计算，感觉需要总结点什么了。以后，我还会在这个文档中添加新的公式。

首先明确本文规定求导使用分子布局约定（numerator layout convention）。在这中布局约定下，函数 f \boldsymbol{f} f (包括标量值函数)关于 x ∈ R n \boldsymbol{x}\in\mathbb{R}^{n} x∈Rn的微分可以用公式 d f ( x ) = ∂ f ∂ x d x \mathrm{d}\boldsymbol{f}(\boldsymbol{x}) = \dfrac{\partial \boldsymbol{f}}{\partial\boldsymbol{x}}\mathrm{d}\boldsymbol{x} df(x)=∂x∂f​dx 计算。向量求导有两种约定形式，分子布局约定（numerator layout convention）和分母布局约定（denominator layout convention）。

接下来给出在分子布局约定下，向量相关求导的定义。

已知 f ( x ) f(\boldsymbol{x}) f(x) 是关于列向量 x = [ x 1 x 2 . . . x n ] T \boldsymbol{x}= [x_{1}\quad x_{2}\quad ...\quad x_{n}]^{T} x=[x1​x2​...xn​]T 的标量函数，则 f ( x ) f(\boldsymbol{x}) f(x)是关于列向量 x \boldsymbol{x} x的导数为 ∂ f ( x ) ∂ x = [ ∂ f ∂ x 1 ∂ f ∂ x 2 . . . ∂ f ∂ x n ] \frac{\partial f(\boldsymbol{x})}{\partial \boldsymbol{x}} = \left[\begin{matrix} \frac{\partial f}{\partial x_{1}} & \frac{\partial f}{\partial x_{2}} &...& \frac{\partial f}{\partial x_{n}} \end{matrix}\right] ∂x∂f(x)​=[∂x1​∂f​​∂x2​∂f​​...​∂xn​∂f​​]

函数 f ( x ) f(\boldsymbol{x}) f(x)的梯度记为 ∇ f ( x ) \nabla f(\boldsymbol{x}) ∇f(x)，则梯度和导数的关系如下 ∇ f ( x ) = [ ∂ f ( x ) ∂ x ] T \nabla f(\boldsymbol{x}) = \left[\frac{\partial f(\boldsymbol{x})}{\partial \boldsymbol{x}}\right]^{T} ∇f(x)=[∂x∂f(x)​]T

函数 f ( x ) f(\boldsymbol{x}) f(x)的Hessian矩阵是 n × n n \times n n×n矩阵，记为 ∇ 2 f ( x ) \nabla^2 f(\boldsymbol{x}) ∇2f(x) ∇ 2 f ( x ) = [ ∂ 2 f ∂ x 1 2 ∂ 2 f ∂ x 1 ∂ x 2 . . . ∂ 2 f ∂ x 1 ∂ x n ∂ 2 f ∂ x 2 ∂ x 1 ∂ 2 f ∂ x 2 2 . . . ∂ 2 f ∂ x 2 ∂ x n . . . . . . . . . . . . ∂ 2 f ∂ x n ∂ x 1 ∂ 2 f ∂ x n ∂ x 2 . . . ∂ 2 f ∂ x n 2 ] \nabla^2 f(\boldsymbol{x}) = \left[\begin{matrix} \frac{\partial^{2} f}{\partial x_{1}^{2}} & \frac{\partial^{2} f}{\partial x_{1}\partial x_{2}} & ... & \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}^{2}} & ...& \frac{\partial^{2} f}{\partial x_{2}\partial x_{n}} \\ ... & ... & ... & ... \\ \frac{\partial^{2} f}{\partial x_{n}\partial x_{1}} & \frac{\partial^{2} f}{\partial x_{n}\partial x_{2}} & ... & \frac{\partial^{2} f}{\partial x_{n}^{2}} \end{matrix}\right] ∇2f(x)= ​∂x12​∂2f​∂x2​∂x1​∂2f​...∂xn​∂x1​∂2f​​∂x1​∂x2​∂2f​∂x22​∂2f​...∂xn​∂x2​∂2f​​............​∂x1​∂xn​∂2f​∂x2​∂xn​∂2f​...∂xn2​∂2f​​ ​

已知 F ( x ) = [ f 1 ( x ) f 2 ( x ) . . . f m ( x ) ] T F(\boldsymbol{x}) = [f_{1}(\boldsymbol{x})\quad f_{2}(\boldsymbol{x})\quad ...\quad f_{m}(\boldsymbol{x})]^{T} F(x)=[f1​(x)f2​(x)...fm​(x)]T 是关于列向量 x \boldsymbol{x} x 的向量值函数，则 F ( x ) F(\boldsymbol{x}) F(x)关于 x \boldsymbol{x} x的导数为 ∂ F ( x ) ∂ x = [ ∂ f 1 ∂ x 1 ∂ f 1 ∂ x 2 . . . ∂ f 1 ∂ x n ∂ f 2 ∂ x 1 ∂ f 2 ∂ x 2 . . . ∂ f 2 ∂ x n . . . . . . . . . . . . ∂ f m ∂ x 1 ∂ f m ∂ x 2 . . . ∂ f m ∂ x n ] \frac{\partial F(\boldsymbol{x})}{\partial \boldsymbol{x}} = \left[\begin{matrix} \frac{\partial f_{1}}{\partial x_{1}} & \frac{\partial f_{1}}{\partial x_{2}} &...& \frac{\partial f_{1}}{\partial x_{n}}\\ \frac{\partial f_{2}}{\partial x_{1}} & \frac{\partial f_{2}}{\partial x_{2}} &...& \frac{\partial f_{2}}{\partial x_{n}} \\ ... & ... & ... & ...\\ \frac{\partial f_{m}}{\partial x_{1}} & \frac{\partial f_{m}}{\partial x_{2}} &...& \frac{\partial f_{m}}{\partial x_{n}} \end{matrix}\right] ∂x∂F(x)​= ​∂x1​∂f1​​∂x1​∂f2​​...∂x1​∂fm​​​∂x2​∂f1​​∂x2​∂f2​​...∂x2​∂fm​​​............​∂xn​∂f1​​∂xn​∂f2​​...∂xn​∂fm​​​ ​

函数 F ( x ) F(\boldsymbol{x}) F(x)的雅克比矩阵记为 J J J，则 J = ∂ F ( x ) / ∂ x J = {\partial F(\boldsymbol{x})}/{\partial \boldsymbol{x}} J=∂F(x)/∂x，雅克比矩阵 J J J等于向量值函数 F ( x ) F(\boldsymbol{x}) F(x)关于 x \boldsymbol{x} x的导数。记函数 F ( x ) F(\boldsymbol{x}) F(x)的梯度记为 ∇ F ( x ) \nabla F(\boldsymbol{x}) ∇F(x)，梯度和导数的关系为 ∂ F ( x ) / ∂ x = ∇ F ( x ) T {\partial F(\boldsymbol{x})}/{\partial \boldsymbol{x}}=\nabla F(\boldsymbol{x})^{T} ∂F(x)/∂x=∇F(x)T。

已知向量 a = [ a 1 a 2 a 3 ] T \boldsymbol{a}= [a_{1}\quad a_{2}\quad a_{3}]^{T} a=[a1​a2​a3​]T，则 a \boldsymbol{a} a 的叉乘矩阵 a × \boldsymbol{a}^{\times} a× 定义如下 a × = [ 0 − a 3 a 2 a 3 0 − a 1 − a 2 a 1 0 ] \boldsymbol{a}^{\times} = \left[\begin{matrix} 0 & -a_{3} & a_{2}\\ a_{3} & 0 & -a_{1}\\ -a_{2} & a_{1} & 0 \end{matrix}\right] a×= ​0a3​−a2​​−a3​0a1​​a2​−a1​0​ ​

有了叉乘矩阵，向量叉乘可以像等式（1）那样表示 a × b = a × b (1) \boldsymbol{a} \times \boldsymbol{b} = \boldsymbol{a}^{\times} \boldsymbol{b} \tag{1} a×b=a×b(1)

在以上前提和定义的基础上，可以总结以下常用的求导公式，用于求函数的微分 d f ( x ) = ∂ f ∂ x d x \mathrm{d}\boldsymbol{f}(\boldsymbol{x}) = \dfrac{\partial \boldsymbol{f}}{\partial\boldsymbol{x}}\mathrm{d}\boldsymbol{x} df(x)=∂x∂f​dx ∂ ∥ a ∥ ∂ a = ∂ a ∂ a = a T a (2) \frac{\partial \lVert \boldsymbol{a}\rVert}{\partial \boldsymbol{a}} = \frac{\partial a}{\partial \boldsymbol{a}} = \frac{\boldsymbol{a}^{T}}{a} \tag{2} ∂a∂∥a∥​=∂a∂a​=aaT​(2)

读者可以将所有向量写成坐标的形式检验下 d a ( a ) \mathrm{d}a(\boldsymbol{a}) da(a)和 ∂ a ∂ a d a \dfrac{\partial a}{\partial\boldsymbol{a}}\mathrm{d}\boldsymbol{a} ∂a∂a​da是否相等，明显 d a ( a ) = a T a d a \mathrm{d}a(\boldsymbol{a}) = \dfrac{\boldsymbol{a}^{T}}{a}\mathrm{d}\boldsymbol{a} da(a)=aaT​da 是成立的 。

∂ a T a ∂ a = ∂ a 2 ∂ a = 2 a T (3) \frac{\partial \boldsymbol{a}^{T}\boldsymbol{a}}{\partial \boldsymbol{a}} = \frac{\partial a^{2}}{\partial \boldsymbol{a}} = 2\boldsymbol{a}^{T}\tag{3} ∂a∂aTa​=∂a∂a2​=2aT(3)

∂ ( A x ) ∂ x = A （雅可比矩阵） (4) \frac{\partial(A \boldsymbol{x})}{\partial \boldsymbol{x}} = A （雅可比矩阵）\tag{4} ∂x∂(Ax)​=A（雅可比矩阵）(4)

∂ x T A x ∂ x = x T ( A + A T ) (5) \frac{\partial \boldsymbol{x}^{T} A \boldsymbol{x}}{\partial \boldsymbol{x}}= \boldsymbol{x}^{T}(A + A^{T}) \tag{5} ∂x∂xTAx​=xT(A+AT)(5)

已知 y = a × b + c \boldsymbol{y} = \boldsymbol{a} \times \boldsymbol{b}+\boldsymbol{c} y=a×b+c，则 ∂ y ∂ b = y T y a × (6) \frac{\partial y}{\partial\boldsymbol{b}} = \frac{\boldsymbol{y}^{T}}{y}\boldsymbol{a}^{\times} \tag{6} ∂b∂y​=yyT​a×(6)

∂ α ∂ r = [ ∂ α ∂ V 1 ∂ α ∂ V 2 ∂ α ∂ V 3 ] [ ∂ V 1 ∂ r 1 ∂ V 1 ∂ r 2 ∂ V 1 ∂ r 3 ∂ V 2 ∂ r 1 ∂ V 2 ∂ r 2 ∂ V 2 ∂ r 3 ∂ V 3 ∂ r 1 ∂ V 3 ∂ r 2 ∂ V 3 ∂ r 3 ] = ∂ α ∂ V ∂ V ∂ r (7) \frac{\partial\alpha}{\partial\boldsymbol{r}} = \left[\begin{matrix} \frac{\partial\alpha}{\partial V_{1}} & \frac{\partial\alpha}{\partial V_{2}} & \frac{\partial\alpha}{\partial V_{3}} \end{matrix}\right] \left[\begin{matrix} \frac{\partial V_{1}}{\partial r_{1}} & \frac{\partial V_{1}}{\partial r_{2}} & \frac{\partial V_{1}}{\partial r_{3}}\\ \frac{\partial V_{2}}{\partial r_{1}} & \frac{\partial V_{2}}{\partial r_{2}} & \frac{\partial V_{2}}{\partial r_{3}}\\ \frac{\partial V_{3}}{\partial r_{1}} & \frac{\partial V_{3}}{\partial r_{2}} & \frac{\partial V_{3}}{\partial r_{3}}\\ \end{matrix}\right]= \frac{\partial\alpha}{\partial\boldsymbol{V}}\frac{\partial\boldsymbol{V}}{\partial\boldsymbol{r}} \tag{7} ∂r∂α​=[∂V1​∂α​​∂V2​∂α​​∂V3​∂α​​] ​∂r1​∂V1​​∂r1​∂V2​​∂r1​∂V3​​​∂r2​∂V1​​∂r2​∂V2​​∂r2​∂V3​​​∂r3​∂V1​​∂r3​∂V2​​∂r3​∂V3​​​ ​=∂V∂α​∂r∂V​(7)

d ( a × b ) d t = d a d t × b + a × d b d t (8) \frac{ d(\boldsymbol{a} \times \boldsymbol{b}) }{dt} = \frac{ d\boldsymbol{a} }{dt} \times \boldsymbol{b} + \boldsymbol{a} \times \frac{ d\boldsymbol{b} }{dt} \tag{8} dtd(a×b)​=dtda​×b+a×dtdb​(8)

∂ [ f ( x ) a ] ∂ x = a ∂ f ( x ) ∂ x (9) \frac{\partial[f(\boldsymbol{x})\boldsymbol{a}]}{\partial\boldsymbol{x}} = \boldsymbol{a} \frac{\partial f(\boldsymbol{x})}{\partial \boldsymbol{x}} \tag{9} ∂x∂[f(x)a]​=a∂x∂f(x)​(9)

已知 ω = [ ω 1 ω 2 ω 3 ] T \boldsymbol{\omega} = \begin{bmatrix} \omega_{1} & \omega_{2} & \omega_{3} \end{bmatrix}^{T} ω=[ω1​​ω2​​ω3​​]T， r = [ r 1 r 2 r 3 ] T \boldsymbol{r} = \begin{bmatrix} r_{1} & r_{2} & r_{3} \end{bmatrix}^{T} r=[r1​​r2​​r3​​]T， I 3 I_{3} I3​ 是 3 × 3 3\times3 3×3 单位矩阵 ∂ [ ( r T ω ) r ] ∂ r = ( r T ω ) I 3 + r ω T (10) \frac{\partial [(\boldsymbol{r}^{T} \boldsymbol{\omega}) \boldsymbol{r}]}{\partial \boldsymbol{r}} = (\boldsymbol{r}^{T} \boldsymbol{\omega})I_{3} + \boldsymbol{r}\boldsymbol{\omega}^{T} \tag{10} ∂r∂[(rTω)r]​=(rTω)I3​+rωT(10)

已知 ω = [ ω 1 ω 2 ω 3 ] T \boldsymbol{\omega} = \begin{bmatrix} \omega_{1} & \omega_{2} & \omega_{3} \end{bmatrix}^{T} ω=[ω1​​ω2​​ω3​​]T， r = [ r 1 r 2 r 3 ] T \boldsymbol{r} = \begin{bmatrix} r_{1} & r_{2} & r_{3} \end{bmatrix}^{T} r=[r1​​r2​​r3​​]T ω × ( ω × r ) = ω × ω × r = [ − ω 2 2 − ω 3 2 ω 1 ω 2 ω 1 ω 3 ω 1 ω 2 − ω 1 2 − ω 3 2 ω 2 ω 3 ω 1 ω 3 ω 2 ω 3 − ω 1 2 − ω 2 2 ] [ r 1 r 2 r 3 ] (11) \begin{aligned} \boldsymbol{\omega} \times (\boldsymbol{\omega}\times \boldsymbol{r}) &= \boldsymbol{\omega}^{\times}\boldsymbol{\omega}^{\times}\boldsymbol{r}\\ &=\begin{bmatrix} -\omega_{2}^{2} - \omega_{3}^{2} & \omega_{1}\omega_{2} & \omega_{1}\omega_{3}\\ \omega_{1}\omega_{2} & -\omega_{1}^{2} - \omega_{3}^{2} & \omega_{2}\omega_{3} \\ \omega_{1}\omega_{3} & \omega_{2}\omega_{3} & -\omega_{1}^{2} - \omega_{2}^{2} \end{bmatrix} \begin{bmatrix} r_{1} \\ r_{2} \\ r_{3} \end{bmatrix} \end{aligned} \tag{11} ω×(ω×r)​=ω×ω×r= ​−ω22​−ω32​ω1​ω2​ω1​ω3​​ω1​ω2​−ω12​−ω32​ω2​ω3​​ω1​ω3​ω2​ω3​−ω12​−ω22​​ ​ ​r1​r2​r3​​ ​​(11) 雅克比恒等式： a × ( b × c ) = b × ( a × c ) + c × ( b × a ) \mathbf{a}\times(\mathbf{b}\times\mathbf{c}) = \mathbf{b}\times(\mathbf{a}\times\mathbf{c}) + \mathbf{c}\times(\mathbf{b}\times\mathbf{a}) a×(b×c)=b×(a×c)+c×(b×a)

拉格朗日公式： a × ( b × c ) = ( a ⋅ c ) b − ( a ⋅ b ) c \mathbf{a}\times(\mathbf{b}\times\mathbf{c}) = (\mathbf{a}\cdot\mathbf{c}) \mathbf{b} - (\mathbf{a}\cdot\mathbf{b}) \mathbf{c} a×(b×c)=(a⋅c)b−(a⋅b)c

可以使用编程语言提供的符号运算功能来求导，例如：Python、Julia、C++和MATLAB等等。下面给出几个简单的例子。比较出名的一个符号运算库是symengine，它是使用C++开发的，很多编程语言都可以使用该符号运算库。

例如，现在要求向量函数 g = − r ∥ r ∥ 3 \boldsymbol{g} = -\dfrac{\boldsymbol{r}}{\Vert\boldsymbol{r}\Vert^3} g=−∥r∥3r​关于 r \boldsymbol{r} r的雅克比矩阵，下面分别用Python和MATLAB实现。

结果： [ 3 r x 2 ( r x 2 + r y 2 + r z 2 ) 5 2 − 1 ( r x 2 + r y 2 + r z 2 ) 3 2 3 r x r y ( r x 2 + r y 2 + r z 2 ) 5 2 3 r x r z ( r x 2 + r y 2 + r z 2 ) 5 2 3 r x r y ( r x 2 + r y 2 + r z 2 ) 5 2 3 r y 2 ( r x 2 + r y 2 + r z 2 ) 5 2 − 1 ( r x 2 + r y 2 + r z 2 ) 3 2 3 r y r z ( r x 2 + r y 2 + r z 2 ) 5 2 3 r x r z ( r x 2 + r y 2 + r z 2 ) 5 2 3 r y r z ( r x 2 + r y 2 + r z 2 ) 5 2 3 r z 2 ( r x 2 + r y 2 + r z 2 ) 5 2 − 1 ( r x 2 + r y 2 + r z 2 ) 3 2 ] \left[\begin{matrix}\frac{3 r_{x}^{2}}{\left(r_{x}^{2} + r_{y}^{2} + r_{z}^{2}\right)^{\frac{5}{2}}} - \frac{1}{\left(r_{x}^{2} + r_{y}^{2} + r_{z}^{2}\right)^{\frac{3}{2}}} & \frac{3 r_{x} r_{y}}{\left(r_{x}^{2} + r_{y}^{2} + r_{z}^{2}\right)^{\frac{5}{2}}} & \frac{3 r_{x} r_{z}}{\left(r_{x}^{2} + r_{y}^{2} + r_{z}^{2}\right)^{\frac{5}{2}}}\\\frac{3 r_{x} r_{y}}{\left(r_{x}^{2} + r_{y}^{2} + r_{z}^{2}\right)^{\frac{5}{2}}} & \frac{3 r_{y}^{2}}{\left(r_{x}^{2} + r_{y}^{2} + r_{z}^{2}\right)^{\frac{5}{2}}} - \frac{1}{\left(r_{x}^{2} + r_{y}^{2} + r_{z}^{2}\right)^{\frac{3}{2}}} & \frac{3 r_{y} r_{z}}{\left(r_{x}^{2} + r_{y}^{2} + r_{z}^{2}\right)^{\frac{5}{2}}}\\\frac{3 r_{x} r_{z}}{\left(r_{x}^{2} + r_{y}^{2} + r_{z}^{2}\right)^{\frac{5}{2}}} & \frac{3 r_{y} r_{z}}{\left(r_{x}^{2} + r_{y}^{2} + r_{z}^{2}\right)^{\frac{5}{2}}} & \frac{3 r_{z}^{2}}{\left(r_{x}^{2} + r_{y}^{2} + r_{z}^{2}\right)^{\frac{5}{2}}} - \frac{1}{\left(r_{x}^{2} + r_{y}^{2} + r_{z}^{2}\right)^{\frac{3}{2}}}\end{matrix}\right] ​(rx2​+ry2​+rz2​)25​3rx2​​−(rx2​+ry2​+rz2​)23​1​(rx2​+ry2​+rz2​)25​3rx​ry​​(rx2​+ry2​+rz2​)25​3rx​rz​​​(rx2​+ry2​+rz2​)25​3rx​ry​​(rx2​+ry2​+rz2​)25​3ry2​​−(rx2​+ry2​+rz2​)23​1​(rx2​+ry2​+rz2​)25​3ry​rz​​​(rx2​+ry2​+rz2​)25​3rx​rz​​(rx2​+ry2​+rz2​)25​3ry​rz​​(rx2​+ry2​+rz2​)25​3rz2​​−(rx2​+ry2​+rz2​)23​1​​ ​

结果： ( − − 2   r 1 2 + r 2 2 + r 3 2 ( r 1 2 + r 2 2 + r 3 2 ) 5 / 2 3   r 1   r 2 ( r 1 2 + r 2 2 + r 3 2 ) 5 / 2 3   r 1   r 3 ( r 1 2 + r 2 2 + r 3 2 ) 5 / 2 3   r 1   r 2 ( r 1 2 + r 2 2 + r 3 2 ) 5 / 2 − r 1 2 − 2   r 2 2 + r 3 2 ( r 1 2 + r 2 2 + r 3 2 ) 5 / 2 3   r 2   r 3 ( r 1 2 + r 2 2 + r 3 2 ) 5 / 2 3   r 1   r 3 ( r 1 2 + r 2 2 + r 3 2 ) 5 / 2 3   r 2   r 3 ( r 1 2 + r 2 2 + r 3 2 ) 5 / 2 − r 1 2 + r 2 2 − 2   r 3 2 ( r 1 2 + r 2 2 + r 3 2 ) 5 / 2 ) \left(\begin{array}{ccc} -\frac{-2\,{r_{1}}^2+{r_{2}}^2+{r_{3}}^2}{{\left({r_{1}}^2+{r_{2}}^2+{r_{3}}^2\right)}^{5/2}} & \frac{3\,r_{1}\,r_{2}}{{\left({r_{1}}^2+{r_{2}}^2+{r_{3}}^2\right)}^{5/2}} & \frac{3\,r_{1}\,r_{3}}{{\left({r_{1}}^2+{r_{2}}^2+{r_{3}}^2\right)}^{5/2}}\\ \frac{3\,r_{1}\,r_{2}}{{\left({r_{1}}^2+{r_{2}}^2+{r_{3}}^2\right)}^{5/2}} & -\frac{{r_{1}}^2-2\,{r_{2}}^2+{r_{3}}^2}{{\left({r_{1}}^2+{r_{2}}^2+{r_{3}}^2\right)}^{5/2}} & \frac{3\,r_{2}\,r_{3}}{{\left({r_{1}}^2+{r_{2}}^2+{r_{3}}^2\right)}^{5/2}}\\ \frac{3\,r_{1}\,r_{3}}{{\left({r_{1}}^2+{r_{2}}^2+{r_{3}}^2\right)}^{5/2}} & \frac{3\,r_{2}\,r_{3}}{{\left({r_{1}}^2+{r_{2}}^2+{r_{3}}^2\right)}^{5/2}} & -\frac{{r_{1}}^2+{r_{2}}^2-2\,{r_{3}}^2}{{\left({r_{1}}^2+{r_{2}}^2+{r_{3}}^2\right)}^{5/2}} \end{array}\right) ​−(r1​2+r2​2+r3​2)5/2−2r1​2+r2​2+r3​2​(r1​2+r2​2+r3​2)5/23r1​r2​​(r1​2+r2​2+r3​2)5/23r1​r3​​​(r1​2+r2​2+r3​2)5/23r1​r2​​−(r1​2+r2​2+r3​2)5/2r1​2−2r2​2+r3​2​(r1​2+r2​2+r3​2)5/23r2​r3​​​(r1​2+r2​2+r3​2)5/23r1​r3​​(r1​2+r2​2+r3​2)5/23r2​r3​​−(r1​2+r2​2+r3​2)5/2r1​2+r2​2−2r3​2​​ ​