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## 数学代写|最优化作业代写optimization theory代考|Hermite Polynomials

Consider a set of $(n+1)$ points $\left(x_i, f\left(x_i\right)\right)$ with $a \leq x_0 \cdots \leq x_n \leq b$. $f$ is a function of type $C^m$ (function having continuous derivatives up to order $m$ ) with $m=\max \left(m_0, \ldots, m_n\right)$ (each $m_i$ is the value of $m$ at $x_i$ ). The osculatory polynomial approximating the function $f$ is the polynomial $P(x)$ of lower degree such that
$$\frac{d^k P\left(x_i\right)}{d x^k}=\frac{d^k f\left(x_i\right)}{d x^k} \quad \forall i=0, \ldots, n \quad \text { and } \quad \forall k=0, \ldots, m_i$$
The polynomial and the function $f$ coincide by the value of the function and its $k$ thorder derivatives at any point $x_i$. More generally, a curve is said to be osculatory to another curve when it touches it at any point and has the same tangent line and curvature at this point.

The case $m_i=1$ corresponds to Hermite polynomials; the polynomial and the function coincide by the value of the function and the first derivative. The corresponding Hermite polynomial of degree $(2 n+1)$ is given by
$$H_{2 n+1}(x)=\sum_{i=0}^n f\left(x_i\right) H_{n, i}(x)+\sum_{i=0}^n f^{\prime}\left(x_i\right) \hat{H}{n, i}(x)$$ with $H{n, i}(x)=\left[1-2\left(x-x_i\right) L_{n, i}^{\prime}\left(x_i\right)\right] L_{n, i}^2(x) \quad$ and $\hat{H}{n, i}(x)=\left(x-x_i\right) L{n, i}^2(x)$ where $L_{n, i}$ is the $i$ th factor $\left(L_i\right.$ of Equation (1.3.28)) of the Lagrange polynomial of degree $n$.
It is possible to show that
$$f(x)-H_{2 n+1}(x)+\frac{\left(x-x_0\right)^2 \ldots\left(x-x_n\right)^2}{(2 n+2) !} f^{(2 n+2)}(\xi) \quad \text { with } \xi \in[a, b]$$

if $f$ is $C^{2 n+2}$ on $[a, b]$. For that purpose, the following properties are useful:
\begin{aligned} &L_{n, i}\left(x_j\right)=0 \quad \text { if } i \neq j \ &L_{n, i}\left(x_i\right)=0 \end{aligned}
hence
$$\begin{array}{ll} H_{n, i}\left(x_j\right)=0 & \text { if } i \neq j \ H_{n, i}\left(x_i\right)=1 & \ \hat{H}{n, i}\left(x_j\right)=0 & \text { if } i \neq j \ \hat{H}{n, i}\left(x_i\right)=0 & \end{array}$$
inducing the coincidence of the polynomial and the functions as well as their respective derivatives
$$H_{2 n+1}\left(x_i\right)=f\left(x_i\right) \quad \text { and } H_{2 n+1}^{\prime}\left(x_i\right)=f^{\prime}\left(x_i\right)$$
resulting in Equation (1.3.52).

## 数学代写|最优化作业代写optimization theory代考|Chebyshev Polynomials and Irregularly Spaced Points

Suppose that the variable $x$ varies in $[-1,1]$. In this case, the interpolation polynomial $P_n(x)$ of $f(x)$ defined as a sum of monomials produces a minimum error when $x$ is close to 0 and maximum when $|x|$ is close to 1 . Thus, the error is irregularly distributed on the interval $[0,1]$. In the general case where $x \in[a, b]$, if we do a change of variable such that
$$t=\frac{2 x-a-b}{b-a}$$
we are brought back to the previous case with $t \in[-1,1]$.
It is interesting to find a series expansion of functions such that the error is better distributed and the maximum error is minimum.

The cosinus functions offer this type of advantage; their values are regularly spread on $[0, \pi]$. Moreover, the extreme values of two functions $\cos j \alpha$ and $\cos k \alpha$ with $j \neq k$ are in general reached for different values of $\alpha$. The cosinus function is numerically evaluated by a series. Indeed, we can use the expression of $\cos n \alpha$ with respect to $x=\cos \alpha$. Thus Chebyshev polynomials $T_i(x)=\cos (i \alpha)$ are defined by
$$T_i(x)=\cos (i \alpha)=2 \cos (\alpha) \cos ((i-1) \alpha)-\cos ((i-2) \alpha), \quad i=0,1, \ldots \quad \text { (1.3.63) }$$
giving the recurrence
$$T_i(x)=2 x T_{i-1}(x)-T_{i-2}(x)$$
In this way, we get
\begin{aligned} T_0(x) &=1 \ T_1(x) &=x \ T_2(x) &=2 x^2-1 \ T_3(x) &=4 x^3-3 x \ T_4(x) &=8 x^4-8 x^2+1 \ \ldots \end{aligned}
and reciprocally the monomials $1, x, x^2, x^3, \ldots$ are simply expressed with respect to Chebyshev polynomials
$$1=T_0$$
\begin{aligned} &x=T_1 \ &x^2=\frac{1}{2}\left(T_0+T_2\right) \ &x^3=\frac{1}{4}\left(3 T_1+T_3\right) \ &x^4=\frac{1}{8}\left(3 T_0+4 T_2+T_4\right) \end{aligned}

# 最优化代写

## 数学代写|最优化作业代写优化理论代考|Hermite多项式

$$\frac{d^k P\left(x_i\right)}{d x^k}=\frac{d^k f\left(x_i\right)}{d x^k} \quad \forall i=0, \ldots, n \quad \text { and } \quad \forall k=0, \ldots, m_i$$

$$H_{2 n+1}(x)=\sum_{i=0}^n f\left(x_i\right) H_{n, i}(x)+\sum_{i=0}^n f^{\prime}\left(x_i\right) \hat{H}{n, i}(x)$$ 用 $H{n, i}(x)=\left[1-2\left(x-x_i\right) L_{n, i}^{\prime}\left(x_i\right)\right] L_{n, i}^2(x) \quad$ 和 $\hat{H}{n, i}(x)=\left(x-x_i\right) L{n, i}^2(x)$ 哪里 $L_{n, i}$ 是 $i$ Th因子 $\left(L_i\right.$ 式(1.3.28))的次拉格朗日多项式的 $n$

$$f(x)-H_{2 n+1}(x)+\frac{\left(x-x_0\right)^2 \ldots\left(x-x_n\right)^2}{(2 n+2) !} f^{(2 n+2)}(\xi) \quad \text { with } \xi \in[a, b]$$

\begin{aligned} &L_{n, i}\left(x_j\right)=0 \quad \text { if } i \neq j \ &L_{n, i}\left(x_i\right)=0 \end{aligned}

$$\begin{array}{ll} H_{n, i}\left(x_j\right)=0 & \text { if } i \neq j \ H_{n, i}\left(x_i\right)=1 & \ \hat{H}{n, i}\left(x_j\right)=0 & \text { if } i \neq j \ \hat{H}{n, i}\left(x_i\right)=0 & \end{array}$$

$$H_{2 n+1}\left(x_i\right)=f\left(x_i\right) \quad \text { and } H_{2 n+1}^{\prime}\left(x_i\right)=f^{\prime}\left(x_i\right)$$

## 数学代写|最优化作业代写优化理论代考|Chebyshev多项式和不规则间隔点

$$t=\frac{2 x-a-b}{b-a}$$
，我们回到前面的情况$t \in[-1,1]$ .

$$T_i(x)=\cos (i \alpha)=2 \cos (\alpha) \cos ((i-1) \alpha)-\cos ((i-2) \alpha), \quad i=0,1, \ldots \quad \text { (1.3.63) }$$

$$T_i(x)=2 x T_{i-1}(x)-T_{i-2}(x)$$

\begin{aligned} T_0(x) &=1 \ T_1(x) &=x \ T_2(x) &=2 x^2-1 \ T_3(x) &=4 x^3-3 x \ T_4(x) &=8 x^4-8 x^2+1 \ \ldots \end{aligned}
，反过来，单项$1, x, x^2, x^3, \ldots$简单地表示为切比雪夫多项式
$$1=T_0$$
\begin{aligned} &x=T_1 \ &x^2=\frac{1}{2}\left(T_0+T_2\right) \ &x^3=\frac{1}{4}\left(3 T_1+T_3\right) \ &x^4=\frac{1}{8}\left(3 T_0+4 T_2+T_4\right) \end{aligned}

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## MATLAB代写

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