深度学习基础 - 从泰勒级数到直线

深度学习基础 - 从泰勒级数到直线

flyfish

该图使用LibreOffice Draw 画

$$\mathrm{f}(\mathrm{x})\approx \mathrm{f}\left(x_0\right)+{\mathrm{f}}^{\prime }\left(x_0\right)\frac{x-x_0}{1!}+f^{\prime \prime }\left(x_0\right)\frac{{\left(x-x_0\right)}^2}{2!}+\cdots +f^{(n)}\left(x_0\right)\frac{{\left(x-x_0\right)}^n}{n!}$$
n如果是正无穷，可换成等号
$${\left(x-x_0\right)}^1\to 0{\left(x-x_0\right)}^2\to 0{\left(x-x_0\right)}^n\to 0$$
一阶展开
$$f(x)=f\left(x_0\right)+f^{\prime }\left(x_0\right)\left(x-x_0\right)$$

一阶展开的由来
$$\begin{array}{l}\Delta y=f(x)-f\left(x_0\right),\Delta x=x-x_0\\ \\ f(x)-f\left(x_0\right)=f^{\prime }\left(x_0\right)\ast \left(x-x_0\right)\\ \\ f(x)=f\left(x_0\right)+f^{\prime }\left(x_0\right)\left(x-x_0\right)\end{array}$$

怎么变成直线的
$$k_1+k_{2}x-k_2{x}_0=k_{2}x+b$$
因为
$f\left(x_0\right)=k_1$,
$f^{\prime }\left(x_0\right)=k_2$,
$k_{2}x+b$就是直线

线性近似

根据直线的点斜式方程得
$y=f(a)+f^{\prime }(a)(x-a).$

$L(x)=f(a)+f^{\prime }(a)(x-a).$

$\sqrt{x}$的线性近似

找到$f(x)=\sqrt{x}$在$x=9$的线性近似，并估算$\sqrt{9.1}$

$L(x)=f(9)+f^{\prime }(9)(x-9).$

$f(x)=\sqrt{x}\Rightarrow f(9)=\sqrt{9}=3$

$f^{\prime }(x)=\frac{1}{2\sqrt{x}}\Rightarrow f^{\prime }(9)=\frac{1}{2\sqrt{9}}=\frac{1}{6}$

$L(x)=3+\frac{1}{6}(x-9)$

$\sqrt{9.1}=f(9.1)\approx L(9.1)=3+\frac{1}{6}(9.1-9)\approx 3.0167$