- Evaluate:
- $\displaystyle\lim_{x\to0^+}\frac{\ln\cos3x}{\ln\cos2x}$, ($5\%$)
- $\displaystyle\int_0^{\pi}\frac{x\sin x}{1+\cos^2x}dx$, ($5\%$)
- $\displaystyle\int_0^1\frac{1-e^{-x^2}}{x^2}dx$, ($10\%$)
- 由羅必達法則可知
$\displaystyle\lim_{x\to0^+}\frac{\ln\cos3x}{\ln\cos2x}=\lim_{x\to0^+}\frac{-3\sin3x/\cos3x}{-2\sin2x/\cos2x}=\frac32\lim_{x\to0^+}\frac{\sin3x}{\sin2x}\cdot\lim_{x\to0^+}\frac{\cos2x}{\cos3x}=\frac32\lim_{x\to0^+}\frac{3\cos3x}{2\cos2x}=\frac94$
- 令 $\displaystyle u=\pi-x$，那麼有
- 當 $x=0$ 時有 $u=\pi$；
- 當 $x=\pi$ 時有 $u=0$；
- 求導有 $du=-dx$，且有 $\sin x=\sin u$，而 $\cos x=-\cos u$。
$\displaystyle\int_0^{\pi}\frac{x\sin x}{1+\cos^2x}dx=\int_{\pi}^0\frac{\left(\pi-u\right)\sin u}{1+\cos^2u}\cdot-du=\int_0^{\pi}\frac{\left(\pi-u\right)\sin u}{1+\sin^2u}du$
透過啞變數變換可知$\displaystyle\int_0^{\pi}\frac{x\sin x}{1+\cos^2x}dx=\int_0^{\pi}\frac{\left(\pi-x\right)\sin x}{1+\cos^2x}dx$
移項便知$\displaystyle\int_0^{\pi}\frac{x\sin x}{1+\cos^2x}dx=\frac\pi2\int_0^{\pi}\frac{\sin x}{1+\cos^2xdx}=\left.-\frac\pi2\tan^{-1}\left(\cos x\right)\right|_0^{\pi}=\frac{\pi^2}4$
- 運用分部積分法可知
$\displaystyle\int_0^1\frac{1-e^{-x^2}}{x^2}dx=\left.-\frac{1-e^{-x^2}}x\right|_0^1+\int_0^1\frac{2xe^{-x^2}}xdx=e^{-1}-1+2\int_0^1e^{-x^2}dx$
而後者的定積分由積分學的經驗可知無法使用初等函數來表達或化簡。 - Prove that if $0<\kappa<1$, ($10\%$)
$\displaystyle K\left(\kappa\right)=\int_0^{\pi/2}\frac{d\theta}{\sqrt{1-\kappa^2\sin^2\theta}}=\frac\pi2\left\{1+\left(\frac12\right)^2\kappa^2+\left(\frac{1\times3}{2\times4}\right)^2\kappa^4+\left(\frac{1\times3\times5}{2\times4\times6}\right)^2\kappa^6+\cdots\right\}.$
- Prove directly that a polynomial function $f:\mathbb{R}^n\to\mathbb{R}$ is uniformly continuous on the set $\left\{\left|x\right|\leq K\right\}$ for any fixed $K$. [First show it is bounded, i.e., $\left|f\left(x\right)\right|<K'$. Then consider $f\left(x+h\right)-f\left(x\right)$.] ($10\%$)
- Suppose that the sequence $\left\{a_n\right\}_{n=1}^{\infty}$ satisfies the following condition: ($10\%$)
There is an $r$, $0<r<1$, such that$\left|a_{n+1}-a_n\right|<br^n$, $n=1,2,\cdots,$
where $b$ is a positive constant. Show that this sequence convergences. - ($10\%$) Find the function whose tangent line has slope $x\ln\sqrt{x}$ for each value of $x>0$ and whose graph passes through the point $\left(2,-5\right)$.
- ($10\%$) Find the volume of the solid generated when the region under the curve $y=\tan x$ over the region $0\leq x\leq\pi/4$ is rotated about the $x$ axis.
- ($10\%$) Find the least squares approximation $h\left(x\right)=a_0+a_0x+a_0x^2$ for $f\left(x\right)=e^x$, $0\leq x\leq1$. (If the answers are with decimal numbers, please round to the nearest hundredth.)
[Hint: Minimize the integral of the squared difference between $f\left(x\right)$ and $h\left(x\right)$.] - ($10\%$) Determine the value of $c$ for which the following series converges.
$\displaystyle\sum_{k=3}^{\infty}\frac1{k\left(\ln k\right)^c}$
- 當 $c>1$ 時有
$\displaystyle\int_3^{\infty}f\left(x\right)dx=\int_3^{\infty}\frac{dx}{x\left(\ln x\right)^c}=\left.\frac{\left(\ln x\right)^{-c+1}}{-c+1}\right|_3^{\infty}=\frac{\left(\ln3\right)^{1-c}}{1-c}<\infty$.
此時給定的級數收斂。 - 當 $c=1$ 時有
$\displaystyle\int_3^{\infty}f\left(x\right)dx=\int_3^{\infty}\frac{dx}{x\left(\ln x\right)}=\ln\left(\ln x\right)\Big|_3^{\infty}=\infty$.
此時給定的級數發散。 - 當 $c<1$ 時有
$\displaystyle\int_3^{\infty}f\left(x\right)dx=\int_3^{\infty}\frac{dx}{x\left(\ln x\right)^c}=\left.\frac{\left(\ln x\right)^{-c+1}}{-c+1}\right|_3^{\infty}=\infty$.
此時給定的級數發散。 - ($10\%$) A company's demand equation is $x=\sqrt{2000-p^2}$, where $p$ is the price in dollars and $x$ is the quantity of demand. Find $dp/dx$ and $d\ln\left(p\right)/d\ln\left(x\right)$ when $p=40$ and interpret your answer.
訣竅
分別運用羅必達法則、變數變換與分部積分法求解即可。解法
訣竅
運用廣義二項式定理將給定的被積分函數進行展開後進行逐項積分即可。解法
首先當 $\theta\in\left[0,\pi/2\right]$ 且 $0<\kappa<1$，容易知道 $0\leq\kappa^2\sin^2\theta\leq\kappa^2<1$，從而給定的被積分函數可透過廣義二項式定理表達如下$\displaystyle\frac1{\sqrt{1-\kappa^2\sin^2\theta}}=\left(1-\kappa^2\sin^2\theta\right)^{-1/2}=1+\sum_{n=1}^{\infty}{-1/2\choose n}\left(-\kappa^2\sin^2\theta\right)^n=1+\sum_{n=1}^{\infty}\frac{\kappa^{2n}\left(2n\right)!}{2^{2n}\left(n!\right)^2}\sin^{2n}\theta$
另一方面，對於高次方正弦函數的積分，我們運用遞迴式計算如下：設 $\displaystyle I_m=\int_0^{\pi/2}\sin^m\theta d\theta$，那麼可知$\displaystyle I_2=\int_0^{\pi/2}\sin^2\theta d\theta=\int_0^{\pi/2}\frac{1-\cos2\theta}2d\theta=\frac\pi4$
再者也有$\displaystyle\begin{aligned}I_{m+2}&=\int_0^{\pi/2}\sin^{m+2}\theta d\theta=\int_0^{\pi/2}\sin^{m+1}\theta\sin\theta d\theta\\&=-\sin^{m+1}\theta\cos\theta\Big|_0^{\pi/2}+\left(m+1\right)\int_0^{\pi/2}\sin^m\theta\cos^2\theta d\theta\\&=\left(m+1\right)\int_0^{\pi/2}\sin^m\theta\left(1-\sin^2\theta\right)d\theta=\left(m+1\right)I_m-\left(m+1\right)I_{m+2}\end{aligned}$
如此便有 $\displaystyle I_{m+2}=\frac{m+1}{m+2}I_m$。故容易發現$\displaystyle I_{2n}=\frac{2n-1}{2n}I_{2n-2}=\frac{\left(2n-1\right)\left(2n-3\right)}{2n\left(2n-2\right)}I_{2n-4}=\cdots=\frac{\left(2n-1\right)\cdots3}{2n\cdots4}I_2=\frac\pi2\frac{\left(2n-1\right)\cdots1}{2n\cdots2}=\frac\pi2\frac{\left(2n\right)!}{2^{2n}\left(n!\right)^2}$
回到所求便有$\displaystyle\begin{aligned}K\left(\kappa\right)&=\int_0^{\pi2}\left[1+\sum_{n=1}^{\infty}\frac{\kappa^{2n}\left(2n\right)!}{2^{2n}\left(n!\right)^2}\sin^{2n}\theta\right]d\theta=\frac\pi2+\sum_{n=1}^{\infty}\frac{\kappa^{2n}\left(2n!\right)}{2^{2n}\left(n!\right)}I_{2n}\\&=\frac\pi2+\frac\pi2\sum_{n=1}^{\infty}\left[\frac{\left(2n\right)!}{2^{2n}\left(n!\right)^2}\right]^2\kappa^{2n}=\frac\pi2\left\{1+\left(\frac12\right)^2\kappa^2+\left(\frac{1\cdot3}{2\cdot4}\right)^2\kappa^4+\cdots\right\}\end{aligned}$
訣竅
注意到多項式函數求導後仍為多項式函數，此外我們須使用均值定理來處理均勻性。解法
首先注意到任給一 $m$ 次多項式 $\displaystyle g\left(x\right)=\sum_{k=0}^ma_kx^k$，那麼容易由三角不等式知道$\displaystyle\left|g\left(x\right)\right|\leq\sum_{k=0}^m\left|a_k\right|\left|x\right|^k\leq\max_{0\leq k\leq m}\left|a_k\right|\sum_{k=0}^m\left|K\right|^m$
故在集合上 $\left\{\left|x\right|\leq K\right\}$ 的任何多項式 $g$ 都有上界（取決於多項式 $g$ 的係數與集合 $\left\{\left|x\right|\leq K\right\}$ 的大小）。現給定正數 $\varepsilon>0$，取 $\displaystyle\delta=\frac{\varepsilon}{\sqrt{n}M}$，此處 $M$ 為
$M:=\max\left\{1,M_1,\dots,M_n\right\}$
而$\displaystyle M_i:=\max_{\left\{\left|x\right|\leq K\right\}}\left|f_{x_i}\left(x\right)\right|$
循此可知當 $\left|x-y\right|<\delta$ 時有$\displaystyle\left|f\left(x\right)-f\left(y\right)\right|=\left|\nabla f\left(c\right)\cdot\left(x-y\right)\right|\leq\sqrt{M_1^2+\cdots+M_n^2}\cdot\frac{\varepsilon}{\sqrt{n}M}\leq\varepsilon$
證明完畢。訣竅
透過條件證明該數列為柯西數列。解法
對給定的正數 $\varepsilon$，可取 $\displaystyle N>\log_r\frac{\left(1-r\right)\varepsilon}b$，那麼當 $n>m>N$ 時有$\displaystyle\left|a_m-a_n\right|\leq\left|a_m-a_{m+1}\right|+\cdots+\left|a_{n-1}-a_n\right|<br^m+\cdots+br^{n-1}=b\sum_{k=m}^{n-1}r^k<\frac{br^m}{1-r}<\frac{br^N}{1-r}<\varepsilon$
這說明了給定的數列為柯西數列，因此該數列收斂。訣竅
按照題意可列出導函數的關係式，運用微積分基本定理求出反導函數並透過通過的座標獲得該函數。解法
由題意可知 $y'\left(x\right)=x\ln\sqrt{x}$，那麼運用微積分基本定理可知$\displaystyle\begin{aligned}y\left(x\right)&=y\left(2\right)+\int_2^xt\ln\sqrt{t}dt=-5+\frac12\int_2^xt\ln tdt=-5+\frac14\left(t^2\ln t\Big|_2^x-\int_2^xtdt\right)\\&=-5+\frac{x^2\ln x-4\ln2}4-\frac{x^2-4}8=\frac{2x^2\ln x-x^2-36-8\ln2}8\end{aligned}$
訣竅
運用旋轉體體積公式即可。解法
使用旋轉體體積公式可知$\displaystyle V=\int_0^{\pi/4}\pi y^2dx=\pi\int_0^{\pi/4}\tan^2xdx=\pi\int_0^{\pi/4}\left(\sec^2x-1\right)dx=\pi\left(\tan x-x\right)\Big|_0^{\pi/4}=\frac{\pi\left(4-\pi\right)}4$
訣竅
按其要求考慮適當的函數並透過微分求其極值。解法
考慮函數$\displaystyle g\left(a_0\right)=\int_0^1\left(a_0+a_0x+a_0x^2-e^x\right)^2dx$
對 $a_0$ 求導可知$\displaystyle g'\left(a_0\right)=2\int_0^1\left(a_0+a_0x+a_0x^2-e^x\right)\left(1+x+x^2\right)dx=2a_0\int_0^1\left(1+x+x^2\right)^2dx-2\int_0^1e^x\left(1+x+x^2\right)dx$
故解方程式 $g'\left(a_0\right)=0$ 有$\displaystyle a_0=\frac{\displaystyle\int_0^1e^x\left(1+x+x^2\right)dx}{\displaystyle\int_0^1\left(1+x+x^2\right)^2dx}:=a_0^*$
再者求二階導函數有 $\displaystyle g''\left(a_0\right)=2\int_0^1\left(1+x+x^2\right)^2dx>0$，這表明 $g$ 在 $a=a_0^*$ 處有絕對最小值。而 $a_0^*$ 的確切值可計算如下：
$\displaystyle\int_0^1\left(1+x+x^2\right)^2dx=\int_0^1\left(1+2x+3x^2+2x^3+x^4\right)dx=\left.x+x^2+x^3+\frac{x^4}4+\frac{x^5}5\right|_0^1=\frac{69}{20}$
$\displaystyle\int_0^1e^x\left(1+x+x^2\right)dx=\left(1+x+x^2\right)e^x\Big|_0^1-\int_0^1e^x\left(1+2x\right)dx=\left(3e-1\right)-e^x\left(1+2x\right)\Big|_0^1+\int_0^12e^xdx=2e^x\Big|_0^1=2e-2$
故 $a_0^*=\frac{10\left(e-1\right)}{69}$，從而在 $\left[0,1\right]$ 上能最佳逼近 $e^x$ 的二次函數為$\displaystyle h\left(x\right)=\frac{10\left(e-1\right)}{69}\left(1+x+x^2\right)$
訣竅
透過積分審歛法來考慮即可。解法
當 $c\in\mathbb{R}$ 時可以知道 $\displaystyle a_k:=\frac1{k\left(\ln k\right)^c}$ 在 $k$ 充分大時遞減且趨於零，那麼設 $\displaystyle f\left(x\right)=\frac1{x\left(\ln x\right)^c}$ 可容易發現訣竅
可改寫需求函數後逕自求導，此外尚需要使用連鎖律計算。解法
整理可知 $p=\sqrt{2000-x^2}$（由於價格恆正，故取正根），故求導有$\displaystyle\frac{dp}{dx}=\frac{-x}{\sqrt{2000-x^2}}$
又當 $p=40$ 時有 $x=\sqrt{2000-40^2}=20$，因此 $\displaystyle\left.\frac{dp}{dx}\right|_{p=40}=-\frac12$。再者，使用連鎖律可知$\displaystyle\frac{d\ln\left(p\right)}{d\ln\left(x\right)}=\frac{d\ln\left(p\right)}{dp}\frac{dp}{dx}\frac{dx}{d\ln\left(x\right)}=\frac1p\cdot\frac{-x}{\sqrt{2000-x^2}}\cdot\frac1{1/x}=\frac{-x^2}{p\sqrt{2000-x^2}}$
故代入 $p=40$ 與 $x=20$ 有 $\displaystyle\left.\frac{d\ln\left(p\right)}{d\ln\left(x\right)}\right|_{p=40}=-\frac14$。以上的計算表示在價格為 $p=40$ 時，當價格上升的瞬間時，需求將以兩單位減少。而在該處的點需求彈性為 $4$，這表明該商品是相對具有彈性的，即當價格改變時其需求是容易變動的。
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