# 计算机代写|机器学习代写machine learning代考|CS7641

## 计算机代写|机器学习代写machine learning代考|An Application of Norms

The problem of solving an inconsistent linear system $A x=b$ often arises in practice. This is a system where $b$ does not belong to the column space of $A$, usually with more equations than variables. Thus, such a system has no solution. Yet we would still like to “solve” such a system, at least approximately.

Such systems often arise when trying to fit some data. For example, we may have a set of $3 \mathrm{D}$ data points
$$\left{p_1, \ldots, p_n\right},$$
and we have reason to believe that these points are nearly coplanar. We would like to find a plane that best fits our data points. Recall that the equation of a plane is
$$\alpha x+\beta y+\gamma z+\delta=0,$$
with $(\alpha, \beta, \gamma) \neq(0,0,0)$. Thus, every plane is either not parallel to the $x$-axis $(\alpha \neq 0)$ or not parallel to the $y$-axis $(\beta \neq 0)$ or not parallel to the $z$-axis $(\gamma \neq 0)$.

Say we have reasons to believe that the plane we are looking for is not parallel to the $z$-axis. If we are wrong, in the least squares solution, one of the coefficients, $\alpha, \beta$, will be very large. If $\gamma \neq 0$, then we may assume that our plane is given by an equation of the form
$$z=a x+b y+d,$$
and we would like this equation to be satisfied for all the $p_i$ ‘s, which leads to a system of $n$ equations in 3 unknowns $a, b, d$, with $p_i=\left(x_i, y_i, z_i\right)$;
$$\begin{gathered} a x_1+b y_1+d=z_1 \ \vdots \vdots \vdots \ a x_n+b y_n+d=z_n . \end{gathered}$$
However, if $n$ is larger than 3 , such a system generally has no solution. Since the above system can’t be solved exactly, we can try to find a solution $(a, b, d)$ that minimizes the least-squares error
$$\sum_{i=1}^n\left(a x_i+b y_i+d-z_i\right)^2 .$$

## 计算机代写|机器学习代写machine learning代考|Limits of Sequences and Series

If $x \in \mathbb{R}$ or $x \in \mathbb{C}$ and if $|x|<1$, it is well known that the sums $\sum_{k=0}^n x^k=1+x+x^2+\cdots+x^n$ converge to the limit $1 /(1-x)$ when $n$ goes to infinity, and we write
$$\sum_{k=0}^{\infty} x^k=\frac{1}{1-x} \text {. }$$

For example,
$$\sum_{k=0}^{\infty} \frac{1}{2^k}=2$$
Similarly, the sums
$$S_n=\sum_{k=0}^n \frac{x^k}{k !}$$
converge to $e^x$ when $n$ goes to infinity, for every $x$ (in $\mathbb{R}$ or $\mathbb{C}$ ). What if we replace $x$ by a real or complex $n \times n$ matrix $A$ ?

The partial sums $\sum_{k=0}^n A^k$ and $\sum_{k=0}^n \frac{A^k}{k !}$ still make sense, but we have to define what is the limit of a sequence of matrices. This can be done in any normed vector space.

Definition 9.12. Let $(E,||)$ be a normed vector space. A sequence $\left(u_n\right)_{n \in \mathbb{N}}$ in $E$ is any function $u: \mathbb{N} \rightarrow E$. For any $v \in E$, the sequence $\left(u_n\right)$ converges to $v$ (and $v$ is the limit of the sequence $\left.\left(u_n\right)\right)$ if for every $\epsilon>0$, there is some integer $N>0$ such that
$$\left|u_n-v\right|<\epsilon \quad \text { for all } n \geq N .$$ Often we assume that a sequence is indexed by $\mathbb{N}-{0}$, that is, its first term is $u_1$ rather than $u_0$. If the sequence $\left(u_n\right)$ converges to $v$, then since by the triangle inequality $$\left|u_m-u_n\right| \leq\left|u_m-v\right|+\left|v-u_n\right|,$$ we see that for every $\epsilon>0$, we can find $N>0$ such that $\left|u_m-v\right|<\epsilon / 2$ and $\left|u_n-v\right|<\epsilon / 2$ for all $m, n \geq N$, and so
$$\left|u_m-u_n\right|<\epsilon \quad \text { for all } m, n \geq N \text {. }$$

# 机器学习代考

## 计算机代写|机器学习代写machine learning代考|An Application of Norms

Veft $\left{p_{-} 1\right.$, \dots, p_nไright $}$,

$$\alpha x+\beta y+\gamma z+\delta=0,$$

$$z=a x+b y+d,$$

$$a x_1+b y_1+d=z_1::: a x_n+b y_n+d=z_n .$$

$$\sum_{i=1}^n\left(a x_i+b y_i+d-z_i\right)^2 .$$

## 计算机代写|机器学习代写machine learning代考|Limits of Sequences and Series

$$\left|u_n-v\right|<\epsilon \quad \text { for all } n \geq N .$$ 通常我们假设一个序列被索引 $\mathbb{N}-0$ ，即它的第一项是 $u_1$ 而不是 $u_0$. 如果序列 $\left(u_n\right)$ 收敛到 $v$ ，那么由于由 三角不等式 $$\left|u_m-u_n\right| \leq\left|u_m-v\right|+\left|v-u_n\right| \text {, }$$ 我们看到，对于每个 $\epsilon>0$ ， 我们可以找 $N>0$ 这样 $\left|u_m-v\right|<\epsilon / 2$ 和 $\left|u_n-v\right|<\epsilon / 2$ 对所有人 $m, n \geq N$ ，所以
$$\left|u_m-u_n\right|<\epsilon \quad \text { for all } m, n \geq N \text {. }$$

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