# 拓扑学I|Topology I代写 MATH 671

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So map each simplex in the first barycentric subdivision of $K$,
$$\sigma_{1}<\sigma_{2}<\cdots<\sigma_{n}, \quad \sigma_{i} \in K$$
to the simplex in nerve $C(U)$,
$$U_{\sigma_{n}} \subseteq U_{\sigma_{n-1}} \subseteq \cdots \subseteq U_{\sigma_{1}}$$

This gives a simplicial map
$K^{\prime} \rightarrow \vec{g}$ nerve $C(U), K^{\prime}=1$ st barycentric subdivision $.$
Now consider the compositions
\begin{aligned} &K^{\prime} \underset{g}{\rightarrow} \text { nerve } C(U) \underset{f}{\rightarrow} K \ &\text { nerve } C(U) \stackrel{f}{\rightarrow} K=K^{\prime} \stackrel{g}{\rightarrow} \text { nerve } C(U) \end{aligned}
One can check for the first composition that a simplex of $K^{\prime}$
$$\sigma_{1}<\cdots<\sigma_{n}$$

## MATH 671 COURSE NOTES ：

(the two sphere, $S^{2}$ ) Let $p$ and $\sigma$ denote two distinct points of $S^{2}$. Consider the category over $S^{2}$ determined by the maps
$$\begin{array}{cc} \text { object } & \text { name } \ S^{2}-p \stackrel{\subseteq}{\longrightarrow} S^{2} & e \ S^{2}-q \subseteq S^{2} & e^{\prime} \ \text { (universal cover } \left.S^{2}-p-q\right) \rightarrow S^{2} & \mathbb{Z} . \end{array}$$
The category has three objects and might be denoted

# 数学概论| Introductory Mathematics 代写 MT1001

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Proof:
For simplicity let us consider the Lindbladian $\mathcal{L}$ associated with an element $r=\sum_{g \in \mathcal{G}} c_{g} U_{g} \in \mathcal{A}$ such that $|r|{2}:=\sum{g \in \mathcal{G}}\left|c_{g}\right||g|^{2}<\infty$. Here $\mathcal{L}$ takes the form

Denoting these two bounded derivations $\left[r_{k}^{}, .\right]$ and $\left[., r_{k}\right]$ on $\mathcal{A}$ by $\delta_{k}^{\dagger}$ and $\delta_{k}$ respectively, $\mathcal{L}(x)=\frac{1}{2} \sum_{k \in \mathbb{Z}^{d}} \delta_{k}^{\dagger}(x) r_{k}+r_{k}^{} \delta_{k}(x)$.
In order to prove (i), for $x \in \mathcal{C}^{1}(\mathcal{A})$, let us estimate the norm of $\mathcal{L}(x)$ :
$$|\mathcal{L}(x)| \leq \frac{1}{2} \sum_{k \in \mathbb{Z}^{d}}\left|\delta_{k}^{\dagger}(x) r_{k}+r_{k}^{*} \delta_{k}(x)\right|$$

## MT1001 COURSE NOTES ：

Summing over $\alpha^{\prime}$, it follows that
$$\left|\left(x_{n}\right)\right|_{1} \leq\left|(\lambda-\Gamma)^{-1}\right||y|_{1}<\infty$$ and hence $x_{n} \in \mathcal{C}^{1}(\mathcal{A})$. Now setting $y_{n}=(\mathcal{L}-\lambda)\left(x_{n}\right)$, we have $$\left|y_{n}-y\right|=\left|\left(\mathcal{L}-\mathcal{L}^{(n)}\right) x_{n}\right|=\sum_{|k|>n} \mathcal{L}{k}\left(x{n}\right)$$

# 数值分析I|Numerical Analysis I代写 MATH 651

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Let us begin by restricting the range to $-1 \leq x \leq 1$ and taking the simplest possible weight function, namely
$$w(x)=1$$
so that equation becomes
$$\frac{d^{2 i+1}}{d x^{2 i=1}}\left[U_{i}(x)\right]=0$$

Since $U_{i}(x)$ is a polynomial of degree $2 i$, an obvious solution which satisfies the boundary conditions is
$$U_{i}(x)=C_{i}\left(x^{2}-1\right)^{i} .$$
Therefore the polynomials that satisfy the orthogonality conditions will be given by
$$\phi_{i}(x)=C_{i} \frac{d^{i}\left(x^{2}-1\right)^{i}}{d x^{i}}$$
If we apply the normalization criterion we get
$$\int_{-1}^{+1} \phi_{i}^{2}(x) d x=1=C_{i} \int_{-1}^{+1}\left[\frac{d^{i}\left(x^{2}-1\right)}{d x^{i}}\right] d x$$
so that
$$\mathrm{C}_{\mathrm{i}}=\left[2^{\mathrm{i}} \mathrm{i} !\right]^{-1}$$

## MATH651 COURSE NOTES ：

Let us begin by considering a collection of $\mathrm{N}$ data points $\left(\mathrm{x}{\mathrm{i}}, \mathrm{Y}{\mathrm{i}}\right)$ which are to be represented by an approximating function $f\left(a_{j}, x\right)$ so that
$$f\left(a_{j}, x_{i}\right)=Y_{i}$$
Here the $(\mathrm{n}+1) a_{j}$ ‘s are the parameters to be determined so that the sum-square of the deviations from $Y_{i}$ are a minimum. We can write the deviation as
$$\varepsilon_{\mathrm{i}}=\mathrm{Y}{\mathrm{i}}-\mathrm{f}\left(\mathrm{a}{\mathrm{j}}, \mathrm{X}{\mathrm{i}}\right)$$ The conditions that the sum-square error be a minimum are just $$\frac{\partial \sum{i}^{N} \varepsilon_{i}^{2}}{\partial a_{i}}=2 \sum_{i=1}^{N}\left[Y_{i}-f\left(a_{j}, x_{i}\right)\right] \frac{\partial f\left(a_{j}, x_{i}\right)}{\partial a_{j}}=0, \quad j=0,1,2, \cdots, n$$

# 实分析1|Real Analysis I代写 MATH 623

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Proof. We can suppose $x<y$. By the ‘additivity’ proved in
$$\int_{a}^{y} f=\int_{a}^{x} f+\int_{x}^{y} f,$$
thus
$$\int_{x}^{y} f=F(y)-F(x) .$$

If $m^{\prime}$ and $M^{\prime}$ are the infimum and supremum of $f$ on the interval $[x, y]$, we have $m \leq m^{\prime} \leq M^{\prime} \leq M$; citing
$$m(y-x) \leq m^{\prime}(y-x) \leq \int_{x}^{y} f \leq M^{\prime}(y-x) \leq M(y-x)$$

## MATH623 COURSE NOTES ：

$$s(\sigma)=\sum_{\nu=1}^{n} f\left(a_{\nu-1}\right) \mathrm{e}{\nu}, \quad S(\sigma)=\sum{\nu=1}^{n} f\left(a_{\nu}\right) e_{\nu}$$
$$S(\sigma)-s(\sigma)=\sum_{\nu=1}^{n}\left[f\left(a_{\nu}\right)-f\left(a_{\nu-1}\right)\right] e_{\nu}$$
Now assume that the points of $\sigma$ are equally spaced, so that
$$e_{\nu}=\frac{1}{n}(b-a) \quad(\nu=1, \ldots, n) ;$$

# 代数1|Algebra I代写 MATH 611

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Proof. Let $M$ be the submodule of $I(G)$ generated by all $x_{i}-1$. We show by induction on the length of the reduced word $x \in G$ that $x-1 \in M$ for all $x \in G$ : indeed, if $x \neq 1$, then either $x=x_{i} y$ or $x=x_{i}^{-1} y$, where $y$ is shorter than $x$, and the induction hypothesis yields either $x_{i} y-1=x_{i}(y-1)+\left(x_{i}-1\right) \in M$ or $x_{i}^{-1} y-1=x_{i}^{-1}(y-1)-x_{i}^{-1}\left(x_{i}-1\right) \in M$. Then $k=\sum_{x \in G} k_{x} x \in I(G)$ implies $k=\sum_{x \in G} k_{x}(x-1) \in M$, since $\sum_{x \in G} k_{x}=0$. Thus $I(G)=M$ is generated by all $x_{i}-1$.

To prove that $\left(x_{i}-1\right){i \in I}$ is linearly independent in $I(G)$ we show that, for every $G$-module $A$ and $a{i} \in A$, there is a module homomorphism $\varphi: I(G) \longrightarrow A$ such that $\varphi\left(x_{i}-1\right)=a_{i}$ for all $i$. In particular, there is a homomorphism $\varphi_{j}: I(G) \longrightarrow \mathbb{Z}[G]$ such that $\varphi_{j}\left(x_{j}-1\right)=1$ and $\varphi_{j}\left(x_{i}-1\right)=0$ for all $i \neq j$; hence $\sum_{i \in I} k_{i}\left(x_{i}-1\right)=0$ implies $k_{j}=\varphi_{j}\left(\sum_{i \in I} k_{i}\left(x_{i}-1\right)\right)=0$ for all $j$. (Alternately, $\left(x_{i}-1\right)_{i \in I}$ has the universal property that characterizes bases.)

## MATH611 COURSE NOTES ：

Proof. Since all $P_{m}$ are projective, Theorem and the exact sequences $0 \longrightarrow K_{0} \longrightarrow P_{0} \longrightarrow A \longrightarrow 0,0 \longrightarrow K_{m} \longrightarrow P_{m} \longrightarrow K_{m-1} \longrightarrow 0$ yield exact sequences that are natural in $B$, for every $k, m \geqq 1$ :
$$\begin{gathered} 0 \longrightarrow \text { Ext }^{n+1}(A, B) \longrightarrow \text { Ext }^{n}\left(K_{0}, B\right) \longrightarrow 0 \ 0 \longrightarrow \text { Ext }^{k+1}\left(K_{m}, B\right) \longrightarrow \text { Ext }^{k}\left(K_{m-1}, B\right) \longrightarrow 0 . \end{gathered}$$
There is also a uniqueness result for syzygies.

# 概率学|Probability代写 MATH 605

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Let $F_{n}$ be a sequence of DFs defined by
$$F_{n}(x)= \begin{cases}0, & x<0 \ 1-\frac{1}{n}, & 0 \leq x<n \ 1, & n \leq x .\end{cases}$$
Clearly $F_{n} \stackrel{w}{\rightarrow} F$, where $F$ is the DF given by
$$F(x)= \begin{cases}0, & x<0 \ 1, & x \geq 0\end{cases}$$

Note that $F_{n}$ is the DF of the RV $X_{n}$ with PMFand $F$ is the DF of the RV $X$ degenerate at 0 . We have
$$E X_{n}^{k}=n^{k}\left(\frac{1}{n}\right)=n^{k-1}$$
where $k$ is a positive integer. Also $E X^{k}=0$. So that
$$E X_{n}^{k} \nrightarrow E X^{k} \quad \text { for any } k \geq 1$$

## MATH605 COURSE NOTES ：

Proof. Since $X$ is an RV, we can, given $\varepsilon>0$, find a constant $k=k(\varepsilon)$ such that
$$P{|X|>k}<\frac{\varepsilon}{2} .$$
Also, $g$ is continuous on $\mathcal{R}$, so that $g$ is uniformly continuous on $[-k, k]$. It follows that there exists a $\delta=\delta(\varepsilon, k)$ such that
$$\left|g\left(x_{n}\right)-g(x)\right|<\varepsilon$$
whenever $|x| \leq k$ and $\left|x_{n}-x\right|<\delta$. Let

# 统计计算|Statistical Computing代写 STAT 535

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Population regression function, or simply, the regression function:

For the Normal linear model
$$\mathrm{E}\left(Y_{i}\right)=\mu_{i}=\mathbf{x}{i}^{T} \boldsymbol{\beta} ; \quad Y{i} \sim \mathrm{N}\left(\mu_{i}, \sigma^{2}\right)$$
for independent random variables $Y_{1}, \ldots, Y_{N}$, the deviance is
$$D=\frac{1}{\sigma^{2}} \sum_{i=1}^{N}\left(y_{i}-\widehat{\mu}_{i}\right)^{2}$$

$$D_{0}=\frac{1}{\sigma^{2}} \sum_{i=1}^{N}\left[y_{i}-\widehat{\mu}{i}(0)\right]^{2}$$ and $$D{1}=\frac{1}{\sigma^{2}} \sum_{i=1}^{N}\left[y_{i}-\widehat{\mu}_{i}(1)\right]^{2} .$$

## STAT535 COURSE NOTES ：

If $\mathrm{E}(\mathbf{y})=\mathbf{X} \boldsymbol{\beta}$ and $\mathrm{E}\left[(\mathbf{y}-\mathbf{X} \boldsymbol{\beta})(\mathbf{y}-\mathbf{X} \boldsymbol{\beta})^{T}\right]=\mathbf{V}$, where $\mathbf{V}$ is known, we can obtain the least squares estimator $\tilde{\beta}$ of $\beta$ without making any further assumptions about the distribution of $\mathbf{y}$. We minimize
$$S_{w}=(\mathbf{y}-\mathbf{X} \boldsymbol{\beta})^{T} \mathbf{V}^{-1}(\mathbf{y}-\mathbf{X}) \boldsymbol{\beta}$$
The solution of
$$\frac{\partial S_{w}}{\partial \beta}=-2 \mathbf{X}^{T} \mathbf{V}^{-1}(\mathbf{y}-\mathbf{X} \beta)=0$$
is
$$\tilde{\boldsymbol{\beta}}=\left(\mathbf{X}^{T} \mathbf{V}^{-1} \mathbf{X}\right)^{-1} \mathbf{X}^{T} \mathbf{V}^{-1} \mathbf{y}$$

# 回归分析|Regression Analysis代写 STAT 525

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Population regression function, or simply, the regression function:
$$\mu_{Y}(x)=\beta_{0}+\beta_{1} x \quad \text { for } a \leq x \leq b$$
Sample regression function:
$$\hat{\mu}{Y}(x)=\hat{\beta}{0}+\hat{\beta}{1} x$$ Population regression model, or simply, the regression model: $$Y{I}=\beta_{0}+\beta_{1} X_{I}+E_{I} \quad \text { for } I=1, \ldots, N$$

Sample regression model:
$$y_{i}=\beta_{0}+\beta_{1} x_{i}+e_{i} \quad \text { for } i=1, \ldots, n$$
A randomly chosen $Y$ value from the subpopulation determined by $X=x$ :
$$\boldsymbol{Y}(\boldsymbol{x})$$
Sample prediction function, or simply, prediction function:
$$\hat{Y}(x)=\hat{\beta}{0}+\hat{\boldsymbol{\beta}}{1} \boldsymbol{x}$$
Note: $\hat{\mu}_{Y}(x)=\hat{Y}(x)$

## STAT525 COURSE NOTES ：

when $X$ and $Y$ are measured using the first system of units. Also suppose the regression function is
$$\mu_{Y^{}}\left(x^{}\right)=\beta_{0}^{}+\beta_{1}^{} x^{}$$ when $X^{}$ and $Y^{}$ are measured using the second system of units. Then it can be proved mathematically that $$\beta_{1}^{}=\frac{d}{b} \beta_{1}$$
and
$$\beta_{0}^{*}=c+\frac{d}{b}\left(b \beta_{0}-a \beta_{1}\right)$$

# 统计学|Statistics代写 STAT 516

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We are being asked to construct a $100(1-\alpha)$ confidence interval estimate, with $\alpha=0.10$ in part (a) and $\alpha=0.01$ in part (b). Now
$$z_{0.05}=1.645 \text { and } z_{0.005}=2.576$$
and so the 90 percent confidence interval estimator is
$$\bar{X} \pm 1.645 \frac{\sigma}{\sqrt{n}}$$

and the 99 percent confidence interval estimator is
$$\bar{X} \pm 2.576 \frac{\sigma}{\sqrt{n}}$$
For the data of Example 8.5, $n=10, \bar{X}=19.3$, and $\sigma=3$. Therefore, the 90 and 99 percent confidence interval estimates for $\mu$ are, respectively,
$$19.3 \pm 1.645 \frac{3}{\sqrt{10}}=19.3 \pm 1.56$$
and
$$19.3 \pm 2.576 \frac{3}{\sqrt{10}}=19.3 \pm 2.44$$

## STAT516 COURSE NOTES ：

$$\bar{X} \pm 1.96 \frac{\sigma}{\sqrt{n}}$$
Since the length of this interval is
$$\text { Length of interval }=2(1.96) \frac{\sigma}{\sqrt{n}}=3.92 \frac{\sigma}{\sqrt{n}}$$
we must choose $n$ so that
$$\frac{3.92 \sigma}{\sqrt{n}} \leq b$$
or, equivalently,
$$\sqrt{n} \geq \frac{3.92 \sigma}{b}$$
Upon squaring both sides we see that the sample size $n$ must be chosen so that
$$n \geq\left(\frac{3.92 \sigma}{b}\right)^{2}$$

# 统计学入门|Introduction to Statistics代写 STAT 515

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To determine the expected value of a chi-squared random variable, note first that for a standard normal random variable $Z$,
\begin{aligned} 1 &=\operatorname{Var}(Z) \ &=E\left[Z^{2}\right]-(E[Z])^{2} \ &=E\left[Z^{2}\right] \quad \text { since } E[Z]=0 \end{aligned}
Hence, $E\left[Z^{2}\right]=1$ and so
$$E\left[\sum_{i=1}^{n} Z_{i}^{2}\right]=\sum_{i=1}^{n} E\left[Z_{i}^{2}\right]=n$$

The expected value of a chi-squared random variable is equal to its number of degrees of freedom.

Suppose now that we have a sample $X_{1}, \ldots, X_{n}$ from a normal population having mean $\mu$ and variance $\sigma^{2}$. Consider the sample variance $S^{2}$ defined by
$$S^{2}=\frac{\sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}}{n-1}$$

## STAT515 COURSE NOTES ：

If the population mean $\mu$ is known, then the appropriate estimator of the population variance $\sigma^{2}$ is
$$\frac{\sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2}}{n}$$
If the population mean $\mu$ is unknown, then the appropriate estimator of the population variance $\sigma^{2}$ is
$$S^{2}=\frac{\sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}}{n-1}$$
$S^{2}$ is an unbiased estimator of $\sigma^{2}$, that is,
$$E\left[S^{2}\right]=\sigma^{2}$$