明治大学 先端数理科学研究科 現象数理学専攻 2023年8月実施 微积分 Author 思齐塾 , 祭音Myyura
Description (I) 関数
f ( x ) = log ( 1 + x ) − 1 3 x 3 + 1 2 x 2 − x f(x) = \log(1+x) - \frac{1}{3}x^3 + \frac{1}{2}x^2 - x f ( x ) = log ( 1 + x ) − 3 1 x 3 + 2 1 x 2 − x について, 極限
lim x → 0 f ( x ) x k \lim_{x \to 0} \frac{f(x)}{x^k} x → 0 lim x k f ( x ) が 0 でない値に収束するような整数 k k k
(II) x y xy x y
f ( x , y ) = x 3 − 12 x y + 8 y 3 f(x, y) = x^3 - 12xy + 8y^3 f ( x , y ) = x 3 − 12 x y + 8 y 3 について, 次の問に答えよ.
(1) 関数 f f f ∂ f ∂ x , ∂ f ∂ y \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} ∂ x ∂ f , ∂ y ∂ f f f f f f f
(III) Ω = { ( x , y , z ) ∈ R 3 ∣ x ≥ 0 , y ≥ 0 , z ≥ 0 , x + y + z ≤ 1 } \Omega = \{(x, y, z) \in \mathbb{R}^3 | x \geq 0, y \geq 0, z \geq 0, x + y + z \leq 1\} Ω = {( x , y , z ) ∈ R 3 ∣ x ≥ 0 , y ≥ 0 , z ≥ 0 , x + y + z ≤ 1 }
∭ Ω d x d y d z \iiint_{\Omega} dx dy dz ∭ Ω d x d y d z を求めよ.
(2)
∭ Ω d x d y d z ( 1 + x + y + z ) 3 \iiint_{\Omega} \frac{dx dy dz}{(1 + x + y + z)^3} ∭ Ω ( 1 + x + y + z ) 3 d x d y d z を求めよ.
题目描述
I. 对函数
f ( x ) = log ( 1 + x ) − 1 3 x 3 + 1 2 x 2 − x , f(x)=\log(1+x)-\frac13x^3+\frac12x^2-x, f ( x ) = log ( 1 + x ) − 3 1 x 3 + 2 1 x 2 − x , 求使极限
lim x → 0 f ( x ) x k \lim_{x\to0}\frac{f(x)}{x^k} x → 0 lim x k f ( x ) 收敛到非零值的整数 k k k
II. 对定义在 x y xy x y
f ( x , y ) = x 3 − 12 x y + 8 y 3 , f(x,y)=x^3-12xy+8y^3, f ( x , y ) = x 3 − 12 x y + 8 y 3 , 回答下列问题。
(1) 求 f f f
∂ f ∂ x , ∂ f ∂ y , \frac{\partial f}{\partial x},
\qquad
\frac{\partial f}{\partial y}, ∂ x ∂ f , ∂ y ∂ f , 以及 Hessian 矩阵。
(2) 求使 f f f 0 0 0 f f f
III. 令
Ω = { ( x , y , z ) ∈ R 3 | x ≥ 0 , y ≥ 0 , z ≥ 0 , x + y + z ≤ 1 } . \Omega=
\left\{(x,y,z)\in\mathbb{R}^3\,\middle|\,
x\geq0,\;y\geq0,\;z\geq0,\;x+y+z\leq1
\right\}. Ω = { ( x , y , z ) ∈ R 3 x ≥ 0 , y ≥ 0 , z ≥ 0 , x + y + z ≤ 1 } . 回答下列问题。
(1) 计算
∭ Ω d x d y d z . \iiint_{\Omega}dx\,dy\,dz. ∭ Ω d x d y d z . (2) 计算
∭ Ω d x d y d z ( 1 + x + y + z ) 3 . \iiint_{\Omega}\frac{dx\,dy\,dz}{(1+x+y+z)^3}. ∭ Ω ( 1 + x + y + z ) 3 d x d y d z . Kai (I)
f ( x ) = log ( 1 + x ) − 1 3 x 3 + 1 2 x 2 − x = ( x − x 2 2 + x 3 3 − x 4 4 + . . . ) − 1 3 x 3 + 1 2 x 2 − x = − x 4 4 + O ( x 5 ) f(x) = \log(1+x) - \frac{1}{3}x^3 + \frac{1}{2}x^2 - x = (x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...) - \frac{1}{3}x^3 + \frac{1}{2}x^2 - x = -\frac{x^4}{4} + O(x^5) f ( x ) = log ( 1 + x ) − 3 1 x 3 + 2 1 x 2 − x = ( x − 2 x 2 + 3 x 3 − 4 x 4 + ... ) − 3 1 x 3 + 2 1 x 2 − x = − 4 x 4 + O ( x 5 ) lim x → 0 f ( x ) x k \lim_{x \to 0} \frac{f(x)}{x^k} lim x → 0 x k f ( x ) k = 4 k=4 k = 4 − 1 / 4 -1/4 − 1/4
(II)
(1) ∂ f ∂ x = 3 x 2 − 12 y \frac{\partial f}{\partial x} = 3x^2 - 12y ∂ x ∂ f = 3 x 2 − 12 y ∂ f ∂ y = − 12 x + 24 y 2 \frac{\partial f}{\partial y} = -12x + 24y^2 ∂ y ∂ f = − 12 x + 24 y 2
∂ 2 f ∂ x 2 = 6 x , ∂ 2 f ∂ x ∂ y = − 12 , ∂ 2 f ∂ y 2 = 48 y \frac{\partial^2 f}{\partial x^2} = 6x, \frac{\partial^2 f}{\partial x \partial y} = -12, \frac{\partial^2 f}{\partial y^2} = 48y ∂ x 2 ∂ 2 f = 6 x , ∂ x ∂ y ∂ 2 f = − 12 , ∂ y 2 ∂ 2 f = 48 y Hesse matrix:
H = ( 6 x − 12 − 12 48 y ) H = \begin{pmatrix} 6x & -12 \\ -12 & 48y \end{pmatrix} H = ( 6 x − 12 − 12 48 y ) (2) ∂ f ∂ x = 0 ⟹ x 2 = 4 y \frac{\partial f}{\partial x} = 0 \implies x^2 = 4y ∂ x ∂ f = 0 ⟹ x 2 = 4 y ∂ f ∂ y = 0 ⟹ x = 2 y 2 \frac{\partial f}{\partial y} = 0 \implies x = 2y^2 ∂ y ∂ f = 0 ⟹ x = 2 y 2 ( 2 y 2 ) 2 = 4 y ⟹ 4 y 4 = 4 y ⟹ y 4 = y ⟹ y ( y 3 − 1 ) = 0 (2y^2)^2 = 4y \implies 4y^4 = 4y \implies y^4 = y \implies y(y^3-1) = 0 ( 2 y 2 ) 2 = 4 y ⟹ 4 y 4 = 4 y ⟹ y 4 = y ⟹ y ( y 3 − 1 ) = 0 y = 0 y=0 y = 0 y = 1 y=1 y = 1 y = 0 y=0 y = 0 x = 0 x = 0 x = 0 y = 1 y=1 y = 1 x = 2 x = 2 x = 2 ( 0 , 0 ) (0, 0) ( 0 , 0 ) ( 2 , 1 ) (2, 1) ( 2 , 1 ) H ( 0 , 0 ) = ( 0 − 12 − 12 0 ) H(0, 0) = \begin{pmatrix} 0 & -12 \\ -12 & 0 \end{pmatrix} H ( 0 , 0 ) = ( 0 − 12 − 12 0 ) det ( H ( 0 , 0 ) ) = − 144 < 0 \det(H(0, 0)) = -144 < 0 det ( H ( 0 , 0 )) = − 144 < 0 ( 0 , 0 ) (0, 0) ( 0 , 0 ) H ( 2 , 1 ) = ( 12 − 12 − 12 48 ) H(2, 1) = \begin{pmatrix} 12 & -12 \\ -12 & 48 \end{pmatrix} H ( 2 , 1 ) = ( 12 − 12 − 12 48 ) det ( H ( 2 , 1 ) ) = 12 ⋅ 48 − 144 = 576 − 144 = 432 > 0 \det(H(2, 1)) = 12 \cdot 48 - 144 = 576 - 144 = 432 > 0 det ( H ( 2 , 1 )) = 12 ⋅ 48 − 144 = 576 − 144 = 432 > 0 ∂ 2 f ∂ x 2 = 12 > 0 \frac{\partial^2 f}{\partial x^2} = 12 > 0 ∂ x 2 ∂ 2 f = 12 > 0 ( 2 , 1 ) (2, 1) ( 2 , 1 )
(III)
(1)
∭ Ω d x d y d z = ∫ 0 1 ∫ 0 1 − x ∫ 0 1 − x − y d z d y d x = ∫ 0 1 ∫ 0 1 − x ( 1 − x − y ) d y d x = ∫ 0 1 [ ( 1 − x ) y − y 2 2 ] y = 0 y = 1 − x d x = ∫ 0 1 ( 1 − x ) 2 − ( 1 − x ) 2 2 d x = ∫ 0 1 ( 1 − x ) 2 2 d x = [ − ( 1 − x ) 3 6 ] x = 0 x = 1 = 1 6 . \begin{aligned}
\iiint_{\Omega} dx\,dy\,dz
&= \int_0^1 \int_0^{1-x} \int_0^{1-x-y} dz\,dy\,dx \\
&= \int_0^1 \int_0^{1-x} (1-x-y)\,dy\,dx \\
&= \int_0^1 \Bigl[(1-x)y - \frac{y^2}{2}\Bigr]_{y=0}^{y=1-x} dx \\
&= \int_0^1 (1-x)^2 - \frac{(1-x)^2}{2}\,dx \\
&= \int_0^1 \frac{(1-x)^2}{2}\,dx \\
&= \Bigl[-\frac{(1-x)^3}{6}\Bigr]_{x=0}^{x=1} \\
&= \frac{1}{6}.
\end{aligned} ∭ Ω d x d y d z = ∫ 0 1 ∫ 0 1 − x ∫ 0 1 − x − y d z d y d x = ∫ 0 1 ∫ 0 1 − x ( 1 − x − y ) d y d x = ∫ 0 1 [ ( 1 − x ) y − 2 y 2 ] y = 0 y = 1 − x d x = ∫ 0 1 ( 1 − x ) 2 − 2 ( 1 − x ) 2 d x = ∫ 0 1 2 ( 1 − x ) 2 d x = [ − 6 ( 1 − x ) 3 ] x = 0 x = 1 = 6 1 . (2） 令 u = 1 + x + y + z u = 1+x+y+z u = 1 + x + y + z x + y + z = u − 1 x+y+z = u-1 x + y + z = u − 1 1 ≤ u ≤ 2 1 \le u \le 2 1 ≤ u ≤ 2
∭ Ω d x d y d z ( 1 + x + y + z ) 3 = ∫ 1 2 ∫ 0 u − 1 ∫ 0 u − 1 − x 1 u 3 d y d x d u = ∫ 1 2 1 u 3 ∫ 0 u − 1 ( u − 1 − x ) d x d u = ∫ 1 2 1 u 3 [ ( u − 1 ) x − x 2 2 ] x = 0 x = u − 1 d u = ∫ 1 2 1 u 3 ⋅ ( u − 1 ) 2 2 d u = 1 2 ∫ 1 2 u 2 − 2 u + 1 u 3 d u = 1 2 ∫ 1 2 ( 1 u − 2 u 2 + 1 u 3 ) d u = 1 2 [ ln u + 2 u − 1 2 u 2 ] u = 1 u = 2 = 1 2 [ ln 2 + 1 − 1 8 − ( 0 + 2 − 1 2 ) ] = 1 2 [ ln 2 − 5 8 ] = ln 2 2 − 5 16 . \begin{aligned}
\iiint_{\Omega} \frac{dx\,dy\,dz}{(1+x+y+z)^3}
&= \int_1^2 \int_0^{u-1} \int_0^{u-1-x} \frac{1}{u^3}\,dy\,dx\,du \\
&= \int_1^2 \frac{1}{u^3}
\int_0^{u-1} (u-1-x)\,dx\,du \\
&= \int_1^2 \frac{1}{u^3}
\Bigl[(u-1)x - \frac{x^2}{2}\Bigr]_{x=0}^{x=u-1} du \\
&= \int_1^2 \frac{1}{u^3} \cdot \frac{(u-1)^2}{2}\,du \\
&= \frac{1}{2} \int_1^2 \frac{u^2 - 2u + 1}{u^3}\,du \\
&= \frac{1}{2} \int_1^2 \left(\frac{1}{u} - \frac{2}{u^2} + \frac{1}{u^3}\right)\,du \\
&= \frac{1}{2} \Bigl[\ln u + \frac{2}{u} - \frac{1}{2u^2}\Bigr]_{u=1}^{u=2} \\
&= \frac{1}{2} \Bigl[\ln 2 + 1 - \frac{1}{8} - (0 + 2 - \tfrac{1}{2})\Bigr] \\
&= \frac{1}{2} \Bigl[\ln 2 - \frac{5}{8}\Bigr] \\
&= \frac{\ln 2}{2} - \frac{5}{16}.
\end{aligned} ∭ Ω ( 1 + x + y + z ) 3 d x d y d z = ∫ 1 2 ∫ 0 u − 1 ∫ 0 u − 1 − x u 3 1 d y d x d u = ∫ 1 2 u 3 1 ∫ 0 u − 1 ( u − 1 − x ) d x d u = ∫ 1 2 u 3 1 [ ( u − 1 ) x − 2 x 2 ] x = 0 x = u − 1 d u = ∫ 1 2 u 3 1 ⋅ 2 ( u − 1 ) 2 d u = 2 1 ∫ 1 2 u 3 u 2 − 2 u + 1 d u = 2 1 ∫ 1 2 ( u 1 − u 2 2 + u 3 1 ) d u = 2 1 [ ln u + u 2 − 2 u 2 1 ] u = 1 u = 2 = 2 1 [ ln 2 + 1 − 8 1 − ( 0 + 2 − 2 1 ) ] = 2 1 [ ln 2 − 8 5 ] = 2 ln 2 − 16 5 .