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B站：DR_CAN
Dr. CAN学习笔记-数学基础Ch0-3线性化Linearization
- 1. 线性系统 Linear System 与 叠加原理 Superposition
- 2. 线性化：Taylor Series
- 3. Summary
1. 线性系统 Linear System 与 叠加原理 Superposition
x ˙ = f ( x ) \dot{x}=f\left( x \right) x˙=f(x)
- x 1 , x 2 x_1,x_2 x1,x2 是解
- x 3 = k 1 x 1 + k 2 x 2 , k 1 , k 2 ∈ R x_3=k_1x_1+k_2x_2,k_1,k_2\in \mathbb{R} x3=k1x1+k2x2,k1,k2∈R
- x 3 x_3 x3 是解
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\ddot{x}+2\dot{x}+\sqrt{2}x=0 √ \\ \ddot{x}+2\dot{x}+\sqrt{2}x^2=0 × \\ \ddot{x}+\sin \dot{x}+\sqrt{2}x=0 ×
x¨+2x˙+2x=0√x¨+2x˙+2x2=0×x¨+sinx˙+2x=0×
2. 线性化：Taylor Series
f ( x ) = f ( x 0 ) + f ′ ( x 0 ) 1 ! ( x − x 0 ) + f ′ ′ ( x 0 ) 2 ! ( x − x 0 ) 2 + ⋯ + f n ( x 0 ) n ! ( x − x 0 ) n f\left( x \right) =f\left( x_0 \right) +\frac{f^{\prime}\left( x_0 \right)}{1!}\left( x-x_0 \right) +\frac{{f^{\prime}}^{\prime}\left( x_0 \right)}{2!}\left( x-x_0 \right) ^2+\cdots +\frac{f^n\left( x_0 \right)}{n!}\left( x-x_0 \right) ^n f(x)=f(x0)+1!f′(x0)(x−x0)+2!f′′(x0)(x−x0)2+⋯+n!fn(x0)(x−x0)n
若 x − x 0 → 0 , ( x − x 0 ) n → 0 x-x_0\rightarrow 0,\left( x-x_0 \right) ^n\rightarrow 0 x−x0→0,(x−x0)n→0，则有： ⇒ f ( x ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) ⇒ f ( x ) = k 1 + k 2 x − k 3 x 0 ⇒ f ( x ) = k 2 x + b \Rightarrow f\left( x \right) =f\left( x_0 \right) +f^{\prime}\left( x_0 \right) \left( x-x_0 \right) \Rightarrow f\left( x \right) =k_1+k_2x-k_3x_0\Rightarrow f\left( x \right) =k_2x+b ⇒f(x)=f(x0)+f′(x0)(x−x0)⇒f(x)=k1+k2x−k3x0⇒f(x)=k2x+b
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3. Summary
- f ( x ) = f ( x 0 ) + f ′ ( x 0 ) 1 ! ( x − x 0 ) , x − x 0 → 0 f\left( x \right) =f\left( x_0 \right) +\frac{f^{\prime}\left( x_0 \right)}{1!}\left( x-x_0 \right) ,x-x_0\rightarrow 0 f(x)=f(x0)+1!f′(x0)(x−x0),x−x0→0
- [ x ˙ 1 d x ˙ 2 d ] = [ ∂ f 1 ∂ x 1 ∂ f 1 ∂ x 2 ∂ f 2 ∂ x 1 ∂ f 2 ∂ x 2 ] ∣ x = x 0 [ x 1 d x 2 d ] \left[ \begin{array}{c} \dot{x}_{1\mathrm{d}}\\ \dot{x}_{2\mathrm{d}}\\ \end{array} \right] =\left. \left[ \begin{matrix} \frac{\partial f_1}{\partial x_1}& \frac{\partial f_1}{\partial x_2}\\ \frac{\partial f_2}{\partial x_1}& \frac{\partial f_2}{\partial x_2}\\ \end{matrix} \right] \right|_{\mathrm{x}=\mathrm{x}_0}\left[ \begin{array}{c} x_{1\mathrm{d}}\\ x_{2\mathrm{d}}\\ \end{array} \right] [x˙1dx˙2d]=[∂x1∂f1∂x1∂f2∂x2∂f1∂x2∂f2] x=x0[x1dx2d]
