Loading... 先设一个函数$f(x)$ $$ f\left( x \right) =\cos \left( x \right) $$ 再设一个多项式函数$g(x)$去`模拟`$f(x)$ $$ \begin{cases} g\left( x \right) =a_0\\ g\left( x \right) =a_0+a_1x\\ g\left( x \right) =a_0+a_1x+a_2x^2\\ g\left( x \right) =a_0+a_1x+a_2x^2+a_3x^3\\ \cdots\\ g\left( x \right) =a_0+a_1x+a_2x^2+a_3x^3+\cdots +a_nx^n\\ \end{cases}\left. \begin{array}{r} g\left( 0 \right) =a_0=f\left( 0 \right)\\ g'\left( 0 \right) =a_1=f'\left( 0 \right)\\ g''\left( 0 \right) =\text{2!}a_2=f''\left( 0 \right)\\ g'''\left( 0 \right) =\text{3!}a_3=f'''\left( 0 \right)\\ \cdots\\ g^{\left( n \right)}\left( 0 \right) =n!a_n=f^{\left( n \right)}\left( 0 \right)\\ \end{array} \right\} $$ 从上可知，为使$g(x)$在$x=0$时尽可能的接近$f(x)$,需要在$x=0$的各阶导数都相等。所以有 $$ f(x) =f\left( 0 \right) +f'\left( 0 \right) x+\frac{f''\left( 0 \right)}{\text{2!}}x^2+\frac{f'''\left( 0 \right)}{\text{3!}}x^3+\cdots +\frac{f^{\left( n \right)}\left( 0 \right)}{n!}x^n $$ 最后修改：2019 年 11 月 14 日 04 : 36 PM © 允许规范转载
