関数 $f(x_1, x_2, x_3) = e^{x_1} \sin x_2 \cos x_3$ が与えられています。 $c = (\frac{\pi}{3}, \frac{\pi}{4}, 0)$、 $h = (h_1, h_2, h_3)$ とします。 (1) $\sum_{\alpha \in \mathbb{Z}_{+}^3, |\alpha|=1} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha}$ を計算します。 (2) $\sum_{\alpha \in \mathbb{Z}_{+}^3, |\alpha|=2} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha}$ を計算します。

## 数学の問題

1. 問題の内容

関数 $f(x_1, x_2, x_3) = e^{x_1} \sin x_2 \cos x_3$ が与えられています。

$c = (\frac{\pi}{3}, \frac{\pi}{4}, 0)$、 $h = (h_1, h_2, h_3)$ とします。

(1) $\sum_{\alpha \in \mathbb{Z}_{+}^3, |\alpha|=1} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha}$ を計算します。

(2) $\sum_{\alpha \in \mathbb{Z}_{+}^3, |\alpha|=2} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha}$ を計算します。

2. 解き方の手順

(1) $\sum_{\alpha \in \mathbb{Z}_{+}^3, |\alpha|=1} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha}$ の計算

$|\alpha| = 1$ なので、$\alpha$ は $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ のいずれかです。したがって、

$\sum_{\alpha \in \mathbb{Z}_{+}^3, |\alpha|=1} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha} = \frac{\partial f}{\partial x_1}(c) h_1 + \frac{\partial f}{\partial x_2}(c) h_2 + \frac{\partial f}{\partial x_3}(c) h_3$

偏微分を計算します。

$\frac{\partial f}{\partial x_1} = e^{x_1} \sin x_2 \cos x_3$

$\frac{\partial f}{\partial x_2} = e^{x_1} \cos x_2 \cos x_3$

$\frac{\partial f}{\partial x_3} = -e^{x_1} \sin x_2 \sin x_3$

$c = (\frac{\pi}{3}, \frac{\pi}{4}, 0)$ を代入すると

$\frac{\partial f}{\partial x_1}(c) = e^{\frac{\pi}{3}} \sin \frac{\pi}{4} \cos 0 = e^{\frac{\pi}{3}} \frac{\sqrt{2}}{2} (1) = \frac{\sqrt{2}}{2}e^{\frac{\pi}{3}}$

$\frac{\partial f}{\partial x_2}(c) = e^{\frac{\pi}{3}} \cos \frac{\pi}{4} \cos 0 = e^{\frac{\pi}{3}} \frac{\sqrt{2}}{2} (1) = \frac{\sqrt{2}}{2}e^{\frac{\pi}{3}}$

$\frac{\partial f}{\partial x_3}(c) = -e^{\frac{\pi}{3}} \sin \frac{\pi}{4} \sin 0 = -e^{\frac{\pi}{3}} \frac{\sqrt{2}}{2} (0) = 0$

よって、

$\frac{\partial f}{\partial x_1}(c) h_1 + \frac{\partial f}{\partial x_2}(c) h_2 + \frac{\partial f}{\partial x_3}(c) h_3 = \frac{\sqrt{2}}{2}e^{\frac{\pi}{3}} h_1 + \frac{\sqrt{2}}{2}e^{\frac{\pi}{3}} h_2 + 0 = \frac{\sqrt{2}}{2}e^{\frac{\pi}{3}} (h_1 + h_2)$

(2) $\sum_{\alpha \in \mathbb{Z}_{+}^3, |\alpha|=2} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha}$ の計算

$|\alpha| = 2$ なので、$\alpha$ は $(2,0,0)$, $(0,2,0)$, $(0,0,2)$, $(1,1,0)$, $(1,0,1)$, $(0,1,1)$ のいずれかです。したがって、

$\sum_{\alpha \in \mathbb{Z}_{+}^3, |\alpha|=2} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha} = \frac{1}{2!} \frac{\partial^2 f}{\partial x_1^2}(c) h_1^2 + \frac{1}{2!} \frac{\partial^2 f}{\partial x_2^2}(c) h_2^2 + \frac{1}{2!} \frac{\partial^2 f}{\partial x_3^2}(c) h_3^2 + \frac{\partial^2 f}{\partial x_1 \partial x_2}(c) h_1 h_2 + \frac{\partial^2 f}{\partial x_1 \partial x_3}(c) h_1 h_3 + \frac{\partial^2 f}{\partial x_2 \partial x_3}(c) h_2 h_3$

二階偏微分を計算します。

$\frac{\partial^2 f}{\partial x_1^2} = e^{x_1} \sin x_2 \cos x_3$

$\frac{\partial^2 f}{\partial x_2^2} = -e^{x_1} \sin x_2 \cos x_3$

$\frac{\partial^2 f}{\partial x_3^2} = -e^{x_1} \sin x_2 \cos x_3$

$\frac{\partial^2 f}{\partial x_1 \partial x_2} = e^{x_1} \cos x_2 \cos x_3$

$\frac{\partial^2 f}{\partial x_1 \partial x_3} = -e^{x_1} \sin x_2 \sin x_3$

$\frac{\partial^2 f}{\partial x_2 \partial x_3} = -e^{x_1} \cos x_2 \sin x_3$

$c = (\frac{\pi}{3}, \frac{\pi}{4}, 0)$ を代入すると

$\frac{\partial^2 f}{\partial x_1^2}(c) = e^{\frac{\pi}{3}} \sin \frac{\pi}{4} \cos 0 = e^{\frac{\pi}{3}} \frac{\sqrt{2}}{2} (1) = \frac{\sqrt{2}}{2}e^{\frac{\pi}{3}}$

$\frac{\partial^2 f}{\partial x_2^2}(c) = -e^{\frac{\pi}{3}} \sin \frac{\pi}{4} \cos 0 = -e^{\frac{\pi}{3}} \frac{\sqrt{2}}{2} (1) = -\frac{\sqrt{2}}{2}e^{\frac{\pi}{3}}$

$\frac{\partial^2 f}{\partial x_3^2}(c) = -e^{\frac{\pi}{3}} \sin \frac{\pi}{4} \cos 0 = -e^{\frac{\pi}{3}} \frac{\sqrt{2}}{2} (1) = -\frac{\sqrt{2}}{2}e^{\frac{\pi}{3}}$

$\frac{\partial^2 f}{\partial x_1 \partial x_2}(c) = e^{\frac{\pi}{3}} \cos \frac{\pi}{4} \cos 0 = e^{\frac{\pi}{3}} \frac{\sqrt{2}}{2} (1) = \frac{\sqrt{2}}{2}e^{\frac{\pi}{3}}$

$\frac{\partial^2 f}{\partial x_1 \partial x_3}(c) = -e^{\frac{\pi}{3}} \sin \frac{\pi}{4} \sin 0 = -e^{\frac{\pi}{3}} \frac{\sqrt{2}}{2} (0) = 0$

$\frac{\partial^2 f}{\partial x_2 \partial x_3}(c) = -e^{\frac{\pi}{3}} \cos \frac{\pi}{4} \sin 0 = -e^{\frac{\pi}{3}} \frac{\sqrt{2}}{2} (0) = 0$

よって、

$\sum_{\alpha \in \mathbb{Z}_{+}^3, |\alpha|=2} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha} = \frac{1}{2} \frac{\sqrt{2}}{2}e^{\frac{\pi}{3}} h_1^2 - \frac{1}{2} \frac{\sqrt{2}}{2}e^{\frac{\pi}{3}} h_2^2 - \frac{1}{2} \frac{\sqrt{2}}{2}e^{\frac{\pi}{3}} h_3^2 + \frac{\sqrt{2}}{2}e^{\frac{\pi}{3}} h_1 h_2 + 0 + 0$

$= \frac{\sqrt{2}}{4}e^{\frac{\pi}{3}} (h_1^2 - h_2^2 - h_3^2 + 2h_1 h_2)$

3. 最終的な答え

(1) $\sum_{\alpha \in \mathbb{Z}_{+}^3, |\alpha|=1} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha} = \frac{\sqrt{2}}{2}e^{\frac{\pi}{3}} (h_1 + h_2)$

(2) $\sum_{\alpha \in \mathbb{Z}_{+}^3, |\alpha|=2} \frac{1}{\alpha!} \partial^{\alpha} f(c) h^{\alpha} = \frac{\sqrt{2}}{4}e^{\frac{\pi}{3}} (h_1^2 - h_2^2 - h_3^2 + 2h_1 h_2)$