求$ \min_x (x_1-2)^2+2(x_1^2-x_2)^2 $的梯度
时间: 2023-06-26 21:06:44 浏览: 47
我们可以先对每个变量求偏导数,然后将它们组合成梯度向量。对于每个变量 $x_i$,我们有:
$$\frac{\partial}{\partial x_i}((x_1-2)^2+2(x_1^2-x_2)^2)$$
对于 $i=1$,我们有:
$$(\frac{\partial}{\partial x_1}(x_1-2)^2)+(\frac{\partial}{\partial x_1}(2(x_1^2-x_2)^2))$$
$$=2(x_1-2)+8x_1(x_1^2-x_2)$$
对于 $i=2$,我们有:
$$-4(x_1^2-x_2)$$
因此,梯度向量为:
$$\nabla f(x)=\begin{bmatrix} 2(x_1-2)+8x_1(x_1^2-x_2) \\ -4(x_1^2-x_2) \end{bmatrix}$$
其中,$f(x)=(x_1-2)^2+2(x_1^2-x_2)^2$
相关问题
求$ \min_x (x_1-1)^2+2(x_1^2-x_2)^2 $的梯度
我们可以先对$x_1$和$x_2$分别求偏导数:
$$\frac{\partial}{\partial x_1}[(x_1-1)^2+2(x_1^2-x_2)^2]=2(x_1-1)+8x_1(x_1^2-x_2)$$
$$\frac{\partial}{\partial x_2}[(x_1-1)^2+2(x_1^2-x_2)^2]=-8(x_1^2-x_2)$$
因此,该函数的梯度向量为:$$\nabla f(x)=\begin{bmatrix}
\frac{\partial}{\partial x_1}[(x_1-1)^2+2(x_1^2-x_2)^2] \\
\frac{\partial}{\partial x_2}[(x_1-1)^2+2(x_1^2-x_2)^2]
\end{bmatrix}=\begin{bmatrix}
2(x_1-1)+8x_1(x_1^2-x_2) \\
-8(x_1^2-x_2)
\end{bmatrix}$$
求$ \min_x (x_1-1)^2+2(x_1^2-x_2)^2 $的Hesse矩阵
首先,计算一阶偏导数:
$$
\begin{aligned}
\frac{\partial}{\partial x_1}[(x_1 - 1)^2 + 2(x_1^2-x_2)^2] &= 2(x_1-1) + 8x_1(x_1^2 - x_2) \\
\frac{\partial}{\partial x_2}[(x_1 - 1)^2 + 2(x_1^2-x_2)^2] &= -8(x_1^2 - x_2)
\end{aligned}
$$
然后,计算二阶偏导数:
$$
\begin{aligned}
&\frac{\partial^2}{\partial x_1^2}[(x_1 - 1)^2 + 2(x_1^2-x_2)^2] = 2 + 8(3x_1^2 - x_2) \\
&\frac{\partial^2}{\partial x_1\partial x_2}[(x_1 - 1)^2 + 2(x_1^2-x_2)^2] = -16x_1 \\
&\frac{\partial^2}{\partial x_2\partial x_1}[(x_1 - 1)^2 + 2(x_1^2-x_2)^2] = -16x_1 \\
&\frac{\partial^2}{\partial x_2^2}[(x_1 - 1)^2 + 2(x_1^2-x_2)^2] = 32
\end{aligned}
$$
因此,该函数的 Hesse 矩阵为:
$$
H = \begin{bmatrix}
2+8(3x_1^2-x_2) & -16x_1 \\
-16x_1 & 32
\end{bmatrix}
$$