115年政治大學精算碩士班-微積分詳解

 國立政治大學115學年度碩士班暨碩士在職專班招生考試試題

考試科目：微積分
系所別：風險管理與保險學系 精算科學組

解答:$$\textbf{(a) }\lim_{x\to 0} f(x) = \lim_{x\to 0}{\sin x\over x}=1 \Rightarrow f(0)=1=a \Rightarrow \bbox[red, 2pt]{a=1} \\\textbf{(b) } f'(0)=\lim_{h\to 0} {f(h)-f(0) \over h} =\lim_{h\to 0} {{\sin h\over h}-1 \over h} = \lim_{h\to 0} {\sin h-h \over h^2} = \lim_{h\to 0} {\cos h-1 \over 2h} =\lim_{h\to 0} {-\sin h \over 2} = 0 \\\qquad \Rightarrow \bbox[red,2pt]{f'(0)\text{ exists, and its value is 0}} $$

解答:$$0\le {x^2\over x^2+y^2} \le 1 \Rightarrow \left| {x^2\over x^2+y^2} \right| \le 1 \Rightarrow  0\le|y|\left| {x^2\over x^2+y^2} \right|\le |y| \Rightarrow 0\le \left| {x^2y \over x^2+y^2} \right| \le |y|\\ \text{Applying the Squeeze Theorem, } \lim_{(x,y)\to(0,0)} |y|=0 \Rightarrow \lim_{(x,y)\to(0,0)} \left| {x^2\over x^2+y^2} \right| =0 \\ \Rightarrow \bbox[red, 2pt]{\text{ The limit exists, and }\lim_{(x,y)\to(0,0)}{x^2\over x^2+y^2} =0}$$

解答:$$\text{By Comparison Test, }\cases{\displaystyle {\ln n\over n} \gt {1\over n}, \text{ for }n\ge 3 \\ \displaystyle \sum{1\over n} \text{diverges}} \Rightarrow \sum_{n=1}^\infty   {\ln n\over n} \bbox[red, 2pt]{\text{ diverges}}$$

解答:$$\lim_{x\to 0} {e^x-1-x\over x^2} =\lim_{x\to 0} {(e^x-1-x)'\over (x^2)'} = \lim_{x\to 0} {e^x-1\over 2x} = \lim_{x\to 0} {(e^x-1)'\over (2x)'} =  \lim_{x\to 0} {e^x\over 2} = \bbox[red, 2pt]{1\over 2}$$

解答:$$x=\tan \theta \Rightarrow dx =\sec^2 \theta \,d\theta \Rightarrow I=\int_0^1 {\ln(1+x) \over 1+x^2}\,dx = \int_0^{\pi/4} {\ln(1+\tan \theta) \over 1+\tan^2 \theta} \sec^2\theta\,d\theta  \\= \int_0^{\pi/4} \ln(1+\tan \theta)\,d \theta = \int_0^{\pi/4} \ln(1+\tan ({\pi\over 4}-\theta))\,d\theta  = \int_0^{\pi /4} \ln(1+{1-\tan \theta\over 1+\tan \theta})\,d\theta \\= \int_0^{\pi/4} \ln{2\over 1+\tan \theta}\,d\theta =\int_0^{\pi/4} \ln 2\,d\theta- \int_0^{\pi /4}\ln(1+\tan \theta)\,d\theta =\int_0^{\pi/4} \ln 2\,d\theta-I \\ \Rightarrow 2I= \int_0^{\pi/4} \ln 2\,d\theta ={\pi\over 4}\ln 2 \Rightarrow I= \bbox[red, 2pt]{{\pi\over 8}\ln 2}$$

解答:$$\iint_D (x+y)\,dA = \int_0^1 \int_0^{1-y} (x+y) \,dx\,dy = \int_0^1 \left. \left[ {1\over 2}x^2+xy \right] \right|_0^{1-y} \,dy ={1\over 2}\int_0^1(1-y^2)\,dy \\={1\over 2} \left. \left[ y-{1\over 3}y^3 \right] \right|_0^1 = \bbox[red, 2pt]{1\over 3}$$

解答:$$\textbf{(a) }\textbf{Case 1: }x=0 \Rightarrow f_n(0) =0 \Rightarrow \lim_{n\to \infty} f_n(0)=0 \\ \textbf{Case 2: }x\ne 0 \Rightarrow \lim_{n\to \infty}f_n(x) =\lim_{n\to \infty}{x\over 1+nx^2}  =\lim_{n\to \infty}{1\over (1/x)+nx} ={1\over \infty}=0 \\\text{Since the limit is 0 for all real numbers }x, \text{ the pointwise limit is: } f(x)=\bbox[red, 2pt]0, \text{ for all }x\in \mathbb R \\\textbf{(b) }f_n(x)= {x\over 1+nx^2} \Rightarrow f'_n(x)= {1-nx^2\over (1+nx^2)^2} =0 \Rightarrow x=\pm {1\over \sqrt n} \\ \Rightarrow \left| f_n \left( \pm{1\over \sqrt n} \right) \right|= \left|   {\pm {1\over \sqrt n} \over 1+n (\pm {1\over \sqrt n})^2} \right| ={1\over 2\sqrt n} \Rightarrow M_n = \text{sup}_{x\in \mathbb R} \left|f_n(x) \right| ={1\over 2\sqrt n} \\ \Rightarrow \lim_{n\to \infty} M_n=  \lim_{n\to \infty} {1\over 2\sqrt n} =0 \Rightarrow \text{ the sequence }\{f_n\} \text{ converges uniformly  to }f(x)=0 \text{ on }\mathbb R \\\textbf{(c) }\text{From (a), we have }f(x) =0 \Rightarrow f'(x)=0 \text{ for all }x\in \mathbb R \\ \text{From (b), we have }f_n'(x) ={1-nx^2 \over (1+nx^2)^2} \Rightarrow f_n'(0) =1 \Rightarrow \lim_{n\to \infty} f_n'(0) \ne f'(0) \\ \Rightarrow \text{The limit and differentiation }\bbox[red, 2pt]{\text{ cannot be interchanged.}}$$

解答:$$\textbf{(a) }J(a)= \int_0^1 x^a\,dx = \left. \left[ {x^{a+1} \over a+1} \right] \right|_0^1={1\over a+1} \\ \Rightarrow J'(a)=\int_0^1 {\partial \over \partial a}x^a\,dx = {d\over da} \left( {1\over a+1} \right) \Rightarrow \int_0^1 x^a \ln x\,dx =-{1\over (a+1)^2} \Rightarrow \bbox[red, 2pt]{I(a)=-{1\over (a+1)^2}} \\\textbf{(b) } \lim_{a\to 0}I(a) =\lim_{a\to 0} \left(-{1\over (a+1)^2} \right)= \bbox[red, 2pt]{-1}$$

解答:$$\int_0^1 \int_y^1 {1\over 1+x^2} \,dx dy = \int_0^1 \int_0^x {1\over 1+x^2} \,dy dx = \int_0^1   {x\over 1+x^2} \,  dx = \left. \left[ {1\over 2}\ln(1+x^2) \right] \right|_0^1 = \bbox[red, 2pt]{{1\over 2}\ln 2}$$

 解答:$$\textbf{(a) } x^{n^2} \ge 0, \text{ for all }x \in [0,1], \text{ the sequence of partial sums }\\S_N(x) =\sum_{n=1}^N x^{n^2} \text{ is monotonically increasing as N increases.} \\\text{By the Monotone Convergence Theorem, the limit and the integral can be interchanged.} \\\textbf{(b) } \int_0^1 x^{n^2}\,dx = \left. \left[ {x^{n^2+1} \over n^2+1} \right] \right|_0^1 ={1\over n^2+1} \\\qquad \text{By Fourier series, we have } \sum_{-\infty}^\infty {1\over n^2+a^2} ={\pi\over a} \coth(a\pi)\\ \Rightarrow \sum_{n=-\infty}^\infty {1\over n^2+1} =1+ 2 \sum_{n=1}^\infty {1\over n^2+1}= \pi \coth(\pi) \Rightarrow \sum_{n=1}^\infty {1\over n^2+1} =\bbox[red, 2pt]{{1\over 2}(\pi \coth(\pi)-1)}$$

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解題僅供參考，碩士班歷年試題及詳解