1. 第三讲(上)_基于优化的IMU与视觉信息融合(上)
第三讲(上): (1)最小二乘问题的求解推导 (2)其中有LM算法的相关推导以及鲁棒核函数的推导
1.1. 最小二乘与非线性优化
1.1.1. 最小二乘
1.1.2. 非线性最小二乘
1.1.2.1. LM算法增量\(\Delta X\)表达式推导
求解如下:
\[ \begin{aligned} &(\boldsymbol{J}^T\boldsymbol{J}+\mu\boldsymbol{I})\boldsymbol{\Delta X}=-\boldsymbol{J}^T\boldsymbol{f} \\ &\Longrightarrow (\boldsymbol{V \Lambda V^T}+\mu \boldsymbol{I})\boldsymbol{\Delta X}=-\boldsymbol{J}^T\boldsymbol{f} \\ &\Longrightarrow(\boldsymbol{V} (\boldsymbol{\Lambda}+\mu \boldsymbol{I})\boldsymbol{V}^T) \boldsymbol{\Delta X}=-\boldsymbol{J}^T\boldsymbol{f} \\ &\Longrightarrow(\boldsymbol{\Lambda}+\mu \boldsymbol{I})\boldsymbol{V}^T\boldsymbol{\Delta X}=-\boldsymbol{V}^T\boldsymbol{J}^T\boldsymbol{f} \\ &\Longrightarrow(\boldsymbol{\Lambda}+\mu \boldsymbol{I})\boldsymbol{V}^T\boldsymbol{\Delta X}=-\boldsymbol{V}^T F^{'}(x)^T \\ &\Longrightarrow\boldsymbol{V}^T\boldsymbol{\Delta X}=-(\boldsymbol{\Lambda}+\mu \boldsymbol{I})^{-1}\boldsymbol{V}^T F^{'}(x)^T \\ \end{aligned} \]
\[ \begin{aligned} \boldsymbol{V}^T\boldsymbol{\Delta X}=- \begin{bmatrix} \frac{1}{\lambda_1+\mu}&0&0& \cdots \\ 0 & \frac{1}{\lambda_2+\mu} & 0 &\cdots \\ 0 & \cdots & & \frac{1}{\lambda_j+\mu} \end{bmatrix} \boldsymbol{V}^T F^{'}(x)^T \end{aligned} \]
\[ \begin{aligned} \boldsymbol{\Delta X}=- \begin{bmatrix} \boldsymbol{v_1} & \cdots &\boldsymbol{v_j} \end{bmatrix} \begin{bmatrix} \frac{1}{\lambda_1+\mu}&0&0& \cdots \\ 0 & \frac{1}{\lambda_2+\mu} & 0 &\cdots \\ 0 & \cdots & & \frac{1}{\lambda_j+\mu} \end{bmatrix} \begin{bmatrix} \boldsymbol{v_1} \\ \vdots \\ \boldsymbol{v_j} \end{bmatrix} F^{'}(x)^T \end{aligned} \]
\[ \begin{aligned} \therefore \boldsymbol{\Delta X}= -\sum_{j=1}^{n}\frac{\boldsymbol{v}_j^T {\boldsymbol{F'}}^T}{\lambda_j+\mu} \boldsymbol{v}_j \end{aligned} \]
从上图可以看到,由于\(\mu\)的值在抖动,使得\(\Delta X\)的步长在来回震荡变化,使得loss值也随着抖动,有一部分的计算是多余的.