115年台師大數學碩士班-基礎數學詳解

國立臺灣師範大學115學年度碩士班招生考試

科目: 基礎數學  適用系所:數學系

Part I: Calculus
一、填充題

解答:$$\lim_{x\to 0}{\tan x-x\over x^3} =\lim_{x\to 0}{{d\over dx}(\tan x-x)\over {d\over dx}x^3} = \lim_{x\to 0}{\sec^2 x-1\over 3x^2} = {1\over 3}\lim_{x\to 0}{\tan^2 x\over x^2} = {1\over 3}\lim_{x\to 0} \left({\tan x\over x} \right)^2 \\={1\over 3} \cdot 1^2= \bbox[red, 2pt]{1\over 3}$$

解答:$$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) \Rightarrow a+1=a^2 \Rightarrow a^2-a-1=0 \Rightarrow a=\bbox[red, 2pt]{1\pm \sqrt 5\over 2}$$

解答:$$ a_n = \frac{(-1)^n(x+1)^n}{2n+1} \Rightarrow  L = \lim_{n \to \infty} \left\vert{} \frac{a_{n+1}}{a_n} \right\vert{}   = \lim_{n \to \infty} \left\vert{} \frac{(-1)^{n+1}(x+1)^{n+1}}{2(n+1)+1} \cdot \frac{2n+1}{(-1)^n(x+1)^n} \right\vert{} \\  = \lim_{n \to \infty} \left\vert{} \frac{(x+1)(2n+1)}{2n+3} \right\vert{}  = \vert{}x+1\vert{} \lim_{n \to \infty} \frac{2n+1}{2n+3} =|x+1| \lt 1 \Rightarrow -2\lt x\lt 0\\ x=-2 \Rightarrow \sum_{n=0}^\infty a_n= \sum_{n=0}^{\infty} \frac{(-1)^n(-2+1)^n}{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n(-1)^n}{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^{2n}}{2n+1} = \sum_{n=0}^{\infty} \frac{1}{2n+1} \\\qquad  \lim_{n\to \infty} {{1\over 2n+1} \over {1\over n}} ={1\over 2}. \text{ Since $0 < \frac{1}{2} < \infty$ and $\sum \frac{1}{n}$ diverges,   series $\sum \frac{1}{2n+1}$ also diverges. } \\ x=0  \Rightarrow \sum_{n=0}^\infty a_n=  \sum_{n=0}^{\infty} \frac{(-1)^n(0+1)^n}{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n(1)^n}{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \\ \qquad \text{Using Alternating Series Test, }  \lim_{n \to \infty} \frac{1}{2n+1} = 0 \text{ and }  \frac{1}{2n+3} \le \frac{1}{2n+1}  \text{ for all }n\ge 0,\\ \qquad \text{series }\sum a_n \text{ converges.}\\ \text{Therefore, the interval of convergence is:}\bbox[red, 2pt]{(-2,0]}$$

二、計算題

解答:$$\textbf{(a) }2x^2-x^3=0 \Rightarrow x^2(2-x)=0 \Rightarrow x=0,2 \Rightarrow \text{The points of intersection are} \bbox[red, 2pt]{(0, 0) \text{ and } (2, 0)} \\\textbf{(b) } V = \int_{a}^{b} 2\pi \cdot (\text{radius}) \cdot (\text{height}) \, dx =  \bbox[red, 2pt]{\int_{0}^{2} 2\pi (2x^3 - x^4) \, dx } \\ \textbf{(c) } V = 2\pi \int_{0}^{2} (2x^3 - x^4) \, dx = 2\pi \left. \left[ {x^4\over 2}-{x^5\over 5} \right] \right|_0^2 =2\pi \left( 8-{32\over 5} \right) = \bbox[red, 2pt]{16\pi\over 5}$$

解答:$$$$

解答:$$ f(x,y) = -x^3 + 4xy - 2y^2 + 1 \Rightarrow \cases{f_x=-3x^2+4y\\f_y=4x-4y} \Rightarrow \cases{f_{xx}=-6x\\ f_{xy} =4\\ f_{yy} =-4} \\ \Rightarrow D(x,y)=f_{xx}f_{yy} - \left( f_{xy} \right)^2= 24x-16 \\ \cases{f_x=0\\f_y=0} \Rightarrow \cases{4y=3x^2\\ x=y} \Rightarrow (x,y)=(0.0), \left( {4\over 3},{4\over 3} \right) \Rightarrow \cases{D(0,0)=-16\lt 0 \Rightarrow (0,0) \text{ is a saddle point} \\D(4/3,4/3)=16\gt 0 \Rightarrow f_{xx}(4/3,4/3)=-8\lt 0} \\ \Rightarrow \text{ there is a relative maximum at $\left(\frac{4}{3}, \frac{4}{3}\right)$} \Rightarrow f(4/3,4/3)= {59\over 27} \\ \Rightarrow \bbox[red, 2pt]{\text{The function has a relative maximum of $\frac{59}{27}$ at the point $\left(\frac{4}{3}, \frac{4}{3}\right)$.}}$$

Part II. Linear Algebra

解答:$$\textbf{(a) }\text{Let }\mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix}, \quad \mathbf{v}_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \end{bmatrix} \text{ and a normal vector }\mathbf n = \begin{bmatrix}a\\b\\c\\d \end{bmatrix} \\ \Rightarrow \cases{ \mathbf n\cdot \mathbf v_1 =0 \\ \mathbf n\cdot \mathbf v_2 =0 \\ \mathbf n\cdot \mathbf v_3 =0 } \Rightarrow \cases{a=0\\ b+c=0\\ c+d=0}\Rightarrow  \mathbf n = c\begin{bmatrix}0\\-1\\ 1\\-1 \end{bmatrix} \Rightarrow \text{Any non-zero scalar multiple of this set is also correct,}\\ \text{such as $a=0, b=-1, c=1, d=-1$} \Rightarrow \bbox[red, 2pt]{\cases{a=0\\b=1\\c=-1\\ d=1}}\\ \textbf{(b) } \mathbf{b} = \begin{bmatrix} 1 \\ 0 \\ -1 \\ 1 \end{bmatrix} \text{ and } \mathbf{n} = \begin{bmatrix} 0 \\ 1 \\ -1 \\ 1 \end{bmatrix} \Rightarrow  \operatorname{proj}_{W^\perp} \mathbf{b} = \frac{\mathbf{b} \cdot \mathbf{n}}{\mathbf{n} \cdot \mathbf{n}} =\frac{2}{3} \begin{bmatrix} 0 \\ 1 \\ -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 2/3 \\ -2/3 \\ 2/3 \end{bmatrix}  \\ \qquad \Rightarrow  \operatorname{proj}_W \mathbf{b} = \mathbf{b} - \operatorname{proj}_{W^\perp} \mathbf{b} =   \begin{bmatrix} 1 \\ 0 \\ -1 \\ 1 \end{bmatrix} - \begin{bmatrix} 0 \\ 2/3 \\ -2/3 \\ 2/3 \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix} 1 \\ -2/3 \\ -1/3 \\ 1/3 \end{bmatrix} } \\\textbf{(c) } \mathbf x\in W \Rightarrow T(\mathbf x)= \mathbf x \Rightarrow \lambda =1 \text{ is an eigenvalue. Because the dimension of $W$ is 3,}\\\quad \text{the eigenspace corresponding to $\lambda = 1$ has a dimension of 3 .} \\ \mathbf x\in W^\bot \Rightarrow T(\mathbf x)=0\mathbf x \Rightarrow \lambda=0 \text{ is an eigenvalue. Because dim($W^\bot$)=1, the eigenspace} \\ \quad \text{ corresponding to $\lambda$=0 has dimension of 1.} \\ \text{Since 3+1=4 equals dim($\mathbb R^4$), there exists a basis for $\mathbb R^4$ consisting entirely of eigenvectors of }T. \\ \Rightarrow \bbox[red, 2pt]{\text{Yes, $T$ is diagonalizable}}$$

解答:$$\textbf{(a) } \text{The given characteristic polynomial is }p(x)=(x-1)^5(x-2) \\ \qquad \Rightarrow n=5+1=\bbox[red, 2pt]6 \\ \textbf{(b) }p(x)=(x-1)^5(x-2) \Rightarrow \text{the exponent for the factor $(x-1)$ is $5$} \\\qquad \Rightarrow \text{the algebraic multiplicity of $\lambda = 1$ is }\bbox[red, 2pt] 5\\ \textbf{(c) }\text{Let $N_k = \text{nullity}((A - I_n)^k)$. The geometric multiplicity of $\lambda = 1$ is exactly $N_1$.}  \text{ Now }N_2=3. \\\qquad d_k= N_k-N_{k-1} \Rightarrow d_1=N_1-0=N_1 \Rightarrow d_2=N_2-N_1=3-N_1 \Rightarrow N_1\ge 3-N_1 \Rightarrow N_1\ge 3/2 \\ \qquad \Rightarrow N_1\ge 2. \text{ If $N_1 = 3$, then $d_1 = 3$ and $d_2 = 0$} \Rightarrow N_2=N_3=3, \text{ but the algebraic multiplicity is 5} \\ \qquad \Rightarrow N_1= \bbox[red, 2pt]2 \\ \textbf{(d) }\cases{N_1=2\\ N_2=3} \Rightarrow \cases{d_1=2\\ d_2=1} \text{ and $d_k$ must be non-increasing} \Rightarrow d_1+d_2+ d_3+d_4=5\\ \qquad \Rightarrow   2+1+1+1=5 \Rightarrow d_3=1 \Rightarrow N_3=N_2+d_3=3+1=4 \Rightarrow  \text{nullity}((A - I_n)^3) =\bbox[red, 2pt]4 \\ \textbf{(e) }p(x)=(x-1)^5(x-2) \Rightarrow  \text{nullity}(A - 2I_n) = 1 \Rightarrow  \text{rank}(A - 2I_n) = n - \text{nullity}(A - 2I_n)\\\qquad  = 6 - 1 = \bbox[red, 2pt]5 \\ \textbf{(f) } \lambda=2 \Rightarrow J_1(2), \lambda=1 \Rightarrow \cases{d_1=2\\ d_2=1\\ d_3=1\\ d_4=1\\ d_5=0} \Rightarrow J = \bbox[red, 2pt]{\begin{bmatrix} 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 \end{bmatrix} }$$

解答:$$\textbf{(a) } M_{p\times q} \Rightarrow \cases{C(M) \subseteq \mathbb R^p\\ N(M) \subseteq \mathbb R^q} \Rightarrow \cases{C(A) \subseteq \mathbb{R}^m\text{ and }N(A) \subseteq \mathbb{R}^k \\C(B) \subseteq \mathbb{R}^k \text{ and } N(B) \subseteq \mathbb{R}^n \\C(AB) \subseteq \mathbb{R}^m \text{ and }N(AB) \subseteq \mathbb{R}^n} \\ \qquad \Rightarrow \bbox[red, 2pt]{C(A)\text{ and }C(AB) \text{ are subspace of }\mathbb R^m} \\ y\in C(AB) \Rightarrow y=(AB)x, x\in \mathbb R^n \Rightarrow y=A(Bx) =A(z), z=Bx \in \mathbb R^k \Rightarrow y\in C(A) \\ \Rightarrow \bbox[red, 2pt]{C(AB) \subseteq C(A)} \\\textbf{(b) } \cases{C(B) \subseteq \mathbb R^k \text{ and }N(B) \subseteq \mathbb R^n \\C(AB) \subseteq \mathbb R^m \text{ and }N(AB) \subseteq \mathbb R^n } \Rightarrow \bbox[red, 2pt]{N(B) \text{ and }N(AB) \text{ are subspaces of }\mathbb R^n} \\ x\in N(B) \Rightarrow Bx=0 \Rightarrow ABx=A(0)=0 \Rightarrow (AB)x=0 \Rightarrow x\in N(AB) \Rightarrow \bbox[red, 2pt]{N(B) \subseteq N(AB)} \\\textbf{(c) } \text{Let $T: C(B) \to \mathbb{R}^m$ be defined by $T(y) = Ay$ for any $y \in C(B)$.} \\ \text{The image of $T$ is the set of all vectors $Ay$ where $y \in C(B)$. Since $y \in C(B)$, $y = Bx$ for some $x \in \mathbb{R}^n$.}\\ \text{ Thus, $Ay = A(Bx) = (AB)x$. The set of all such $(AB)x$ is exactly $C(AB)$.}\\ \text {Therefore, $\text{Im}(T) = C(AB)$, and $\dim(\text{Im}(T)) = \dim(C(AB)) = \text{rank}(AB)$.} \\ \\ \text{The kernel of $T$ consists of all $y \in C(B)$ such that $T(y) = 0$.} \\\text{Since $T(y) = Ay$, this means $Ay = 0$, so $y \in N(A)$.} \\ \text{
Therefore, the kernel of $T$ is the set of vectors that are in both $C(B)$ and $N(A)$,}\\\text{ which is exactly $N(A) \cap C(B)$.Thus, $\text{Ker}(T) = N(A) \cap C(B)$.} \\ \\ \text{According to the Rank-Nullity Theorem:$\dim(\text{Domain}) = \dim(\text{Im}(T)) + \dim(\text{Ker}(T))$} \\\text{ Substituting our findings:$\dim(C(B)) = \dim(C(AB)) + \dim(N(A) \cap C(B))$} \\ \Rightarrow \text{ $\text{rank}(B) = \text{rank}(AB) + \dim(N(A) \cap C(B))$}\\ \Rightarrow \text{rank}(AB) = \text{rank}(B) \iff \dim(N(A) \cap C(B)) = 0 \qquad \bbox[red, 2pt]{QED.} \\\textbf{(d) }    \text{From (c), we have } \text{rank}(B) = \text{rank}(AB) + \dim(N(A) \cap C(B))  \\\qquad \Rightarrow  \dim(N(A) \cap C(B)) = \text{rank}(B) - \text{rank}(AB) =\text{rank}(B) - \text{rank}(A) \\ \text{Conversely, assume $\dim(N(A) \cap C(B)) = \text{rank}(B) - \text{rank}(A)$} \\ \Rightarrow  \text{rank}(B) = \text{rank}(AB) + (\text{rank}(B) - \text{rank}(A)) \Rightarrow  \text{rank}(B) - \text{rank}(AB) =0=  \text{rank}(B) - \text{rank}(A)  \\ \Rightarrow \text{rank}(B) = \text{rank}(A) \qquad \bbox[red, 2pt]{QED.}$$

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解題僅供參考，碩士班歷年試題及詳解