### 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Interpolation and Approximation Theory

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Lagrange Form of Polynomial Interpolation

The simplest case is to find a straight line
$$p(x)=a_{1} x+a_{0}$$
through a pair of points given by $\left(x_{0}, f_{0}\right)$ and $\left(x_{1}, f_{1}\right)$. This means solving 2 equations, one for each data point. Thus we have 2 degrees of freedom. For a quadratic curve there are 3 degrees of freedom, fitting a cubic curve we have 4 degrees of freedom, etc.

Let $\mathbb{P}{n}[x]$ denote the linear space of all real polynomials of degree at most $n$. Each $p \in \mathbb{P}{n}[x]$ is uniquely defined by its $n+1$ coefficients. This gives $n+1$ degrees of freedom, while interpolating $x_{0}, x_{1}, \ldots, x_{n}$ gives rise to $n+1$ conditions.

As we have mentioned above, in determining the polynomial interpolant we can solve a linear system of equations. However, this can be done more easily.
Definition 3.1 (Lagrange cardinal polynomials). These are given by
$$L_{k}(x):=\prod_{\substack{l=0 \ l \neq k}}^{n} \frac{x-x l}{x_{k}-x_{l}}, \quad x \in \mathbb{R}$$

Note that the Lagrange cardinal polynomials lie in $\mathbb{P}{n}[x]$. It is easy to verify that $L{k}\left(x_{k}\right)=1$ and $L_{k}\left(x_{j}\right)=0$ for $j \neq k$. The interpolant is then given by the Lagrange formula
$$p(x)=\sum_{k=0}^{n} f_{k} L_{k}(x)=\sum_{k=0}^{n} f_{k} \prod_{\substack{l=0 \ l \neq k}}^{n} \frac{x-x_{l}}{x_{k}-x_{l}} .$$
Exercise 3.1. Let the function values $f(0), f(1), f(2)$, and $f(3)$ be given. We want to estimate
$$f(-1), f^{\prime}(1) \text { and } \int_{0}^{3} f(x) d x .$$
To this end, we let $p$ be the cubic polynomial that interpolates these function values, and then approximate by
$$p(-1), p^{\prime}(1) \text { and } \int_{0}^{3} p(x) d x$$
Using the Lagrange formula, show that every approximation is a linear combination of the function values with constant coefficients and calculate these coefficients. Show that the approximations are exact if $f$ is any cubic polynomial.

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Newton Form of Polynomial Interpolation

Another way to describe the polynomial interpolant was introduced by Newton. First we need to introduce some concepts, however.

Definition $3.2$ (Divided difference). Given pairwise distinct points $x_{0}, x_{1}, \ldots, x_{n} \in[a, b]$, let $p \in \mathbb{P}{n}[x]$ interpolate $f \in C^{m}[a, b]$ at these points. The coefficient of $x^{n}$ in $p$ is called the divided difference of degree $n$ and denoted by $f\left[x{0}, x_{1}, \ldots, x_{n}\right]$.
From the Lagrange formula we see that
$$f\left[x_{0}, x_{1}, \ldots, x_{n}\right]=\sum_{k=0}^{n} f\left(x_{k}\right) \prod_{\substack{l=0 \ l \neq k}}^{n} \frac{1}{x_{k}-x_{l}}$$
Theorem 3.2. There exists $\xi \in[a, b]$ such that
$$f\left[x_{0}, x_{1}, \ldots, x_{n}\right]=\frac{1}{n !} f^{(n)}(\xi) .$$
Proof. Let $p$ be the polynomial interpolant. The difference $f-p$ has at least $n+1$ zeros. By applying Rolle’s theorem $n$ times, it follows that the $n^{\text {th }}$ derivative $f^{(n)}-p^{(n)}$ is zero at some $\xi \in[a, b]$. Since $p$ is of degree $n, p^{(n)}$ is constant, say $c$, and we have $f^{(n)}(\xi)=c$. On the other hand the coefficient of $x^{n}$ in $p$ is given by $\frac{1}{n !} c$, since the $n^{\text {th }}$ derivative of $x^{n}$ is $n !$. Hence we have
$$f\left[x_{0}, x_{1}, \ldots, x_{n}\right]=\frac{1}{n !} c=\frac{1}{n !} f^{(n)}(\xi)$$
Thus, divided differences can be used to approximate derivatives.

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Polynomial Best Approximations

We now turn our attention to best approximations where best is defined by a norm (possibly introduced by a scalar product) which we try to minimize. Recall that a scalar or inner product is any function $\mathbb{V} \times \mathbb{R}$, where $\mathbb{V}$ is a real vector space, subject to the three axioms
Symmetry:
$$\langle x, y\rangle=\langle y, x\rangle \text { for all } x, y \in \mathbb{V},$$
Non-negativity:
$\langle x, x\rangle \geq 0$ for all $x \in \mathbb{V}$ and $\langle x, x\rangle=0$ if and only if $x=0$, and
Linearity:
$$\langle a x+b y, z\rangle=a\langle x, z\rangle+b\langle y, z\rangle \text { for all } x, y, z \in \mathbb{V}, a, b \in \mathbb{R} \text {. }$$
We already encountered the vector space $\mathbb{R}^{n}$ and its scalar product with the QR factorization of matrices. Another example of a vector space is the space of polynomials of degree $n, \mathbb{P}_{n}[x]$, but no scalar product has been defined for it so far.

Once a scalar product is defined, we can define orthogonality: $x, y \in \mathbb{V}$ are orthogonal if $\langle x, y\rangle=0$. A norm can be defined by
$$|x|=\sqrt{\langle x, x\rangle} \quad x \in \mathbb{V} .$$
For $\mathbb{V}=C[a, b]$, the space of all continuous functions on the interval $[a, b]$, we can define a scalar product using a fixed positive function $w \in C[a, b]$, the weight function, in the following way
$$\langle f, g\rangle:=\int_{a}^{b} w(x) f(x) g(x) d x \text { for all } f, g \in C[a, b] .$$
All three axioms are easily verified for this scalar product. The associated norm is
$$|f|_{2}=\sqrt{\langle f, f\rangle}=\sqrt{\int_{a}^{b} w(x) f^{2}(x) d x} .$$

For $w(x) \equiv 1$ this is known as the $L_{2}$-norm. Note that $\mathbb{P}{n}$ is a subspace of $C[a, b]$. Generally $L{p}$ norms are defined by
$$|f|_{p}=\left(\int_{a}^{b}|f(x)|^{p} d x\right)^{1 / p}$$
The vector space of functions for which this integral exists is denoted by $L_{p}$. Unless $p=2$, this is a normed space, but not an inner product space, because this norm does not satisfy the parallelogram equality given by
$$2|f|_{p}^{2}+2|g|_{p}^{2}=|f+g|_{p}^{2}+|f-g|_{p}^{2}$$
required for a norm to have an associated inner product.
Let $g$ be an approximation to the function $f$. Often $g$ is chosen to lie in a certain subspace, for example the space of polynomials of a certain degree. The best approximation chooses $g$ such that the norm of the error $|f-g|$ is minimized. Different choices of norm give different approximations.
For $p \rightarrow \infty$ the norm becomes
$$|f|_{\infty}=\max {x \in[a, b]}|f(x)|{.}$$
We actually have already seen the best $L_{\infty}$ approximation from $\mathbb{P}{n}[x]$. It is the interpolating polynomial where the interpolation points are chosen to be the Chebyshev points. This is why the best approximation with regards to the $L{\infty}$ norm is sometimes called the Chebyshev approximation. It is also known as Minimax approximation, since the problem can be rephrased as finding $g$ such that
$$\min {g} \max {x \in[a, b]}|f(x)-g(x)| .$$

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Lagrange Form of Polynomial Interpolation

p(X)=一种1X+一种0

p(X)=∑ķ=0nFķ大号ķ(X)=∑ķ=0nFķ∏l=0 l≠ķnX−XlXķ−Xl.

F(−1),F′(1) 和 ∫03F(X)dX.

p(−1),p′(1) 和 ∫03p(X)dX

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Newton Form of Polynomial Interpolation

F[X0,X1,…,Xn]=∑ķ=0nF(Xķ)∏l=0 l≠ķn1Xķ−Xl

F[X0,X1,…,Xn]=1n!F(n)(X).

F[X0,X1,…,Xn]=1n!C=1n!F(n)(X)

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Polynomial Best Approximations

⟨X,是⟩=⟨是,X⟩ 对全部 X,是∈在,

⟨X,X⟩≥0对全部X∈在和⟨X,X⟩=0当且仅当X=0，和

⟨一种X+b是,和⟩=一种⟨X,和⟩+b⟨是,和⟩ 对全部 X,是,和∈在,一种,b∈R.

|X|=⟨X,X⟩X∈在.

⟨F,G⟩:=∫一种b在(X)F(X)G(X)dX 对全部 F,G∈C[一种,b].

|F|2=⟨F,F⟩=∫一种b在(X)F2(X)dX.

|F|p=(∫一种b|F(X)|pdX)1/p

2|F|p2+2|G|p2=|F+G|p2+|F−G|p2

|F|∞=最大限度X∈[一种,b]|F(X)|.

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