# 数学代写|数值分析代写numerical analysis代考|MATHS7104

## 数学代写|数值分析代写numerical analysis代考|Lemma

Suppose that $v \in V$ and the cluster point $u^$ form a minimum sequence. If $u^ \in T$, then it is the best approximation of $v$ out of $T$.
Proof: Suppose that $\left(u_i\right)$ is a minimum sequence that
$$\lim {i \rightarrow \infty}\left|v-u_i\right|=E_T(v)$$ Also assume that subsequence $\left(u{j(i)}\right)$ converges to $u^* \in T$. In this case, given that
$$\lim {i \rightarrow \infty}\left|v-u_i\right|=E_T(v), \quad \lim {j \rightarrow \infty}\left|u_i-u^{+}\right|=0$$
we can say that for every $j$, we have:
$$\left|u-u^2\right| \leq\left|v-u_i\right|+\left|u_j-u^2\right|, \quad\left|v-u^{+}\right| \leq E_T(v)$$
For every $u \in T$, we have
$$E_T(v) \leq|v-u|$$
So, it can be concluded that
$$\left|v-u^2\right|=E_T(v)$$
and $u^*$ is the best approximation.

## 数学代写|数值分析代写numerical analysis代考|Types of Splines

In this section, the types of splines in vector space $S_l\left(\Omega_n\right)$ for $l=2 m-1$ and $m \geq 2$ are discussed.
It is obvious that
$$\operatorname{dim}\left(S_l\left(\Omega_n\right)\right)=n+2 m-1$$
Given that
$$\forall S\left(S \in S_l\left(\Omega_n\right) \Rightarrow S\left(x_i\right)=f_i, \quad i=0, \ldots, n\right)$$
To obtain this number of equations, we have the following conditions:

1. Suppose that $f \in c^m[a, b]$ and $2 \leq m \leq n+1$ and
$$S^{(\mu)}(a)=S^{(\mu)}(b)=0, \quad \mu=m, \ldots, 2 m-2$$
In this case, a natural spline is obtained.
2. Assuming $f \in c^m[a, b]$ and $2 \leq m \leq n+1$ and
$$S^{(\mu)}(a)=f^{(\mu)}(a), \quad S^{(\mu)}(b)=f^{(\mu)}(b), \quad \mu=1, \ldots, m-1$$
we have a bounded (Hermite) spline.
3. Assuming $f \in c^m[a, b]$ and
\begin{aligned} &f^{(k)}(a)=f^{(k)}(b), \quad k=0,1, \ldots, m-1 \ &S^{(\mu)}(a)=S^{(\mu)}(b), \quad \mu=1, \ldots, 2 m-2 \end{aligned}
we have periodic splines.

# 数值分析代考

## 数学代写|数值分析代写numerical analysis代考|Rational Function Interpolation

$$\frac{P_n(x)}{Q_d(x)}=\frac{p_n x^n+p_{n-1} x^{n-1}+\ldots+p_1 x+p_0}{q_d x^d+q_{d-1} x^{d-1}+\ldots+q_1 x+q_0}$$

## 数学代写|数值分析代写numerical analysis代考|The Best Approximation Problem

$$\max |f(x)-g(x)|$$

$$\int_a^b|f(x)-g(x)| d x$$

$$\left(\int_a^b(f(x)-g(x))^2 d x\right)^{1 / 2}$$

$$\max \left|f_1(x)-f_2(x)\right|=1000$$

$$\int_a^b\left|f_1(x)-f_2(x)\right| d x \doteq 49.5$$
(曲线之间的面积只有大约 50 个单位) 。如果我们希望使用 $f_2$ 近似值 $f_1$ 那么我们必须预料到错误的顺序 $10^3$ ，但如果我们希望使用 $f_2$ 近似积分 $f_1$ 那么我们应该预计会出现大约 $\$ 10^{\wedge} 1\$ 的错误。

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