2変数関数のマクローリン展開の問題

■問題

次の関数のマクローリン展開式を第3次の項まで求めよ．

${\displaystyle f\left(x,y\right)=e^x\sin y}$

■答

${\displaystyle f\left(x,y\right)=y+xy+\frac{1}{2}{x}^{2}y-\frac{1}{6}{y}^3+\cdot \cdot \cdot }$

■ヒント

2変数のマクローリンの定理

■解き方

${\displaystyle f\left(0,0\right)=0}$

${\displaystyle f_x=e^x\sin y}$    ${\displaystyle \therefore f_x\left(0,0\right)=0}$

${\displaystyle f_y=e^x\cos y}$    ${\displaystyle \therefore f_y(0,0)=1}$

${\displaystyle f_{xx}=e^x\sin y}$    ${\displaystyle \therefore f_{xx}\left(0,0\right)=0}$

${\displaystyle f_{yy}=-e^x\sin y}$    ${\displaystyle \therefore f_{yy}(0,0)=0}$

${\displaystyle f_{yx}=e^x\cos y}$    ${\displaystyle \therefore f_{yx}(0,0)=1}$

${\displaystyle f_{xxx}=e^x\sin y}$    ${\displaystyle \therefore f_{xxx}\left(0,0\right)=0}$

${\displaystyle f_{yyy}=-e^x\cos y}$     ${\displaystyle \therefore f_{yyy}(0,0)=-1}$

${\displaystyle f_{yxx}=e^x\cos y}$    ${\displaystyle \therefore f_{yxx}(0,0)=1}$

${\displaystyle f_{yyx}=-e^x\sin y}$    ${\displaystyle \therefore f_{yyx}(0,0)=0}$

よって

${\displaystyle =y+\frac{1}{2!}\left(2xy\right)+\frac{1}{3!}(3x^{2}y-y^3)+\cdot \cdot \cdot }$

${\displaystyle =y+xy+\frac{1}{2}{x}^{2}y-\frac{1}{6}{y}^3+\cdot \cdot \cdot }$

　

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最終更新日： 2024年5月28日