2変数関数のマクローリン展開（原点周りのテイラー展開）を3次の項まで求める問題です。 (1) $z = e^{x+y}$ (2) $z = e^x \cos y$

1. 問題の内容

2変数関数のマクローリン展開（原点周りのテイラー展開）を3次の項まで求める問題です。

(1) $z = e^{x+y}$

(2) $z = e^x \cos y$

2. 解き方の手順

(1) $z = e^{x+y}$の場合

マクローリン展開は、

$f(x,y) = f(0,0) + f_x(0,0)x + f_y(0,0)y + \frac{1}{2}(f_{xx}(0,0)x^2 + 2f_{xy}(0,0)xy + f_{yy}(0,0)y^2) + \frac{1}{6}(f_{xxx}(0,0)x^3 + 3f_{xxy}(0,0)x^2y + 3f_{xyy}(0,0)xy^2 + f_{yyy}(0,0)y^3) + \cdots$

で与えられます。

$f(x,y) = e^{x+y}$

$f(0,0) = e^0 = 1$

$f_x(x,y) = e^{x+y}$

$f_x(0,0) = 1$

$f_y(x,y) = e^{x+y}$

$f_y(0,0) = 1$

$f_{xx}(x,y) = e^{x+y}$

$f_{xx}(0,0) = 1$

$f_{xy}(x,y) = e^{x+y}$

$f_{xy}(0,0) = 1$

$f_{yy}(x,y) = e^{x+y}$

$f_{yy}(0,0) = 1$

$f_{xxx}(x,y) = e^{x+y}$

$f_{xxx}(0,0) = 1$

$f_{xxy}(x,y) = e^{x+y}$

$f_{xxy}(0,0) = 1$

$f_{xyy}(x,y) = e^{x+y}$

$f_{xyy}(0,0) = 1$

$f_{yyy}(x,y) = e^{x+y}$

$f_{yyy}(0,0) = 1$

したがって、

$e^{x+y} = 1 + x + y + \frac{1}{2}(x^2 + 2xy + y^2) + \frac{1}{6}(x^3 + 3x^2y + 3xy^2 + y^3) + \cdots$

$e^{x+y} = 1 + (x+y) + \frac{1}{2}(x+y)^2 + \frac{1}{6}(x+y)^3 + \cdots$

(2) $z = e^x \cos y$の場合

$f(x,y) = e^x \cos y$

$f(0,0) = e^0 \cos 0 = 1$

$f_x(x,y) = e^x \cos y$

$f_x(0,0) = 1$

$f_y(x,y) = -e^x \sin y$

$f_y(0,0) = 0$

$f_{xx}(x,y) = e^x \cos y$

$f_{xx}(0,0) = 1$

$f_{xy}(x,y) = -e^x \sin y$

$f_{xy}(0,0) = 0$

$f_{yy}(x,y) = -e^x \cos y$

$f_{yy}(0,0) = -1$

$f_{xxx}(x,y) = e^x \cos y$

$f_{xxx}(0,0) = 1$

$f_{xxy}(x,y) = -e^x \sin y$

$f_{xxy}(0,0) = 0$

$f_{xyy}(x,y) = -e^x \cos y$

$f_{xyy}(0,0) = -1$

$f_{yyy}(x,y) = e^x \sin y$

$f_{yyy}(0,0) = 0$

したがって、

$e^x \cos y = 1 + x + \frac{1}{2}(x^2 - y^2) + \frac{1}{6}(x^3 - 3xy^2) + \cdots$

3. 最終的な答え

(1) $e^{x+y} = 1 + x + y + \frac{1}{2}(x+y)^2 + \frac{1}{6}(x+y)^3 + \cdots$

(2) $e^x \cos y = 1 + x + \frac{1}{2}(x^2 - y^2) + \frac{1}{6}(x^3 - 3xy^2) + \cdots$