SLAM 14 notes

SLAM 14 notes"/>

SLAM 14 notes

4.23 推导

f ( x ) f(x) f(x)在点a处的泰勒展开
f ( x ) = ∑ n = 0 ∞ f ( n ) a n ! ( x − a ) n f(x) = \sum_{n=0}^\infty \frac{f^{(n)}a}{n!}(x-a)^n f(x)=∑n=0∞​n!f(n)a​(x−a)n

l n x lnx lnx的n阶导数
l n ( n ) x = ( − 1 ) n − 1 ( n − 1 ) ! x n ln^{(n)}x = \frac{(-1)^{n-1}(n-1)!}{x^n} ln(n)x=xn(−1)n−1(n−1)!​

l n ( R ) ln(R) ln(R)在 I I I处展开
l n ( R ) = ∑ n = 0 ∞ l n ( n ) I ( R − I ) n n ! = ∑ n = 1 ∞ l n ( n ) I ( R − I ) n n ! = ∑ n = 1 ∞ ( − 1 ) n − 1 ( R − I ) n n = ∑ n = 0 ∞ ( − 1 ) n ( R − I ) n + 1 n + 1 \begin{aligned} ln(R) & =\sum_{n=0}^\infty\frac{ln^{(n)}I(R-I)^n}{n!} \\ & = \sum_{n=1}^\infty\frac{ln^{(n)}I(R-I)^n}{n!} \\ & = \sum_{n=1}^\infty\frac{(-1)^{n-1}(R-I)^n}{n}\\ & = \sum_{n=0}^\infty\frac{(-1)^{n}(R-I)^{n+1}}{n+1} \end{aligned} ln(R)​=n=0∑∞​n!ln(n)I(R−I)n​=n=1∑∞​n!ln(n)I(R−I)n​=n=1∑∞​n(−1)n−1(R−I)n​=n=0∑∞​n+1(−1)n(R−I)n+1​​

4.24 推导

ξ ∧ = [ ϕ ∧ ρ 0 T 0 ] \begin{aligned} \xi ^ \wedge = \begin{bmatrix} \phi ^ \wedge & \rho\\ 0^T & 0 \end{bmatrix} \end{aligned} ξ∧=[ϕ∧0T​ρ0​]​

e x p ( ξ ∧ ) = ∑ n = 0 ∞ 1 n ! [ ϕ ∧ ρ 0 T 0 ] n = I + ∑ n = 1 ∞ 1 n ! [ ϕ ∧ ρ 0 T 0 ] n = [ I 0 0 T 1 ] + ∑ n = 1 ∞ 1 n ! [ ( ϕ ∧ ) n ( ϕ ∧ ) n − 1 ρ 0 T 0 ] = [ I + ∑ n = 1 ∞ 1 n ! ( ϕ ∧ ) n ∑ n = 1 ∞ 1 n ! ( ϕ ∧ ) n − 1 ρ 0 T 1 ] = [ ∑ n = 0 ∞ 1 n ! ( ϕ ∧ ) n ∑ n = 0 ∞ 1 ( n + 1 ) ! ( ϕ ∧ ) n ρ 0 T 1 ] \begin{aligned} exp(\xi ^ \wedge) &= \sum_{n=0}^{\infty}\frac{1}{n!}\begin{bmatrix} \phi ^ \wedge & \rho\\ 0^T & 0 \end{bmatrix}^n=I + \sum_{n=1}^{\infty}\frac{1}{n!}\begin{bmatrix} \phi ^ \wedge & \rho\\ 0^T & 0 \end{bmatrix}^n = \begin{bmatrix}I & 0\\ 0^T & 1 \end{bmatrix} + \sum_{n=1}^{\infty}\frac{1}{n!}\begin{bmatrix} (\phi ^ \wedge)^n & (\phi ^ \wedge)^{n-1}\rho\\ 0^T & 0 \end{bmatrix} \\ &=\begin{bmatrix}I + \sum_{n=1}^{\infty}\frac{1}{n!}(\phi ^ \wedge)^n & \sum_{n=1}^{\infty}\frac{1}{n!}(\phi ^ \wedge)^{n-1}\rho\\ 0^T & 1 \end{bmatrix} \\ &=\begin{bmatrix}\sum_{n=0}^{\infty}\frac{1}{n!}(\phi ^ \wedge)^n & \sum_{n=0}^{\infty}\frac{1}{(n+1)!}(\phi ^ \wedge)^{n}\rho\\ 0^T & 1 \end{bmatrix} \end{aligned} exp(ξ∧)​=n=0∑∞​n!1​[ϕ∧0T​ρ0​]n=I+n=1∑∞​n!1​[ϕ∧0T​ρ0​]n=[I0T​01​]+n=1∑∞​n!1​[(ϕ∧)n0T​(ϕ∧)n−1ρ0​]=[I+∑n=1∞​n!1​(ϕ∧)n0T​∑n=1∞​n!1​(ϕ∧)n−1ρ1​]=[∑n=0∞​n!1​(ϕ∧)n0T​∑n=0∞​(n+1)!1​(ϕ∧)nρ1​]​