# 应用数学入门 Introduction to Applied Mathematics MAT00003C

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Let $R=K\left[x_{1}, \ldots, x_{n}\right]$ be a polynomial ring over a field $K$ and let $I$ be a monomial ideal of $R$ generated by a finite set of monomials $\left{x^{v_{1}}, \ldots, x^{v_{q}}\right}$. As usual we use $x^{a}$ as an abbreviation for $x_{1}^{a_{1}} \cdots x_{n}^{a_{n}}$, where $a=\left(a_{1}, \ldots, a_{n}\right)$ is in $\mathbb{N}^{n}$. The three central objects of study here are the following blowup algebras: (a) the Rees algebra
$$R[I t]:=R \oplus I t \oplus \cdots \oplus I^{i} t^{i} \oplus \cdots \subset R[t],$$
where $t$ is a new variable, (b) the associated graded ring
$$\operatorname{gr}{I}(R):=R / I \oplus I / I^{2} \oplus \cdots \oplus I^{i} / I^{i+1} \oplus \cdots \simeq R[I t] \otimes{R}(R / I),$$

with multiplication
$$\left(a+I^{i+1}\right)\left(b+I^{j+1}\right)=a b+I^{i+j+1} \quad\left(a \in I^{i}, b \in I^{j}\right),$$
and (c) the symbolic Rees algebra
$$R_{s}(I):=R+I^{(1)} t+I^{(2)} t^{2}+\cdots+I^{(t)} t^{i}+\cdots \subset R[t],$$
where $I^{(i)}$ is the ith symbolic power of $I$.

## MAT00003C COURSE NOTES ：

Let $R=k\left[x_{1}, \ldots, x_{n}\right]$ be a polynomial ring over a field $k$. Suppose $M=x_{1} a_{1} \ldots x_{n} a_{n}$ is a monomial in $R$. Then we define the polarization of $M$ to be the square-free monomial
$$\mathscr{P}(M)=x_{1,1} x_{1,2} \ldots x_{1, a_{1}} x_{2,1} \ldots x_{2, a_{2}} \ldots x_{n, 1} \ldots x_{n, a_{n}}$$
in the polynomial ring $S=k\left[x_{i, j} \mid 1 \leq i \leq n, 1 \leq j \leq a_{i}\right]$.
If $I$ is an ideal of $R$ generated by monomials $M_{1}, \ldots, M_{q}$, then the polarization of $I$ is defined as:
$$P(I)=\left(P\left(M_{1}\right), \ldots, P\left(M_{q}\right)\right)$$

# 实分析 Real Analysis MAT00005C

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{Condition (b) is Riemann’s original formulation of integrability. 1 }
(a) $\Rightarrow$ (b): Assuming $f$ is Riemann-integrable, let
$$\lambda=\int_{a}^{b} f$$

For each $\nu=1, \ldots, n$, choose a sequence $\left(x_{\nu}^{k}\right)$ in $\left[a_{\nu-1}, a_{\nu}\right]$ such that
$$f\left(x_{\nu}^{k}\right) \rightarrow M_{\nu} \text { as } k \rightarrow \infty$$
then
$$\sum_{\nu=1}^{n} f\left(x_{\nu}^{k}\right) e_{\nu} \rightarrow \sum_{\nu=1}^{n} M_{\nu} e_{\nu}=S(\sigma)$$
so by $()$ we have (*)
$$\lambda-\epsilon \leq S(\sigma) \leq \lambda+\epsilon .$$

## MAT00005C COURSE NOTES ：

Similarly,
$$\lambda-\epsilon \leq s(\sigma) \leq \lambda+\epsilon \text {; }$$
thus $S(\sigma)$ and $s(\sigma)$ both belong to the interval $[\lambda-\epsilon, \lambda+\epsilon]$, therefore
$$W_{f}(\sigma)=S(\sigma)-s(\sigma) \leq 2 \epsilon .$$
This proves that $f$ is Riemann-integrable and since
$$S(\sigma) \rightarrow \int_{a}^{b} f \text { as } \mathrm{N}(\sigma) \rightarrow 0,$$
it is clear from $\left({ }^{* *}\right)$ that
$$\lambda=\int_{a}^{b} f . \diamond$$

# 概率与统计学简介 Introduction to Probability & Statistics MAT00004C

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We know from integral calculus that for $0 \leq a \leq b \leq 1$
$$\int_{a}^{b} f(x) \mathrm{d} x=\int_{a}^{b} \frac{1}{2 \sqrt{x}} \mathrm{~d} x=\sqrt{b}-\sqrt{a}$$
Hence $\int_{-\infty}^{\infty} f(x) \mathrm{d} x=\int_{0}^{1} 1 /(2 \sqrt{x}) \mathrm{d} x=1$ (so $f$ is a probability density function-nonnegativity being obvious), and

\begin{aligned}
\mathrm{P}\left(10^{-4} \leq X \leq 10^{-2}\right) &=\int_{10^{-4}}^{10^{-2}} \frac{1}{2 \sqrt{x}} \mathrm{~d} x \
&=\sqrt{10^{-2}}-\sqrt{10^{-4}}=10^{-1}-10^{-2}=0.09
\end{aligned}

## MAT00004C COURSE NOTES ：

Suppose $U$ has a $U(0,1)$ distribution. To construct a $\operatorname{Ber}(p)$ random variable for some $0<p<1$, we define
$$X= \begin{cases}1 & \text { if } U<p \ 0 & \text { if } U \geq p\end{cases}$$
so that
\begin{aligned} &\mathrm{P}(X=1)=\mathrm{P}(U<p)=p \ &\mathrm{P}(X=0)=\mathrm{P}(U \geq p)=1-p \end{aligned}
This random variable $X$ has a Bernoulli distribution with parameter $p$.
QUICK EXERCISE 6.2 A random variable $Y$ has outcomes 1,3 , and 4 with the following probabilities: $\mathrm{P}(Y=1)=3 / 5, \mathrm{P}(Y=3)=1 / 5$, and $\mathrm{P}(Y=4)=$ 1/5. Describe how to construct $Y$ from a $U(0,1)$ random variable.

# 数学技能 I: 推理与交流 Mathematical Skills I: Reasoning & Communication MAT00011C

0 数学技能 I: 推理与交流 Mathematical Skills I: Reasoning & Communication MAT00011C

the problem was $\frac{10^{2}}{10^{5}}$. Both bases have positive indices so to divide, we subtract the indices. Therefore
$$\frac{10^{2}}{10^{5}}=10^{2-5}$$

We cannot normally subtract 5 from 2 , but, just as we have negative indices, we can subtract and have a negative or minus result as an answer, We say that $2-5=-3$. Check it the long way:
$$\frac{10^{2}}{10^{5}}=\frac{100}{100000}=\frac{1}{1000} \text { or } 10^{-3}$$

## MAT00011C COURSE NOTES ：

We have now worked out that:
$$10^{2} \div 10^{4}=10^{-2}$$
If you are told that:
$$10^{-4} \div 10^{2}=10^{-6}$$
and that:
$$10^{-3} \div 10^{2}=10^{-5}$$
can you see that the subtractions of the indices of the dividing bases are as follows:
\begin{aligned} &10^{2} \div 10^{4}=10^{2-4}=10^{-2} \ &10^{-4} \div 10^{2}=10^{-4-2}=10^{-6} \ &10^{-3} \div 10^{2}=10^{-3-2}=10^{-5} \end{aligned}

# 代数 Algebra MAT00010C

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first box. Since the third box is formed from $x_{1}\left(\begin{array}{l}1 \ 3\end{array}\right)+x_{2}\left(\begin{array}{l}2 \ 1\end{array}\right)=\left(\begin{array}{l}6 \ 8\end{array}\right)$ and $\left(\begin{array}{l}2 \ 1\end{array}\right)$, and the determinant is unchanged by adding $x_{2}$ times the second column to the first column, the size of the third box equals that of the second. We have this.
$$\left|\begin{array}{ll} 6 & 2 \ 8 & 1 \end{array}\right|=\left|\begin{array}{ll} x_{1} \cdot 1 & 2 \ x_{1} \cdot 3 & 1 \end{array}\right|=x_{1} \cdot\left|\begin{array}{ll} 1 & 2 \ 3 & 1 \end{array}\right|$$

Solving gives the value of one of the variables.
$$x_{1}=\frac{\left|\begin{array}{ll} 6 & 2 \ 8 & 1 \end{array}\right|}{\left|\begin{array}{ll} 1 & 2 \ 3 & 1 \end{array}\right|}=\frac{-10}{-5}=2$$
The theorem that generalizes this example, Cramer’s Rule, is: if $|A| \neq 0$ then the system $A \vec{x}=\vec{b}$ has the unique solution $x_{i}=\left|B_{i}\right| /|A|$ where the matrix $B_{i}$ is formed from $A$ by replacing column $i$ with the vector $\vec{b}$. Exercise 3 asks for a proof.
For instance, to solve this system for $x_{2}$
$$\left(\begin{array}{ccc} 1 & 0 & 4 \ 2 & 1 & -1 \ 1 & 0 & 1 \end{array}\right)\left(\begin{array}{l} x_{1} \ x_{2} \ x_{3} \end{array}\right)=\left(\begin{array}{c} 2 \ 1 \ -1 \end{array}\right)$$

## MAT00010C COURSE NOTES ：

Recall the definitions of the complex number addition
$$(a+b i)+(c+d i)=(a+c)+(b+d) i$$
and multiplication.
\begin{aligned} (a+b i)(c+d i) &=a c+a d i+b c i+b d(-1) \ &=(a c-b d)+(a d+b c) i \end{aligned}

# 微积分 Calculus MAT00001C

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This is
$$\left[\left[\begin{array}{c} R \circ g \ -Q \circ g \ P \circ g \end{array}\right] \cdot\left[\left[\frac{\partial g}{\partial u}\right] \times\left[\frac{\partial g}{\partial v}\right]\right]\right] d u \wedge d v$$
(The permutation of the $P, Q, R$ (and the minus sign) come from the way the $d x \wedge d y$ acts on a piece of surface normal to the $(d) z$ direction.)

We can rewrite this as
$$\left|\frac{\partial g}{\partial u} \times \frac{\partial g}{\partial v}\right|\left[\left[\begin{array}{c} R \circ g \ -Q \circ g \ P \circ g \end{array}\right] \cdot \hat{\boldsymbol{n}}[u, v]\right] \quad d u \wedge d v$$
where $\hat{\boldsymbol{n}}[u, v]$ is the unit normal to the surface at $g\left[\begin{array}{l}u \ v\end{array}\right]$, and
$$\left|\frac{\partial g}{\partial u} \times \frac{\partial g}{\partial v}\right|$$
is the “area stretching factor”.
We have that
$$\int_{g\left(I^{2}\right)} \omega$$

is the limit of the sums of values of $\omega$ on small elements of the surface $g\left(I^{2}\right)$. Suppose $g$ takes a rectangle $\triangle u \times \triangle v$ in $I^{2}$ to a (small) piece of the surface. $\omega$ at $g\left[\begin{array}{l}u \ v\end{array}\right]$ is, say,
$$P d x \wedge d y+Q d x \wedge d z+R d y \wedge d z$$
and the unit normal to the surface is $\hat{\boldsymbol{n}}[u, v]$ (located at $\left.g\left[\begin{array}{l}u \ v\end{array}\right]\right)$.
Write $\hat{\boldsymbol{n}}[u, v]$ as
$$\left[\begin{array}{l} \hat{n} x \ \hat{n} y \ \hat{n} z \end{array}\right]$$