明治大学 先端数理科学研究科 現象数理学専攻 2021年8月実施 微积分 Author 思齐塾 , 祭音Myyura
Description I. 実数全体の集合 R \mathbb{R} R f , g f, g f , g f ∗ g f * g f ∗ g
( f ∗ g ) ( x ) = ∫ 0 x f ( x − y ) g ( y ) d y ( x ∈ R ) (f * g)(x) = \int_0^x f(x - y)g(y) dy \quad (x \in \mathbb{R}) ( f ∗ g ) ( x ) = ∫ 0 x f ( x − y ) g ( y ) d y ( x ∈ R ) で定める。次の問に答えよ。
(1) 関数 e n ( x ) e_n(x) e n ( x )
e 0 ( x ) = 1 , e n + 1 ( x ) = ( e n ∗ e 0 ) ( x ) ( n = 0 , 1 , 2 , … ) e_0(x) = 1, \quad e_{n+1}(x) = (e_n * e_0)(x) \quad (n = 0, 1, 2, \dots) e 0 ( x ) = 1 , e n + 1 ( x ) = ( e n ∗ e 0 ) ( x ) ( n = 0 , 1 , 2 , … ) で定める。関数 e n ( x ) e_n(x) e n ( x )
(2) R \mathbb{R} R f , g f, g f , g f ∗ g = g ∗ f f * g = g * f f ∗ g = g ∗ f
II. 2変数関数 h ( x , y ) h(x, y) h ( x , y )
h ( x , y ) = e − x 2 + y 2 h(x, y) = e^{-\sqrt{x^2 + y^2}} h ( x , y ) = e − x 2 + y 2 で定める。次の問に答えよ。
(1) 偏導関数
∂ h ∂ x , ∂ h ∂ y \frac{\partial h}{\partial x}, \quad \frac{\partial h}{\partial y} ∂ x ∂ h , ∂ y ∂ h をそれぞれ求めよ。
(2) D D D x y xy x y
x 2 + y 2 ≤ 1 x^2 + y^2 \leq 1 x 2 + y 2 ≤ 1 で定まる領域とする。 D D D
∬ D 1 x 2 + y 2 ( x ∂ h ∂ x + y ∂ h ∂ y ) d x d y \iint_D \frac{1}{x^2 + y^2} \left( x \frac{\partial h}{\partial x} + y \frac{\partial h}{\partial y} \right) dxdy ∬ D x 2 + y 2 1 ( x ∂ x ∂ h + y ∂ y ∂ h ) d x d y を計算せよ。
题目描述
I. 对定义在全体实数集 R \mathbb{R} R f , g f,g f , g f ∗ g f*g f ∗ g
( f ∗ g ) ( x ) = ∫ 0 x f ( x − y ) g ( y ) d y ( x ∈ R ) . (f*g)(x)=\int_0^x f(x-y)g(y)\,dy
\qquad(x\in\mathbb{R}). ( f ∗ g ) ( x ) = ∫ 0 x f ( x − y ) g ( y ) d y ( x ∈ R ) . 回答下列问题。
(1) 定义函数 e n ( x ) e_n(x) e n ( x )
e 0 ( x ) = 1 , e n + 1 ( x ) = ( e n ∗ e 0 ) ( x ) ( n = 0 , 1 , 2 , … ) . e_0(x)=1,\qquad
e_{n+1}(x)=(e_n*e_0)(x)
\quad(n=0,1,2,\dots). e 0 ( x ) = 1 , e n + 1 ( x ) = ( e n ∗ e 0 ) ( x ) ( n = 0 , 1 , 2 , … ) . 求 e n ( x ) e_n(x) e n ( x )
(2) 证明：对任意定义在 R \mathbb{R} R f , g f,g f , g
II. 定义二元函数
h ( x , y ) = e − x 2 + y 2 . h(x,y)=e^{-\sqrt{x^2+y^2}}. h ( x , y ) = e − x 2 + y 2 . 回答下列问题。
(1) 分别求偏导数
∂ h ∂ x , ∂ h ∂ y . \frac{\partial h}{\partial x},
\qquad
\frac{\partial h}{\partial y}. ∂ x ∂ h , ∂ y ∂ h . (2) 设 D D D x y xy x y
x 2 + y 2 ≤ 1. x^2+y^2\leq1. x 2 + y 2 ≤ 1. 计算 D D D
∬ D 1 x 2 + y 2 ( x ∂ h ∂ x + y ∂ h ∂ y ) d x d y . \iint_D\frac{1}{x^2+y^2}
\left(
x\frac{\partial h}{\partial x}
+y\frac{\partial h}{\partial y}
\right)\,dx\,dy. ∬ D x 2 + y 2 1 ( x ∂ x ∂ h + y ∂ y ∂ h ) d x d y . Kai I. (1) e 0 ( x ) = 1 e_0(x) = 1 e 0 ( x ) = 1 e 1 ( x ) = ( e 0 ∗ e 0 ) ( x ) = ∫ 0 x e 0 ( x − y ) e 0 ( y ) d y = ∫ 0 x 1 ⋅ 1 d y = x e_1(x) = (e_0 * e_0)(x) = \int_0^x e_0(x - y) e_0(y) dy = \int_0^x 1 \cdot 1 dy = x e 1 ( x ) = ( e 0 ∗ e 0 ) ( x ) = ∫ 0 x e 0 ( x − y ) e 0 ( y ) d y = ∫ 0 x 1 ⋅ 1 d y = x e 2 ( x ) = ( e 1 ∗ e 0 ) ( x ) = ∫ 0 x e 1 ( x − y ) e 0 ( y ) d y = ∫ 0 x ( x − y ) ⋅ 1 d y = ∫ 0 x ( x − y ) d y = [ x y − 1 2 y 2 ] 0 x = x 2 − 1 2 x 2 = x 2 2 e_2(x) = (e_1 * e_0)(x) = \int_0^x e_1(x - y) e_0(y) dy = \int_0^x (x - y) \cdot 1 dy = \int_0^x (x - y) dy = [xy - \frac{1}{2}y^2]_0^x = x^2 - \frac{1}{2}x^2 = \frac{x^2}{2} e 2 ( x ) = ( e 1 ∗ e 0 ) ( x ) = ∫ 0 x e 1 ( x − y ) e 0 ( y ) d y = ∫ 0 x ( x − y ) ⋅ 1 d y = ∫ 0 x ( x − y ) d y = [ x y − 2 1 y 2 ] 0 x = x 2 − 2 1 x 2 = 2 x 2 e 3 ( x ) = ( e 2 ∗ e 0 ) ( x ) = ∫ 0 x e 2 ( x − y ) e 0 ( y ) d y = ∫ 0 x ( x − y ) 2 2 ⋅ 1 d y = 1 2 ∫ 0 x ( x − y ) 2 d y = 1 2 [ − ( x − y ) 3 3 ] 0 x = 1 2 ( 0 − ( − x 3 3 ) ) = x 3 6 e_3(x) = (e_2 * e_0)(x) = \int_0^x e_2(x - y) e_0(y) dy = \int_0^x \frac{(x - y)^2}{2} \cdot 1 dy = \frac{1}{2} \int_0^x (x - y)^2 dy = \frac{1}{2} [-\frac{(x - y)^3}{3}]_0^x = \frac{1}{2} (0 - (-\frac{x^3}{3})) = \frac{x^3}{6} e 3 ( x ) = ( e 2 ∗ e 0 ) ( x ) = ∫ 0 x e 2 ( x − y ) e 0 ( y ) d y = ∫ 0 x 2 ( x − y ) 2 ⋅ 1 d y = 2 1 ∫ 0 x ( x − y ) 2 d y = 2 1 [ − 3 ( x − y ) 3 ] 0 x = 2 1 ( 0 − ( − 3 x 3 )) = 6 x 3
Mathematical Induction:
Hypothesis: Assume e k ( x ) = x k k ! e_k(x) = \frac{x^k}{k!} e k ( x ) = k ! x k k ≥ 0 k \ge 0 k ≥ 0 n = k + 1 n = k+1 n = k + 1 e k + 1 ( x ) = ( e k ∗ e 0 ) ( x ) = ∫ 0 x e k ( x − y ) e 0 ( y ) d y e_{k+1}(x) = (e_k * e_0)(x) = \int_0^x e_k(x - y) e_0(y) dy e k + 1 ( x ) = ( e k ∗ e 0 ) ( x ) = ∫ 0 x e k ( x − y ) e 0 ( y ) d y = ∫ 0 x ( x − y ) k k ! ⋅ 1 d y = \int_0^x \frac{(x - y)^k}{k!} \cdot 1 dy = ∫ 0 x k ! ( x − y ) k ⋅ 1 d y u = x − y u = x - y u = x − y d u = − d y du = -dy d u = − d y 0 → x 0 \to x 0 → x x → 0 x \to 0 x → 0 = ∫ x 0 u k k ! ( − d u ) = ∫ 0 x u k k ! d u = \int_x^0 \frac{u^k}{k!} (-du) = \int_0^x \frac{u^k}{k!} du = ∫ x 0 k ! u k ( − d u ) = ∫ 0 x k ! u k d u = 1 k ! [ u k + 1 k + 1 ] 0 x = 1 k ! ⋅ x k + 1 k + 1 = x k + 1 ( k + 1 ) ! = \frac{1}{k!} [\frac{u^{k+1}}{k+1}]_0^x = \frac{1}{k!} \cdot \frac{x^{k+1}}{k+1} = \frac{x^{k+1}}{(k+1)!} = k ! 1 [ k + 1 u k + 1 ] 0 x = k ! 1 ⋅ k + 1 x k + 1 = ( k + 1 )! x k + 1 e n ( x ) = x n n ! e_n(x) = \frac{x^n}{n!} e n ( x ) = n ! x n n ≥ 0 n \ge 0 n ≥ 0
(2) To prove: f ∗ g = g ∗ f f*g = g*f f ∗ g = g ∗ f
( f ∗ g ) ( x ) = ∫ 0 x f ( x − y ) g ( y ) d y (f * g)(x) = \int_0^x f(x - y) g(y) dy ( f ∗ g ) ( x ) = ∫ 0 x f ( x − y ) g ( y ) d y Let u = x − y u = x - y u = x − y y = x − u y = x - u y = x − u d y = − d u dy = -du d y = − d u
( f ∗ g ) ( x ) = ∫ x 0 f ( u ) g ( x − u ) ( − d u ) = ∫ 0 x g ( x − u ) f ( u ) d u = ( g ∗ f ) ( x ) (f * g)(x) = \int_x^0 f(u) g(x - u) (-du) = \int_0^x g(x - u) f(u) du = (g * f)(x) ( f ∗ g ) ( x ) = ∫ x 0 f ( u ) g ( x − u ) ( − d u ) = ∫ 0 x g ( x − u ) f ( u ) d u = ( g ∗ f ) ( x ) Hence, f ∗ g = g ∗ f f * g = g * f f ∗ g = g ∗ f
II. (1)
h ( x , y ) = e − x 2 + y 2 h(x, y) = e^{-\sqrt{x^2 + y^2}} h ( x , y ) = e − x 2 + y 2 ∂ h ∂ x = e − x 2 + y 2 ⋅ ( − 1 2 x 2 + y 2 ) ⋅ 2 x = − x x 2 + y 2 e − x 2 + y 2 \frac{\partial h}{\partial x} = e^{-\sqrt{x^2 + y^2}} \cdot (-\frac{1}{2\sqrt{x^2 + y^2}}) \cdot 2x = -\frac{x}{\sqrt{x^2 + y^2}} e^{-\sqrt{x^2 + y^2}} ∂ x ∂ h = e − x 2 + y 2 ⋅ ( − 2 x 2 + y 2 1 ) ⋅ 2 x = − x 2 + y 2 x e − x 2 + y 2 ∂ h ∂ y = e − x 2 + y 2 ⋅ ( − 1 2 x 2 + y 2 ) ⋅ 2 y = − y x 2 + y 2 e − x 2 + y 2 \frac{\partial h}{\partial y} = e^{-\sqrt{x^2 + y^2}} \cdot (-\frac{1}{2\sqrt{x^2 + y^2}}) \cdot 2y = -\frac{y}{\sqrt{x^2 + y^2}} e^{-\sqrt{x^2 + y^2}} ∂ y ∂ h = e − x 2 + y 2 ⋅ ( − 2 x 2 + y 2 1 ) ⋅ 2 y = − x 2 + y 2 y e − x 2 + y 2 These formulas hold for ( x , y ) ≠ ( 0 , 0 ) (x,y)\ne(0,0) ( x , y ) = ( 0 , 0 )
h ( t , 0 ) − h ( 0 , 0 ) t = e − ∣ t ∣ − 1 t \frac{h(t,0)-h(0,0)}{t}
=\frac{e^{-|t|}-1}{t} t h ( t , 0 ) − h ( 0 , 0 ) = t e − ∣ t ∣ − 1 tends to − 1 -1 − 1 t → 0 + t\to0^+ t → 0 + 1 1 1 t → 0 − t\to0^- t → 0 − y y y
(2)
∬ D 1 x 2 + y 2 ( x ∂ h ∂ x + y ∂ h ∂ y ) d x d y \iint_D \frac{1}{x^2 + y^2} \left( x \frac{\partial h}{\partial x} + y \frac{\partial h}{\partial y} \right) dxdy ∬ D x 2 + y 2 1 ( x ∂ x ∂ h + y ∂ y ∂ h ) d x d y The integrand is therefore understood on D ∖ { ( 0 , 0 ) } D\setminus\{(0,0)\} D ∖ {( 0 , 0 )}
1 x 2 + y 2 ( x ∂ h ∂ x + y ∂ h ∂ y ) = 1 x 2 + y 2 ( x ( − x x 2 + y 2 e − x 2 + y 2 ) + y ( − y x 2 + y 2 e − x 2 + y 2 ) ) \frac{1}{x^2 + y^2} \left( x \frac{\partial h}{\partial x} + y \frac{\partial h}{\partial y} \right) = \frac{1}{x^2 + y^2} \left( x (-\frac{x}{\sqrt{x^2 + y^2}} e^{-\sqrt{x^2 + y^2}}) + y (-\frac{y}{\sqrt{x^2 + y^2}} e^{-\sqrt{x^2 + y^2}}) \right) x 2 + y 2 1 ( x ∂ x ∂ h + y ∂ y ∂ h ) = x 2 + y 2 1 ( x ( − x 2 + y 2 x e − x 2 + y 2 ) + y ( − x 2 + y 2 y e − x 2 + y 2 ) ) = 1 x 2 + y 2 ( − x 2 + y 2 x 2 + y 2 e − x 2 + y 2 ) = − e − x 2 + y 2 x 2 + y 2 = \frac{1}{x^2 + y^2} \left( -\frac{x^2 + y^2}{\sqrt{x^2 + y^2}} e^{-\sqrt{x^2 + y^2}} \right) = -\frac{e^{-\sqrt{x^2 + y^2}}}{\sqrt{x^2 + y^2}} = x 2 + y 2 1 ( − x 2 + y 2 x 2 + y 2 e − x 2 + y 2 ) = − x 2 + y 2 e − x 2 + y 2 Let x = r cos θ x = r \cos \theta x = r cos θ y = r sin θ y = r \sin \theta y = r sin θ x 2 + y 2 = r 2 x^2 + y^2 = r^2 x 2 + y 2 = r 2 d x d y = r d r d θ dxdy = r dr d\theta d x d y = r d r d θ
D : x 2 + y 2 ≤ 1 ⇒ r 2 ≤ 1 ⇒ 0 ≤ r ≤ 1 D: x^2 + y^2 \leq 1 \Rightarrow r^2 \leq 1 \Rightarrow 0 \leq r \leq 1 D : x 2 + y 2 ≤ 1 ⇒ r 2 ≤ 1 ⇒ 0 ≤ r ≤ 1 0 ≤ θ ≤ 2 π 0 \leq \theta \leq 2\pi 0 ≤ θ ≤ 2 π ∬ D − e − x 2 + y 2 x 2 + y 2 d x d y = ∫ 0 2 π ∫ 0 1 − e − r r r d r d θ = ∫ 0 2 π ∫ 0 1 − e − r d r d θ \iint_D -\frac{e^{-\sqrt{x^2 + y^2}}}{\sqrt{x^2 + y^2}} dxdy = \int_0^{2\pi} \int_0^1 -\frac{e^{-r}}{r} r dr d\theta = \int_0^{2\pi} \int_0^1 -e^{-r} dr d\theta ∬ D − x 2 + y 2 e − x 2 + y 2 d x d y = ∫ 0 2 π ∫ 0 1 − r e − r r d r d θ = ∫ 0 2 π ∫ 0 1 − e − r d r d θ = ∫ 0 2 π [ e − r ] 0 1 d θ = ∫ 0 2 π ( e − 1 − 1 ) d θ = ( e − 1 − 1 ) [ θ ] 0 2 π = 2 π ( e − 1 − 1 ) = 2 π ( 1 e − 1 ) = 2 π ( 1 − e e ) = \int_0^{2\pi} [e^{-r}]_0^1 d\theta = \int_0^{2\pi} (e^{-1} - 1) d\theta = (e^{-1} - 1) [\theta]_0^{2\pi} = 2\pi (e^{-1} - 1) = 2\pi (\frac{1}{e} - 1) = 2\pi (\frac{1 - e}{e}) = ∫ 0 2 π [ e − r ] 0 1 d θ = ∫ 0 2 π ( e − 1 − 1 ) d θ = ( e − 1 − 1 ) [ θ ] 0 2 π = 2 π ( e − 1 − 1 ) = 2 π ( e 1 − 1 ) = 2 π ( e 1 − e )