東京大学 新領域創成科学研究科 複雑理工学専攻 2017年8月実施 専門基礎科目 第1問 Author 之遥
Description 関数 f ( x , y ) = ( x + y ) e − ( x 2 + y 2 ) f(x,y) = (x + y)e^{-(x^2 + y^2)} f ( x , y ) = ( x + y ) e − ( x 2 + y 2 ) x , y x,y x , y e e e
(問1) ∂ f ∂ x \frac{\partial f}{\partial x} ∂ x ∂ f
(問2) ∂ 2 f ∂ x 2 \frac{\partial ^2 f}{\partial x^2} ∂ x 2 ∂ 2 f ∂ 2 f ∂ x ∂ y \frac{\partial ^2f}{\partial x\partial y} ∂ x ∂ y ∂ 2 f
(問3) 関数 f f f ( x , y ) = ( 1 , − 1 ) (x,y) = (1,-1) ( x , y ) = ( 1 , − 1 )
(問4) 関数 f f f ( x , y ) = ( 0.5 , 0.5 ) (x,y) = (0.5,0.5) ( x , y ) = ( 0.5 , 0.5 )
(問5) x 2 + y 2 = 1 , x , y ≥ 0 x^2 + y^2 = 1 ,x,y \ge 0 x 2 + y 2 = 1 , x , y ≥ 0 f f f
Kai (問1)
∂ f ∂ x = e − ( x 2 + y 2 ) + ( x + y ) e − ( x 2 + y 2 ) ⋅ ( − 2 x ) = ( 1 − 2 x 2 − 2 x y ) e − ( x 2 + y 2 ) \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)} ∂ x ∂ f = e − ( x 2 + y 2 ) + ( x + y ) e − ( x 2 + y 2 ) ⋅ ( − 2 x ) = ( 1 − 2 x 2 − 2 x y ) e − ( x 2 + y 2 ) (問2)
∂ 2 f ∂ x 2 = ( 4 x 3 + 4 x 2 y − 6 x − 2 y ) e − ( x 2 + y 2 ) ∂ 2 f ∂ x ∂ y = ( 4 x 2 y + 4 x y 2 − 2 x − 2 y ) e − ( x 2 + y 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} ∂ x 2 ∂ 2 f = ( 4 x 3 + 4 x 2 y − 6 x − 2 y ) e − ( x 2 + y 2 ) ∂ x ∂ y ∂ 2 f = ( 4 x 2 y + 4 x y 2 − 2 x − 2 y ) e − ( x 2 + y 2 ) (問3)
∂ f ∂ y = ( 1 − 2 y 2 − 2 x y ) e − ( x 2 + y 2 ) ∂ 2 f ∂ y 2 = ( 4 y 3 + 4 x y 2 − 6 y − 2 x ) e − ( x 2 + y 2 ) ∂ f ∂ x ∣ ( 1 , − 1 ) = e − 2 , ∂ 2 f ∂ x 2 ∣ ( 1 , − 1 ) = − 4 e − 2 , ∂ f ∂ y ∣ ( 1 , − 1 ) = e − 2 , ∂ 2 f ∂ y 2 ∣ ( 1 , − 1 ) = 4 e − 2 , ∂ 2 f ∂ x ∂ y ∣ ( 1 , − 1 ) = 0 f ( x , y ) ≈ 0 + e − 2 1 ! ( x − 1 ) + − 4 e − 2 2 ! + e − 2 1 ! ( y + 1 ) + 4 e − 2 2 ! ( y + 1 ) 2 + 0 1 ! ⋅ 1 ! ( x − 1 ) ( y + 1 ) = e − 2 ( x − 1 ) − 2 e − 2 ( x − 1 ) 2 + e − 2 ( y + 1 ) + 2 e − 2 ( y + 1 ) 2 \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} ∂ y ∂ f = ( 1 − 2 y 2 − 2 x y ) e − ( x 2 + y 2 ) ∂ y 2 ∂ 2 f = ( 4 y 3 + 4 x y 2 − 6 y − 2 x ) e − ( x 2 + y 2 ) ∂ x ∂ f ( 1 , − 1 ) = e − 2 , ∂ x 2 ∂ 2 f ( 1 , − 1 ) = − 4 e − 2 , ∂ y ∂ f ( 1 , − 1 ) = e − 2 , ∂ y 2 ∂ 2 f ( 1 , − 1 ) = 4 e − 2 , ∂ x ∂ y ∂ 2 f ( 1 , − 1 ) = 0 f ( x , y ) ≈ 0 + 1 ! e − 2 ( x − 1 ) + 2 ! − 4 e − 2 + 1 ! e − 2 ( y + 1 ) + 2 ! 4 e − 2 ( y + 1 ) 2 + 1 ! ⋅ 1 ! 0 ( x − 1 ) ( y + 1 ) = e − 2 ( x − 1 ) − 2 e − 2 ( x − 1 ) 2 + e − 2 ( y + 1 ) + 2 e − 2 ( y + 1 ) 2 (問4)
∂ f ∂ x ∣ ( 0.5 , 0.5 ) = ∂ f ∂ y ∣ ( 0.5 , 0.5 ) = 0 \frac{\partial f}{\partial x}\bigg|_{(0.5,0.5)} =
\frac{\partial f}{\partial y}\bigg|_{(0.5,0.5)} = 0 ∂ x ∂ f ( 0.5 , 0.5 ) = ∂ y ∂ f ( 0.5 , 0.5 ) = 0 ∵ ∣ ∂ 2 f ∂ x 2 ∂ 2 f ∂ x ∂ y ∂ 2 f ∂ x ∂ y ∂ 2 f ∂ y 2 ∣ ( 0.5 , 0.5 ) = ∣ − 3 e − 0.5 − e − 0.5 − e − 0.5 − 3 e − 0.5 ∣ = e 8 > 0 , ∂ 2 f ∂ x 2 ∣ ( 0.5 , 0.5 ) = − 3 e − 0.5 < 0 ∴ f obtain its local maximum at point ( 0.5 , 0.5 ) . \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} ∵ ∂ x 2 ∂ 2 f ∂ x ∂ y ∂ 2 f ∂ x ∂ y ∂ 2 f ∂ y 2 ∂ 2 f ( 0.5 , 0.5 ) = − 3 e − 0.5 − e − 0.5 − e − 0.5 − 3 e − 0.5 = 8 e > 0 , ∂ x 2 ∂ 2 f ( 0.5 , 0.5 ) = − 3 e − 0.5 < 0 ∴ f obtain its local maximum at point ( 0.5 , 0.5 ) . (問5)
Let g ( x , y ) = x 2 + y 2 − 1 g(x,y) = x^2 + y^2 - 1 g ( x , y ) = x 2 + y 2 − 1
{ ∇ f = λ ∇ g g = 0 ⇒ { ( 1 − 2 x 2 − 2 x y ) / e = 2 λ x ( 1 − 2 y 2 − 2 x y ) / e = 2 λ y x 2 + y 2 = 1 \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. { ∇ f = λ ∇ g g = 0 ⇒ ⎩ ⎨ ⎧ ( 1 − 2 x 2 − 2 x y ) / e = 2 λ x ( 1 − 2 y 2 − 2 x y ) / e = 2 λ y x 2 + y 2 = 1 (i) When x = 0 x = 0 x = 0
y = 1 , f ( 0 , 1 ) = 1 e y = 1,f(0,1) = \frac{1}{e} y = 1 , f ( 0 , 1 ) = e 1 (ii) When y = 0 y = 0 y = 0
x = 1 , f ( 1 , 0 ) = 1 e x = 1,f(1,0) = \frac{1}{e} x = 1 , f ( 1 , 0 ) = e 1 (iii) When x , y ≠ 0 x,y \neq 0 x , y = 0
{ ( 1 / x − 2 x − 2 y ) / e = 2 λ ( 1 / y − 2 y − 2 x ) / e = 2 λ x 2 + y 2 = 1 x = y = 2 2 ≥ 0 , f ( 2 2 , 2 2 ) = 2 e \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} ⎩ ⎨ ⎧ ( 1/ x − 2 x − 2 y ) / e = 2 λ ( 1/ y − 2 y − 2 x ) / e = 2 λ x 2 + y 2 = 1 x = y = 2 2 ≥ 0 , f ( 2 2 , 2 2 ) = e 2 Therefore, f f f 1 e \frac{1}{e} e 1 ( 0 , 1 ) (0,1) ( 0 , 1 ) ( 1 , 0 ) (1,0) ( 1 , 0 ) 2 e \frac{\sqrt{2}}{e} e 2 ( 2 2 , 2 2 ) (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}) ( 2 2 , 2 2 )