题目
试确定A,B,C的值,使得e^x(1+Bx+Cx^2)=1+Ax+o(x^3),其中o(x^3)是当xarrow 0时比x^3高阶的无穷小.
试确定$A$,$B$,$C$的值,使得$e^{x}\left(1+Bx+Cx^{2}\right)=1+Ax+o\left(x^{3}\right)$,其中$o\left(x^{3}\right)$是当$x\rightarrow 0$时比$x^{3}$高阶的无穷小.
题目解答
答案
利用皮亚诺型泰勒公式可得:
$e^{x}=1+x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{6}+o\left(x^{3}\right)$,
代入已知等式得:
$\left[1+x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{6}+o\left(x^{3}\right)\right]\left(1+bx+Cx^{2}\right)=1+Ax+o\left(x^{3}\right)$,
整理得:
$1+\left(B+1\right)x+\left(C+B+\dfrac{1}{2}\right)x^{2}+\left(\dfrac{B}{2}+C+\dfrac{1}{6}\right)+o\left(x^{3}\right)=1+Ax+o\left(x^{3}\right)$,
比较两边同次幂函数得:
$B+1=A $①
$C+B+\dfrac{1}{2}=0 $②
$\dfrac{B}{2}+C+\dfrac{1}{6}=0 $③
式②$-$③得:
$\dfrac{B}{2}+\dfrac{1}{3}=0$,
则:$B=-\dfrac{2}{3}$,
代入①得:$A=\dfrac{1}{3}$,
代入②得:$C=\dfrac{1}{6}$.
故:$A=\dfrac{1}{3}$,$B=-\dfrac{2}{3}$,$C=\dfrac{1}{6}$.