國立政治大學112學年度碩士班招生考試
考試科目:基礎數學 系所別:統計學系
Part I: Calculus
解答:$$\textbf{State:}\\\text{Let }f \text{ be continuous on }[a,b] \text{ and differentiable on }(a,b). \text{ Then }\exists c\in (a,b) \text{ such that }\\ \qquad \qquad f'(c)=\cfrac{f(b) -f(a)}{b-a} \\ \textbf{Prove:}\\ \qquad \text{Let }\cases{A(a,f(a))\\ B(b,f(b))}, \text{ then }\overline{AB}: y={f(b)-f(a)\over b-a}(x-a)+ f(a)\\ \qquad \text{And let }g(x)=f(x)-\left[{f(b)-f(a)\over b-a}(x-a)+ f(a) \right]\\ \qquad \text{By Rolle's theorem, }\exists\, c \in (a,b) \ni g'(c)=0.\text{ That is,}\\ \qquad g'(x)=f'(x)-{f(b)-f(a)\over b-a} \Rightarrow g'(c)=f'(c)-{f(b)-f(a)\over b-a} =0 \\ \qquad \Rightarrow f'(c)={f(b)-f(a)\over b-a}, \bbox[red, 2pt]{Q.E.D}$$
解答:$$\text{Let }\epsilon \gt 0, \text{ we choose }\delta={1\over \sqrt \epsilon}. \text{ Then, we have }\forall x\gt \delta= {1\over \sqrt\epsilon} \\ \Rightarrow \left| {1\over x^2}-0\right|={1\over x^2}\lt \epsilon \;(\because x\gt {1\over \sqrt \epsilon} \Rightarrow {1\over x^2}\lt \epsilon) \Rightarrow \lim_{x\to \infty}{1\over x^2}=0,\bbox[red, 2pt]{Q.E.D.}$$
解答:$$\textbf{(a)}\; \int{2x^2-x+4\over x^3+4x}\,dx = \int \left( {1\over x}+{x-1\over x^2+4}\right)\,dx =\int {1\over x}\,dx +\int {x\over x^2+4}\,dx -{1\over 4}\int {1\over (x/2)^2+1}\,dx \\\quad = \bbox[red, 2pt]{\ln x+ {1\over 2}\ln (x^2+4)-{1\over 2}\tan^{-1}({x\over 2})+C} \\\textbf{(b)}\; f(x)＝{1\over x^4} \Rightarrow f(0)不存在 \Rightarrow \int_{-2}^3 f(x)\,dx \bbox[red, 2pt]{不存在} \\ \textbf{(c)}\; \int_0^a f(x)\,dx =\int_0^a f(a-x)\,dx\\\quad \Rightarrow I=\int_0^{\pi/2} {\sin^n x\over \sin^n x+ \cos^n x}\,dx = \int_0^{\pi/2} {\sin^n (\pi/2-x)\over \sin^n (\pi/2-x)+ \cos^n (\pi/2-x)}\,dx \\\quad = \int_0^{\pi/2} {\cos^n x\over \cos^n x+ \sin^n x}\,dx \Rightarrow I+I=2I= \int_0^{\pi/2} {\sin^n x+ \cos^n x\over \sin^n x+ \cos^n x}\,dx = \int_0^{\pi/2} 1\,dx ={\pi\over 2}\\\quad \Rightarrow I=\bbox[red, 2pt]{\pi \over 4}\\\textbf{(d)}\; I=\int_0^\infty e^{-x^2}\, dx \Rightarrow I^2 =\left(\int_0^\infty e^{-x^2}\, dx \right) \left(\int_0^\infty e^{-y^2}\, dy \right) =\int_0^\infty \int_0^\infty e^{-(x^2+y^2)} \,dxdy \\\quad \text{Let }\cases{x=r\cos \theta\\ y=r \sin \theta},\text{ then }I^2 =\int_0^{\pi/2} \int_0^\infty re^{-r^2}\,drd\theta =\int_0^{\pi/2} {1\over 2}\,d\theta= {\pi \over 4 } \Rightarrow I=\sqrt{\pi \over 4} =\bbox[red, 2pt]{\sqrt \pi \over 2} \\\textbf{(e)}\; I=\int \sin^5 x\cos^2 x\,dx = \int \sin x(1-\cos^2 x)^2 \cos^2 x\,dx\\\quad \text{Let }u=\cos x, \text{ then }du=-\sin xdx \Rightarrow I=\int -(1-u^2)^2 u^2\,du =\int (-u^6+2u^4-u^2) \,du \\\quad =-{1\over 7}u^7+{2\over 5}u^5-{1\over 3}u^3+C =\bbox[red, 2pt]{-{1\over 7} \cos^7 x+ {2\over 5}\cos^5 x-{1\over 3}\cos^3x +C}$$
解答:$$\left(\tan{x\over y} \right)'=(x+y)' \Rightarrow \sec^2 {x\over y} \cdot \left({1\over y} -{xy'\over y^2} \right) =1+y' \Rightarrow {1\over y}\sec^2{x\over y}-1= \left(1+{x\over y^2}\sec^2{x \over y} \right)y' \\ \Rightarrow \frac{d y}{dx}=y'= \bbox[red, 2pt] {\cfrac{{1\over y}\sec^2{x\over y}-1}{1+{x\over y^2}\sec^2{x \over y}}}$$
解答:$$x^2+y^2=1 \Rightarrow \cases{x=\cos \theta\\ y=\sin \theta} \Rightarrow z=1-x+y= 1-\cos \theta+\sin \theta \\ \Rightarrow f(x,y,z)=x+2y+3z= \cos \theta+2 \sin \theta+3-3\cos \theta+3\sin \theta= 3-2\cos \theta+5\sin \theta \\ =3-\sqrt{29}\left( {2\over \sqrt{29}}\cos \theta-{5\over \sqrt{29}} \sin \theta\right) =3-\sqrt{29} \sin(\alpha -\theta) \Rightarrow \bbox[red, 2pt]{\cases{\max(f)= 3+\sqrt{29} \\ \min(f)=3-\sqrt{29}}}$$
Part II: Linear Algebra
解答:$$\textbf{(a)}\; \lambda v =Av = A^2v=A(Av)=A(\lambda v)= \lambda( Av)= \lambda ^2 v \\\Rightarrow \lambda^2v-\lambda v=0 \Rightarrow \lambda(\lambda-1)v=0 \Rightarrow \lambda= \bbox[red, 2pt]{0,1} \\\textbf{(b)}\; \text{By (a), the eigenvalues are 0 or 1, then rank}(A)= \text{number of 1}\\\qquad \Rightarrow rank(A)=\bbox[red, 2pt]{tr(A)}$$
解答:$$Q \text{ is orthogonal } \Rightarrow Q^{-1}=Q^T \Rightarrow QQ^T=I \Rightarrow \det(QQ^T)=\det(Q)\det(Q^T) =(\det(Q))^2=1\\ \Rightarrow \det(Q)= \bbox[cyan,2pt]\pm 1,\bbox[red, 2pt]{Q.E.D.}$$
解答:$$N(T)=\{ v\in V\mid T(v)=0\}\\ (a)\; T(0)=0 \Rightarrow 0\in N(T)\\(b) \text{Suppose that }\cases{v_1\in N(T)\\ v_2\in N(T)} \Rightarrow \cases{T(v_1)=0\\ T(v_2)=0} \Rightarrow T(v_1+v_2)= T(v_1)+ T(v_2)=0 \\\qquad \Rightarrow v_1+v_2 \in N(T) \\(c)\; \text{Suppose that } v\in N(T) \Rightarrow T(v)=0 \Rightarrow T(cv)=cT(v)=0 \Rightarrow cv \in N(T)\\ \text{From (a),(b), and (c), we have }N(T)\text{ is a subspace of }V\\ R(T)=\{w\in W\mid w=T(v) \;\exists v\in V\} \\(a) T(0)=0 \Rightarrow 0 \in R(T)\\(b)\; \text{Suppose that }\cases{w_1\in R(T)\\ w_2\in R(T)} \Rightarrow \exists v_1,v_2\in V \ni \cases{T(v_1)=w_1\\ T(v_2)=w_2} \\\qquad \Rightarrow w_1+w_2= T(v_1)+T(v_2)= T(v_1+v_2) \Rightarrow w_1+w_2 \in R(T) \\(c)\; \text{Suppose that } w\in R(T) \Rightarrow \exists v \in V \ni T(v)=w \Rightarrow T(cv)=cT(v)=cw \Rightarrow cw \in R(T)\\\text{From (a),(b), and (c), we have }R(T) \text{ is a subspace of }W. \bbox[red, 2pt]{Q.E.D.}$$
解答:$$A=\left( \begin{matrix} 0 & -2 \\1 & 3 \end{matrix} \right) =\left( \begin{matrix} -2 & -1 \\1 & 1 \end{matrix} \right) \left( \begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix} \right) \left( \begin{matrix}-1 & -1 \\1 & 2\end{matrix} \right) \\ \Rightarrow A^n =\left( \begin{matrix} 0 & -2 \\1 & 3 \end{matrix} \right) =\left( \begin{matrix} -2 & -1 \\1 & 1 \end{matrix} \right) \left( \begin{matrix} 1^n & 0 \\ 0 & 2^n \end{matrix} \right) \left( \begin{matrix}-1 & -1 \\1 & 2\end{matrix} \right) = \bbox[red, 2pt] {\left( \begin{matrix}-2^n+2 & -2^{n+1}+2 \\2^n-1 & 2^{n+1}-1 \end{matrix} \right)}$$
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