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# 数学代写|数值分析代写Numerical analysis代考|MAT12004 Hermite interpolation

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## 数学代写数值分析代写Numerical analysis代考|Hermite interpolation

The construction is similar to that of the Lagrange interpolation polynomial, but now requires two sets of polynomials $H_k$ and $K_k$ with $k=0, \ldots, n$; these will be defined in the proof of the next theorem.
Theorem $6.3$ (Hermite Interpolation Theorem) Let $n \geq 0$, and suppose that $x_i, i=0, \ldots, n$, are distinct real numbers. Then, given two sets of real numbers $y_i, i=0, \ldots, n$, and $z_i, i=0, \ldots, n$, there is a unique polynomial $p_{2 n+1}$ in $\mathcal{P}{2 n+1}$ such that $$p{2 n+1}\left(x_i\right)=y_i, \quad p_{2 n+1}^{\prime}\left(x_i\right)=z_i, \quad i=0, \ldots, n .$$
Proof Let us begin by supposing that $n \geq 1$. As in the case of Lagrange interpolation, we start by constructing a set of auxiliary polynomials; we consider the polynomials $H_k$ and $K_k, k=0,1, \ldots, n$, defined by
\begin{aligned} & H_k(x)=\left[L_k(x)\right]^2\left(1-2 L_k^{\prime}\left(x_k\right)\left(x-x_k\right)\right), \ & K_k(x)=\left[L_k(x)\right]^2\left(x-x_k\right), \end{aligned}
where
$$L_k(x)=\prod_{\substack{i=0 \ i \neq k}}^n \frac{x-x_i}{x_k-x_i}$$

## 数学代写|数值分析代写Numerical analysis代考|Differentiation

From the Lagrange interpolation polynomial $p_n$, defined by $(6.7)$, which is an approximation to $f$, it is easy to obtain the polynomial $p_n^{\prime}$, which is an approximation to the derivative $f^{\prime}$. The polynomial $p_n^{\prime}$ is given by
$$p_n^{\prime}(x)=\sum_{k=0}^n L_k^{\prime}(x) f\left(x_k\right), \quad n \geq 1 .$$
The degree of the polynomial $p_n^{\prime}$ is clearly at most $n-1 ; p_n^{\prime}$ is a linear combination of the derivatives of the polynomials $L_k \in \mathcal{P}n$, the coefficients being the values of $f$ at the interpolation points $x_k, k=0,1, \ldots, n$. In order to find an expression for the difference between $f^{\prime}(x)$ and the approximation $p_n^{\prime}(x)$, we might simply differentiate (6.8) to give $$f^{\prime}(x)-p_n^{\prime}(x)=\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{f^{(n+1)}(\xi(x))}{(n+1) !} \pi{n+1}(x)\right) .$$
However, the result is not helpful: on application of the chain rule, the right-hand side involves the derivative $\mathrm{d} \xi / \mathrm{d} x$; the value of $\xi$ depends on $x$, but not in any simple manner. In fact, it is not a priori clear that the function $x \mapsto \xi(x)$ is continuous, let alone differentiable. An alternative approach is given by the following theorem.

## MATLAB代写

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