関数 $z = f(x, y) = \sin x \cos y$ の2変数のマクローリン展開を、$x, y$ について3次の項まで求める問題です。

1. 問題の内容

関数 $z = f(x, y) = \sin x \cos y$ の2変数のマクローリン展開を、$x, y$ について3次の項まで求める問題です。

2. 解き方の手順

マクローリン展開は、原点$(0, 0)$のまわりのテイラー展開のことです。2変数のマクローリン展開は以下の式で与えられます。

$f(x, y) = f(0, 0) + \frac{\partial f}{\partial x}(0, 0)x + \frac{\partial f}{\partial y}(0, 0)y + \frac{1}{2!}\left(\frac{\partial^2 f}{\partial x^2}(0, 0)x^2 + 2\frac{\partial^2 f}{\partial x \partial y}(0, 0)xy + \frac{\partial^2 f}{\partial y^2}(0, 0)y^2\right) + \frac{1}{3!}\left(\frac{\partial^3 f}{\partial x^3}(0, 0)x^3 + 3\frac{\partial^3 f}{\partial x^2 \partial y}(0, 0)x^2y + 3\frac{\partial^3 f}{\partial x \partial y^2}(0, 0)xy^2 + \frac{\partial^3 f}{\partial y^3}(0, 0)y^3\right) + \dots$

与えられた関数 $f(x, y) = \sin x \cos y$ について、必要な偏微分を計算します。

* $f(0, 0) = \sin 0 \cos 0 = 0 \cdot 1 = 0$

* $\frac{\partial f}{\partial x} = \cos x \cos y$

$\frac{\partial f}{\partial x}(0, 0) = \cos 0 \cos 0 = 1 \cdot 1 = 1$

* $\frac{\partial f}{\partial y} = -\sin x \sin y$

$\frac{\partial f}{\partial y}(0, 0) = -\sin 0 \sin 0 = -0 \cdot 0 = 0$

* $\frac{\partial^2 f}{\partial x^2} = -\sin x \cos y$

$\frac{\partial^2 f}{\partial x^2}(0, 0) = -\sin 0 \cos 0 = -0 \cdot 1 = 0$

* $\frac{\partial^2 f}{\partial x \partial y} = -\cos x \sin y$

$\frac{\partial^2 f}{\partial x \partial y}(0, 0) = -\cos 0 \sin 0 = -1 \cdot 0 = 0$

* $\frac{\partial^2 f}{\partial y^2} = -\sin x \cos y$

$\frac{\partial^2 f}{\partial y^2}(0, 0) = -\sin 0 \cos 0 = -0 \cdot 1 = 0$

* $\frac{\partial^3 f}{\partial x^3} = -\cos x \cos y$

$\frac{\partial^3 f}{\partial x^3}(0, 0) = -\cos 0 \cos 0 = -1 \cdot 1 = -1$

* $\frac{\partial^3 f}{\partial x^2 \partial y} = \sin x \sin y$

$\frac{\partial^3 f}{\partial x^2 \partial y}(0, 0) = \sin 0 \sin 0 = 0 \cdot 0 = 0$

* $\frac{\partial^3 f}{\partial x \partial y^2} = -\cos x \cos y$

$\frac{\partial^3 f}{\partial x \partial y^2}(0, 0) = \cos 0 \cos 0 = -1 \cdot 1 = -1$

* $\frac{\partial^3 f}{\partial y^3} = \sin x \sin y$

$\frac{\partial^3 f}{\partial y^3}(0, 0) = \sin 0 \sin 0 = 0 \cdot 0 = 0$

これらの値をマクローリン展開の式に代入します。

$f(x, y) = 0 + 1 \cdot x + 0 \cdot y + \frac{1}{2}(0 \cdot x^2 + 2 \cdot 0 \cdot xy + 0 \cdot y^2) + \frac{1}{6}(-1 \cdot x^3 + 3 \cdot 0 \cdot x^2y + 3 \cdot (-1) \cdot xy^2 + 0 \cdot y^3) + \dots$

整理すると、

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

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

3. 最終的な答え

$f(x, y) \approx x - \frac{1}{6}x^3 - \frac{1}{2}xy^2$