114年政大應數轉學考-微積分詳解

 國立政治大學114學年度學士班轉學生招生考試

科目:微積分
系所別:應用數學系三年級

解答:$$\textbf{(a) } \lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{(\ln x)'}{(1/x)'} =\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0 \\ \Rightarrow  \lim_{x \to 0^+} x^x = \lim_{x \to 0^+} e^{x \ln x} = e^0 = 1 \Rightarrow \bbox[red, 2pt]{\text{Yes, it is possible.}} \text{ We define }f(0)=1, \text{then } \\\qquad f\text{ becomes a continuous function.} \\ \textbf{(b) }f(x) =x^x =e^{x\ln x} \Rightarrow  f'(x) = \frac{d}{dx} (e^{x \ln x}) = e^{x \ln x} \cdot \frac{d}{dx} (x \ln x) = e^{x\ln x} (\ln x+1) \\\qquad \Rightarrow \bbox[red, 2pt]{f'(x)=x^x(\ln x+1) }\Rightarrow  f''(x) = \frac{d}{dx} [x^x (\ln x + 1)] = \left(\frac{d}{dx} x^x\right)(\ln x + 1) + x^x \left(\frac{d}{dx}(\ln x + 1)\right)  \\ \qquad =  [x^x(\ln x + 1)](\ln x + 1) + x^x \left( \frac{1}{x}\right) \Rightarrow \bbox[red, 2pt]{f''(x)=x^x (\ln x+1)+x^{x-1}} \\\textbf{(c) } f'(x)=0 \Rightarrow x^x(\ln x+1)=0 \Rightarrow \ln x=-1 \Rightarrow x={1\over e}  \Rightarrow f''({1/e}) =(1/e)^{1/e-1} \gt 0\\ \Rightarrow \text{local minimum: }f(1/e)=(1/e)^{1/e}=e^{-1/e} \Rightarrow \begin{cases}f'(x)\lt 0,& 0\lt x\lt 1/e\\f'(x)\gt 0, & x\gt 1/e \end{cases} \\\Rightarrow f(1/e) \text{ is global minimum}; \lim_{x\to \infty}f(x) =\infty \Rightarrow \text{no maximum}\\ \Rightarrow \bbox[red, 2pt]{\text{Yes, $f$ has a minimum at $x=1/e, $ no maximum. And $f(1/e)$ is a global minimum.}}$$

解答:$$y=\sin^{-1}(x) \Rightarrow \sin y=x \Rightarrow {d\over dx} \sin y={d\over dx}x \Rightarrow {dy\over dx} \cos y=1 \\\Rightarrow {dy\over dx}={1\over \cos y} ={1\over \sqrt{1-\sin^2 y}} = \bbox[red, 2pt]{1\over \sqrt{1-x^2}}$$

解答:$$\text{Let }x_n={1\over n\pi}, n\in \mathbb N \Rightarrow \lim_{n\to \infty} x_n=0 \Rightarrow  \lim_{n\to\infty} f(x_n) = \lim_{n\to\infty} \sin(n\pi) = 0\\ \text{Let } y_n = \frac{1}{2n\pi + \frac{\pi}{2}} , n\in \mathbb N \Rightarrow \lim_{n\to \infty} y_n=0 \Rightarrow  \lim_{n\to\infty} f(y_n) = \lim_{n\to\infty} \sin \left( 2n\pi+{\pi\over 2} \right)=1\\ \text{Therefore, }\lim_{n\to\infty} f(x_n) \ne \lim_{n\to\infty} f(y_n)  \Rightarrow \lim_{x\to 0} \sin \left( {1\over x} \right) \text{ does not exist} \quad \bbox[red, 2pt]{QED.}$$

解答:$$\textbf{(a) }  \text{Let $x_0 = \frac{p_0}{q_0} \in (0,1)$ be an arbitrary rational number with $\gcd(p_0, q_0) = 1$}. \\ \quad \text{Then } f(x_0) = \frac{1}{q_0} > 0 . \text{ Since the irrational numbers are dense in $\mathbb{R}$, for every positive integer $n$,}\\ \quad \text{ there exists an irrational number $x_n \in \left(x_0 - \frac{1}{n}, x_0 + \frac{1}{n}\right) \cap (0, 1)$.}\\ \quad \text{By the Squeeze Theorem, }\lim_{n\to \infty} x_n =x_0. \text{ Since each $x_n$ is irrational,  } f(x_n) = 0 \quad \text{for all } n \in \mathbb{N} \\ \text{Then we have } \lim_{n \to \infty} f(x_n) = 0 \ne f(x_0)={1\over q_0}\gt 0.  \text{ Since the limit along this sequence does not }\\ \quad \text{equal $f(x_0)$, $f$ is not continuous at any rational point $x \in (0,1)$.}  \quad \bbox[red, 2pt]{QED.} \\ \textbf{(b) }  \text{Let $x_0 \in (0,1)$ be an arbitrary irrational number }\Rightarrow f(x_0)=0\\ \text{By the Archimedean property, choose a positive integer $N$ such that $\frac{1}{N} < \varepsilon$.} \\\quad \text{Let } S_N = \left\{ \frac{p}{q} \in (0,1) \;\middle\vert{}\; p, q \in \mathbb{N},\, \gcd(p,q)=1,\, q \le N \right\} , \text{ then $S_N$ contains only a finite} \\ \text{ number of elements. Define } \delta = \min \left\{ \left\vert{} x_0 - \frac{p}{q} \right\vert{} : \frac{p}{q} \in S_N \right\} \gt 0\\ \text{Now, for any $x \in (0,1)$ satisfying $\vert{}x - x_0\vert{} < \delta$: }\cases{x \text{ is irrational} \Rightarrow f(x)=0 \lt \varepsilon \\ x\text{ is rational} \Rightarrow x=p/q, \text{ gcd}(p,q)=1\\\qquad   \vert{}x - x_0\vert{} < \delta\Rightarrow x\not \in S_N \Rightarrow q\gt N \Rightarrow \\ \qquad  f(x) = \frac{1}{q} < \frac{1}{N} < \varepsilon } \\ \quad \Rightarrow   \vert{}f(x) - f(x_0)\vert{} = f(x) < \varepsilon  \Rightarrow \text{Hence, by the $\varepsilon$-$\delta$ definition of continuity,}\\ \qquad \text{ $f$ is continuous at every irrational point $x \in (0,1)$. } \quad \bbox[red, 2pt]{QED.}$$

解答:$$\cases{f(x, y, z) = xyz \\ g(x, y, z) = x^2 + y^2 + z^2 - 1 = 0 } \Rightarrow  L(x, y, z, \lambda) = xyz - \lambda(x^2 + y^2 + z^2 - 1) \\ \Rightarrow \cases{ \frac{\partial L}{\partial x} = yz - 2\lambda x = 0 \Rightarrow yz = 2\lambda x \\ \frac{\partial L}{\partial y} = xz - 2\lambda y = 0 \Rightarrow xz = 2\lambda y \\ \frac{\partial L}{\partial z} = xy - 2\lambda z = 0 \Rightarrow xy = 2\lambda z } \Rightarrow \cases{xyz=2\lambda x^2\\ xyz=2\lambda y^2 \\ xyz= 2\lambda z^2} \Rightarrow \lambda x^2= \lambda z^2= \lambda z^2 \\ \lambda=0 \Rightarrow  xyz=0 \Rightarrow f(x,y,z)=0\\ \lambda \ne 0 \Rightarrow x^2=y^2=z^2 \Rightarrow  x^2 + x^2 + x^2 = 1 \Rightarrow 3x^2 = 1 \Rightarrow x^2 = \frac{1}{3}\\ \Rightarrow(x,y,z)= \left( \pm {1\over \sqrt 3}, \pm {1\over \sqrt 3}, \pm {1\over \sqrt 3} \right) \Rightarrow \cases{f_{max} ={1\over \sqrt 3} \cdot {1\over \sqrt 3} \cdot {1\over \sqrt 3}={\sqrt 3\over 9} \\ f_{min}= -{1\over \sqrt 3} \cdot {1\over \sqrt 3} \cdot {1\over \sqrt 3}=-{\sqrt 3\over 9}} \Rightarrow \bbox[red, 2pt]{\cases{\max: \sqrt 3/9\\ \min: -\sqrt 3/9}}$$

解答:$$\textbf{(a) } \text{If $ab=0$, then $\frac{1}{p}a^p + \frac{1}{q}b^q \ge ab=0$,trivially. }\\ \text{Jensen inequality is  used for  $a\gt 0$ and $b\gt 0$, }\\  \ln(tx_1 + (1-t)x_2) \ge t\ln(x_1) + (1-t)\ln(x_2) , \text{ for all }x_1,x_2\gt 0\\ \text{Let }\cases{x_1= a^p\\ x_2=b^q\\ t=1/p} \Rightarrow  \ln\left(\frac{1}{p}a^p + \frac{1}{q}b^q\right) \ge \frac{1}{p}\ln(a^p) + \frac{1}{q}\ln(b^q)  = \frac{1}{p}(p\ln a) + \frac{1}{q}(q\ln b) = \ln a + \ln b = \ln(ab) \\ \Rightarrow  \ln\left(\frac{1}{p}a^p + \frac{1}{q}b^q\right) \ge \ln(ab)  \Rightarrow  \frac{1}{p}a^p + \frac{1}{q}b^q \ge ab \quad \bbox[red, 2pt]{QED.} \\\textbf{(b) } \text{Let } A = \left(\int_0^1 \vert{}f(x)\vert{}^p dx\right)^{1/p} \text{ and }B = \left(\int_0^1 \vert{}g(x)\vert{}^q dx\right)^{1/q}.\\ \text{If $A$=0 or $B$=0, then }0\le 0\cdot 0. \text{ Holder inequality  holds.}\\ \text{Suppose $A\gt 0$ and $ B\gt 0$, let } u(x) = \frac{\vert{}f(x) \vert{}}{A}, \quad v(x) = \frac{\vert{}g(x)\vert{}}{B} . \text{ By (a), we have} \\ u(x)v(x) \le \frac{1}{p}u(x)^p + \frac{1}{q}v(x)^q \Rightarrow  \frac{\vert{}f(x)\vert{}}{A} \cdot \frac{\vert{}g(x)\vert{}}{B} \le \frac{1}{p} \frac{ \vert{}f(x) \vert{}^p}{A^p} + \frac{1}{q}\frac{\vert{}g(x)\vert{}^q}{B^q} \\ \Rightarrow  \int_0^1 \frac{\vert{}f(x)g(x)\vert{}}{AB} dx \le \int_0^1 \left( \frac{1}{p}\frac{\vert{}f(x)\vert{}^p}{A^p} + \frac{1}{q}\frac{ \vert{}g(x)\vert{}^q}{B^q} \right) dx \\ \Rightarrow  \frac{1}{AB} \int_0^1 \vert{}f(x)g(x)\vert{} dx \le \frac{1}{pA^p}\int_0^1 \vert{}f(x)\vert{}^p dx + \frac{1}{qB^q} \int_0^1 \vert{}g(x)\vert{}^q dx  \\ \Rightarrow  \frac{1}{AB} \int_0^1 \vert{}f(x)g(x)\vert{} dx \le \frac{1}{pA^p}(A^p) + \frac{1}{qB^q}(B^q) = \frac{1}{p} + \frac{1}{q} =1 \\ \Rightarrow  \frac{1}{AB} \int_0^1 \vert{}f(x)g(x)\vert{} dx \le 1 \\\Rightarrow  \int_0^1 \vert{}f(x)g(x) \vert{} dx \le AB = \left(\int_0^1 \vert{}f(x)\vert{}^p dx\right)^{1/p} \left(\int_0^1 \vert{}g(x)\vert{}^q dx\right)^{1/q} \quad \bbox[red, 2pt]{QED.} \\\textbf{(c) }  \vert{}\vert{}f + g\vert{}\vert{}_{L^p} = 0  \Rightarrow \text{ Minkowski inequality holds.}\\   \vert{}\vert{}f + g\vert{}\vert{}_{L^p} > 0 \Rightarrow  \int_0^1 \vert{}f(x) + g(x)\vert{}^p dx = \int_0^1 \vert{}f(x) + g(x)\vert{} \cdot \vert{}f(x) + g(x)\vert{}^{p-1} dx \\ \qquad  \le \int_0^1 (\vert{}f(x)\vert{} + \vert{}g(x)\vert{}) \cdot \vert{}f(x) + g(x)\vert{}^{p-1} dx \\ \qquad  = \int_0^1 \vert{}f(x)\vert{} \vert{}f(x) + g(x)\vert{}^{p-1} dx + \int_0^1 \vert{}g(x)\vert{} \vert{}f(x) + g(x)\vert{}^{p-1} dx \\ \text{By (b), we have }\int_0^1 \vert{}f(x)\vert{} \vert{}f(x) + g(x)\vert{}^{p-1} dx \le \left(\int_0^1 \vert{}f(x)\vert{}^p dx\right)^{1/p} \left(\int_0^1 (\vert{}f(x) + g(x)\vert{}^{p-1})^q dx\right)^{1/q} \\ \text{ and } \int_0^1 \vert{}g(x)\vert{} \vert{}f(x) + g(x)\vert{}^{p-1} dx \le \left(\int_0^1 \vert{}g(x)\vert{}^p dx\right)^{1/p} \left(\int_0^1 (\vert{}f(x) + g(x)\vert{}^{p-1})^q dx\right)^{1/q} \\ \Rightarrow  \int_0^1 \vert{}f+g\vert{}^p dx \le \vert{}\vert{}f \vert{}\vert{}_{L^p} \left(\int_0^1 \vert{}f+g\vert{}^p dx\right)^{1/q} + \vert{}\vert{}g \vert{}\vert{}_{L^p} \left(\int_0^1 \vert{}f+g\vert{}^p dx\right)^{1/q}  \\ \Rightarrow \int_0^1 \vert{}f+g\vert{}^p dx \le (\vert{}\vert{}f\vert{}\vert{}_{L^p} + \vert{}\vert{}g \vert{}\vert{}_{L^p}) \left(\int_0^1 \vert{}f+g\vert{}^p dx\right)^{1/q} \\ \Rightarrow  \vert{}\vert{}f+g\vert{}\vert{}_{L^p}^p \le (\vert{}\vert{}f \vert{} \vert{}_{L^p} + \vert{}\vert{}g\vert{}\vert{}_{L^p}) \cdot \vert{} \vert{}f+g\vert{}\vert{}_{L^p}^{p/q} \Rightarrow \vert{}\vert{}f +g\vert{}\vert{}_{L^p}^{p - p/q} \le \vert{} \vert{}f\vert{} \vert{}_{L^p} + \vert{} \vert{}g\vert{}\vert{}_{L^p} \\ \Rightarrow  \vert{} \vert{}f +g\vert{}\vert{}_{L^p(0,1)} \le \vert{} \vert{}f \vert{} \vert{}_{L^p(0,1)} + \vert{} \vert{}g\vert{}\vert{}_{L^p(0,1)} \quad \bbox[red, 2pt]{QED.}$$

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