東京大学 新領域創成科学研究科 複雑理工学専攻 2018年度 専門基礎科目 第1問
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関数 \(f(x,y) = (x + y)e^{-(x^2 + y^2)}\) について, 以下の問に答えよ。ただし, \(x,y\) は実数であり, \(e\) は自然対数の底とする。
(問1) \(\frac{\partial f}{\partial x}\) を求めよ。
(問2) \(\frac{\partial ^2 f}{\partial x^2}\) および \(\frac{\partial ^2f}{\partial x\partial y}\) を求めよ。
(問3) 関数 \(f\) を \((x,y) = (1,-1)\) のまわりで二次の項までテイラー展開せよ。
(問4) 関数 \(f\) が \((x,y) = (0.5,0.5)\) において極大値をとることを示せ。
(問5) \(x^2 + y^2 = 1 ,x,y \ge 0\) の条件下で, 関数 \(f\) の極値を求めよ。
Kai
(問1)
\[
\frac{\partial f}{\partial x} = e^{-(x^2 + y^2)} + (x + y)e^{-(x^2 + y^2)} \cdot (-2x) = (1 - 2x^2 - 2xy)e^{-(x^2 + y^2)}
\]
(問2)
\[
\begin{aligned}
&\frac{\partial ^2f}{\partial x^2} = (4x^3 + 4x^2y - 6x - 2y)e^{-(x^2 + y^2)} \\
&\frac{\partial ^2f}{\partial x\partial y} = (4x^2y + 4xy^2 - 2x - 2y)e^{-(x^2 + y^2)} \\
\end{aligned}
\]
(問3)
\[
\begin{aligned}
&\frac{\partial f}{\partial y} = (1 - 2y^2 - 2xy)e^{-(x^2 + y^2)} \\
&\frac{\partial ^2f}{\partial y^2} = (4y^3 + 4xy^2 - 6y - 2x)e^{-(x^2 + y^2)} \\
&\frac{\partial f}{\partial x}\bigg|_{(1,-1)} = e^{-2},
\frac{\partial ^2f}{\partial x^2}\bigg|_{(1,-1)} = -4e^{-2},
\frac{\partial f}{\partial y}\bigg|_{(1,-1)} = e^{-2},
\frac{\partial ^2f}{\partial y^2}\bigg|_{(1,-1)} = 4e^{-2},
\frac{\partial ^2f}{\partial x\partial y}\bigg|_{(1,-1)} = 0 \\
&f(x,y) \approx 0 + \frac{e^{-2}}{1!}(x - 1) + \frac{-4e^{-2}}{2!} + \frac{e^{-2}}{1!}(y + 1) + \frac{4e^{-2}}{2!}(y + 1)^2 + \frac{0}{1! \cdot 1!}(x - 1)(y + 1) \\
&= e^{-2}(x - 1) - 2e^{-2}(x - 1)^2 + e^{-2}(y + 1) + 2e^{-2}(y + 1)^2
\end{aligned}
\]
(問4)
\[
\frac{\partial f}{\partial x}\bigg|_{(0.5,0.5)} =
\frac{\partial f}{\partial y}\bigg|_{(0.5,0.5)} = 0
\]
\[
\begin{aligned}
&\because
\begin{vmatrix}
\frac{\partial ^2f}{\partial x^2} & \frac{\partial ^2f}{\partial x\partial y} \\
\frac{\partial ^2f}{\partial x\partial y} & \frac{\partial ^2f}{\partial y^2}
\end{vmatrix}_{(0.5,0.5)} =
\begin{vmatrix}
-3e^{-0.5} & -e^{-0.5} \\
-e^{-0.5} & -3e^{-0.5} \\
\end{vmatrix} = \frac{e}{8} > 0 ,\frac{\partial ^2f}{\partial x^2}\bigg|_{(0.5,0.5)} = -3e^{-0.5} < 0 \\
&\therefore f \text{ obtain its local maximum at point } (0.5,0.5).
\end{aligned}
\]
(問5)
Let \(g(x,y) = x^2 + y^2 - 1\), then we have
\[
\left\{
\begin{aligned}
&\nabla f = \lambda \nabla g\\
&g = 0
\end{aligned}
\right.
\Rightarrow
\left\{
\begin{aligned}
&(1 - 2x^2 - 2xy) / e = 2\lambda x \\
&(1 - 2y^2 - 2xy) / e = 2\lambda y \\
&x^2 + y^2 = 1
\end{aligned}
\right.
\]
(i) When \(x = 0\),
\[
y = 1,f(0,1) = \frac{1}{e}
\]
(ii) When \(y = 0\),
\[
x = 1,f(1,0) = \frac{1}{e}
\]
(iii) When \(x,y \neq 0\),
\[
\left\{
\begin{aligned}
&(1/x - 2x - 2y)/e = 2\lambda \\
&(1/y - 2y - 2x)/e = 2\lambda \\
&x^2 + y^2 = 1
\end{aligned}
\right. \qquad
x = y = \frac{\sqrt{2}}{2}\ge 0 ,f(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}) = \frac{\sqrt{2}}{e}
\]
Therefore, \(f\) obtains its global minimum \(\frac{1}{e}\) at point \((0,1)\) and \((1,0)\), and obtains its global maximum \(\frac{\sqrt{2}}{e}\) at point \((\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\)