- Consider a sequence of closed intervals $\left\{I_n;~n\in\mathbb{Z}_+\right\}$, where $\mathbb{Z}_+$ is the set of positive integers (natural numbers). Suppose that for all $n\in\mathbb{Z}_+$, $I_{n+1}\subset I_n$. Let $I_n=\left[a_n,b_n\right]$. Suppose further that
$\displaystyle\lim_{n\to\infty}\left(b_n-a_n\right)=0$.
Show that the set$\displaystyle\bigcap_{n\in\mathbb{Z}_+}I_n$
contains exactly one element. - A function $f:\mathbb{R}\to\mathbb{R}$ is convex if for all $x,y\in\mathbb{R}$ and all $\pi\in\left(0,1\right)$,
$f\left(\pi x+\left(1-\pi\right)y\right)\leq\pi f\left(x\right)+\left(1-\pi\right)f\left(y\right)$.
Show that for all $x\in\mathbb{R}$, the left-hand derivative of $f$ at $x$,$\displaystyle f_-'\left(x\right)\equiv\lim_{h\downarrow0}\frac{f\left(x\right)-f\left(x-h\right)}h$
exists. (Hint: Show by the definition of convex functions that for all $a<x<b$, $a,x,b\in\mathbb{R}$,$\displaystyle\frac{f\left(b\right)-f\left(a\right)}{b-a}\leq\frac{f\left(b\right)-f\left(x\right)}{b-x}$
so that with $b$ fixed, the function$\displaystyle g\left(a\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$
is increasing in $a$; an increasing sequence which is bounded above always has a limit.) - Evaluate
$\displaystyle\iiint_c\sqrt{x^2+y^2+z^2}dV$,
where $c$ is the ice cream cone, and $\displaystyle\left\{\left(x,y,z\right)|x^2+y^2+z^2\leq1,~x^2+y^2\leq\frac{z^2}3,~z\geq0\right\}$. - Determine whether $\displaystyle\iint_D\frac1{x^2\left(1+y^2\right)}dA$ is convergent or divergent if $D=\left[1,+\infty\right)\times\left[0,+\infty\right)$.
- Suppose that $u=x^2+y^2$, $v=x^2-y^2$, $w=x^2y^2$, and $Z=u^3+v^3+w^3$, $Q=uvw$. Find $\displaystyle\frac{\partial Z}{\partial y}$, $\displaystyle\frac{\partial Q}{\partial x}$.
訣竅
此為區間套定理，為實數完備性的公設之一。觀察這些區間套的端點使用單調有界定理說明其收斂之值落於所有區間之中；最後運用區間長度趨於零來說明該點唯一。解法
由於 $I_{n+1}\subset I_n$，且按設定有 $\left[a_{n+1},b_{n+1}\right]\subset\left[a_n,b_n\right]$，這便表明 $a_{n+1}\geq a_n$ 且 $b_{n+1}\leq b_n$，從而說明數列 $\left\{a_n\right\}_{n=1}^{\infty}$ 遞增而 $\left\{b_n\right\}_{n=1}^{\infty}$ 遞減。再者 $a_n\leq b_n\leq b_1$，故數列 $\left\{a_n\right\}_{n=1}^{\infty}$ 遞增且有界，因而收斂，記此收斂之點為 $c$。另一方面，由 $\displaystyle\lim_{n\to\infty}\left(b_n-a_n\right)=0$ 搭配極限的四則運算定理可知
$\displaystyle\lim_{n\to\infty}b_n=\lim_{n\to\infty}\left[\left(b_n-a_n\right)+a_n\right]=0+c=c$
由兩數列的單調性，我們也有 $a_n\leq c\leq b_n$，這表明對所有 $n\in\mathbb{Z}_+$ 恆有 $c\in I_n$。至此我們說明了 $\displaystyle c\in\bigcap_{n\in\mathbb{Z}_+}I_n$。為了說明恰有一點落於其中，我們運用反證法，假設有兩相異點 $x\neq y$ 皆屬於 $\displaystyle\bigcap_{n\in\mathbb{Z}_+}I_n$。那麼可以知道對於所有 $n$ 恆有
$\left|x-y\right|\leq\left|b_n-a_n\right|$
兩邊同時取 $n$ 趨於無窮便有 $\left|x-y\right|\leq0$，這就表明了 $x=y$。此與反證法的假設矛盾，因而該集合僅恰有一元素。訣竅
沿著題目所給定的提示出發證明該不等式，隨後應用單調有界函數有單邊極限並適當的改寫為題目所欲證之形式即可。解法
由於 $x\in\left(a,b\right)$，因此取 $\displaystyle\pi=\frac{b-x}{b-a}\in\left(0,1\right)$ 可滿足 $x=\pi a+\left(1-\pi\right)b$。如此應用凸函數的定義立即可以注意到
$\displaystyle f\left(x\right)\leq\frac{b-x}{b-a}f\left(a\right)+\frac{x-a}{b-a}f\left(b\right)=\frac{b-x}{b-a}\left(f\left(a\right)-f\left(b\right)\right)+f\left(b\right)$
如此移項可得$\displaystyle\frac{b-x}{b-a}\left(f\left(b\right)-f\left(a\right)\right)\leq f\left(b\right)-f\left(x\right)$
同除以 $b-x$ 便得所欲證的不等式。那麼考慮函數$\displaystyle g\left(x\right)=\frac{f\left(b\right)-f\left(x\right)}{b-x}$
容易看出當 $x_1<x_2<b$ 時有不等式 $g\left(x_1\right)\leq g\left(x_2\right)$，這便說明了 $g$ 對 $x$ 遞增。而在 $\mathbb{R}$ 上單調函數的函數皆有單邊極限，即對於任意給定的 $x_0\in\mathbb{R}$ 取 $b=x_0$，左極限 $\displaystyle\lim_{x\to x_0^-}g\left(x\right)=\lim_{x\to x_0^-}\frac{f\left(x_0\right)-f\left(x\right)}{x_0-x}$ 存在。命 $h=x_0-x$，那麼便有
$\displaystyle\lim_{h\downarrow0}\frac{f\left(x_0\right)-f\left(x_0-h\right)}{h}$
存在，此即所欲證的。訣竅
將積分區域適當的表達後使用球座標變換處理之。解法
運用球面座標變換，令 $\left\{\begin{aligned}&x=\rho\cos\theta\sin\phi\\&y=\rho\sin\theta\sin\phi\\&z=\rho\cos\phi\end{aligned}\right.$，那麼由 $x^2+y^2+z^2\leq1$ 則給出 $0\leq\rho\leq1$，$\displaystyle x^2+y^2\leq\frac{z^2}3$ 則可寫成 $\displaystyle\rho^2\sin^2\phi\leq\frac{\rho^2\cos^2\phi}3$，即 $\displaystyle\tan^2\phi\leq\frac13$，最後搭配 $z\geq0$ 可知 $\displaystyle0\leq\phi\leq\frac\pi6$。如此所求的三重積分可改寫並計算如下$\displaystyle\begin{aligned}\iint_c\sqrt{x^2+y^2+z^2}dV&=\int_0^{2\pi}\int_0^{\frac\pi6}\int_0^1\rho\cdot\rho^2\sin\phi d\rho d\phi d\theta\\&=\left(\int_0^{2\pi}d\theta\right)\left(\int_0^{\frac\pi6}\sin\phi d\phi\right)\left(\int_0^1\rho^3d\rho\right)=2\pi\cdot\left(1-\frac{\sqrt3}2\right)\cdot\frac14=\frac{\left(2-\sqrt3\right)\pi}4\end{aligned}$
訣竅
應嚴密的使用瑕重積分的定義處理而不應直接分離變量逕自計算之，因為尚未確認其收斂性下不應當直接計算。解法
首先先回憶起：定義在 $\mathbb{R}^2$ 上的被積分函數非負時，其歛散性由下列極限所決定$\displaystyle\lim_{a\to\infty}\iint_{\left[-a,a\right]^2}f\left(x,y\right)dA$
為了使用此定義，我們可先將 $f$ 擴充至整個 $\mathbb{R}^2$ 如下：$\tilde{f}\left(x,y\right)=\begin{cases}f\left(x,y\right),&\left(x,y\right)\in D,\\0,&\left(x,y\right)\in\mathbb{R}^2-D.\end{cases}$
那麼我們所欲探求的瑕積分所對應的極限可計算如下$\displaystyle\begin{aligned}\iint_Df\left(x,y\right)dA&=\iint_{\mathbb{R}^2}\tilde{f}\left(x,y\right)dA=\lim_{a\to\infty}\iint_{\left[-a,a\right]^2}\tilde{f}\left(x,y\right)dA\\&=\lim_{a\to\infty}\int_1^a\int_0^a\frac1{x^2\left(1+y^2\right)}dydx=\lim_{a\to\infty}\left(\int_1^a\frac{dx}{x^2}\right)\left(\int_0^a\frac{dy}{1+y^2}\right)\\&=\lim_{a\to\infty}\left(1-\frac1a\right)\tan^{-1}a=\frac\pi2\end{aligned}$
因此該瑕重積分收斂。訣竅
運用多變數函數的連鎖律求解即可。解法
運用連鎖律可知$\displaystyle\begin{aligned}\frac{\partial Z}{\partial y}&=\frac{\partial}{\partial y}\left(u^3+v^3+w^3\right)=3u^2\frac{\partial u}{\partial y}+3v^2\frac{\partial v}{\partial y}+3w^2\frac{\partial w}{\partial y}\\&=3u^2\cdot 2y+3v^2\cdot\left(-2y\right)+3w^2\cdot2x^2y=6\left(x^2+y^2\right)^2y-6\left(x^2-y^2\right)^2y+6\left(x^2y^2\right)^2x^2y=24x^2y^3+6x^6y^5\end{aligned}$
以及$\displaystyle\begin{aligned}\frac{\partial Q}{\partial x}&=\frac{\partial u}{\partial x}vw+u\frac{\partial v}{\partial x}w+uv\frac{\partial w}{\partial x}\\&=2x\left(x^2-y^2\right)x^2y^2+\left(x^2+y^2\right)\cdot2x\cdot x^2y^2+\left(x^2+y^2\right)\left(x^2-y^2\right)\cdot2xy^2\\&=6x^5y^2-4x^3y^4-2xy^6\end{aligned}$
沒有留言:
張貼留言