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# 数学代写|数值分析代写Numerical analysis代考|AMATH352 General Properties of Rational Interpolation

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## 数学代写|数值分析代写Numerical analysis代考|General Properties of Rational Interpolation

Consider again a given set of support points $\left(x_i, f_i\right), i=0,1, \ldots$. We will now examine the use of rational functions
$$\Phi^{\mu, \nu}(x) \equiv \frac{P^{\mu, \nu}(x)}{Q^{\mu, \nu}(x)} \equiv \frac{a_0+a_1 x+\cdots+a_\mu x^\mu}{b_0+b_1 x+\cdots+b_\nu x^\nu}$$
for interpolating these support points. Here the integers $\mu$ and $\nu$ denote the maximum degrees of the polynomials in the numerator and denominator, respectively. We call the pair of integers $(\mu, \nu)$ the degree type of the rational interpolation problem.

The rational function $\Phi^{\mu, \nu}$ is determined by its $\mu+\nu+2$ coefficients
$$a_0, a_1, \ldots, a_\mu, b_0, b_1, \ldots, b_\nu .$$
On the other hand, $\Phi^{\mu, \nu}$ determines these coefficients only up to a common factor $\rho \neq 0$. This suggests that $\Phi^{\mu, \nu}$ is fully determined by the $\mu+\nu+1$ interpolation conditions

$(2.2 .1 .1) \quad \Phi^{\mu, \nu}\left(x_i\right)=f_i, \quad i=0,1, \ldots, \mu+\nu$
We denote by $A^{\mu, \nu}$ the problem of calculating the rational function $\Phi^{\mu, \nu}$ from (2.2.1.1).

It is clearly necessary that the coefficients $a_r, b_s$, of $\Phi^{\mu, \nu}$ solve the homogeneous system of linear equations
(2.2.1.2) $\quad P^{\mu, \nu}\left(x_i\right)-f_i Q^{\mu, \nu}\left(x_i\right)=0, \quad i=0,1, \ldots, \mu+\nu$,
or written out in full,
$a_0+a_1 x_i+\cdots+a_\mu x_i^\mu-f_i\left(b_0+b_1 x_i+\cdots+b_\nu x_i^\nu\right)=0, \quad i=0,1, \ldots, \mu+\nu$.
We denote this linear system by $S^{\mu, \nu}$.

## 数学代写|数值分析代写Numerical analysis代考|Inverse and Reciprocal Differences. Thiele’s Continued Fraction

The algorithms to be described in this section calculate rational expressions along the main diagonal of the $(\mu, \nu)$-plane:

Starting from the support points $\left(x_i, f_i\right), i=0,1, \ldots$, we build the following tableau of inverse differences:

The inverse differences are defined recursively as follows:
$$\varphi\left(x_i, x_j\right)=\frac{x_i-x_j}{f_i-f_j},$$
\begin{aligned} &\varphi\left(x_i, x_j, x_k\right)=\frac{x_j-x_k}{\varphi\left(x_i, x_j\right)-\varphi\left(x_i, x_k\right)}, \ &\varphi\left(x_i, \ldots, x_l, x_m, x_n\right)=\frac{x_m-x_n}{\varphi\left(x_i, \ldots, x_l, x_m\right)-\varphi\left(x_i, \ldots, x_l, x_n\right)} . \end{aligned}
On occasion, certain inverse differences become $\infty$ because the denominators in (2.2.2.2) vanish.

Note that the inverse differences are, in general, not symmetric functions of their arguments.

Let $P^\mu, Q^\nu$ be polynomials whose degree is bounded by $\mu$ and $\nu$, respectively. We will now try to use inverse differences in order to find a rational expression
$$\Phi^{n, n}(x)=\frac{P^n(x)}{Q^n(x)}$$
with
$$\Phi^{n, n}\left(x_i\right)=f_i \quad \text { for } i=0,1, \ldots, 2 n \text {. }$$

## 数学代写|数值分析代写Numerical analysis代考|General Properties of Rational Interpolation

$$\Phi^{\mu, \nu}(x) \equiv \frac{P^{\mu, \nu}(x)}{Q^{\mu, \nu}(x)} \equiv \frac{a_0+a_1 x+\cdots+a_\mu x^\mu}{b_0+b_1 x+\cdots+b_\nu x^\nu}$$

$$a_0, a_1, \ldots, a_\mu, b_0, b_1, \ldots, b_\nu .$$

(2.2.1.1) $\quad \Phi^{\mu, \nu}\left(x_i\right)=f_i, \quad i=0,1, \ldots, \mu+\nu$

$$\text { (2.2.1.2) } \quad P^{\mu, \nu}\left(x_i\right)-f_i Q^{\mu, \nu}\left(x_i\right)=0, \quad i=0,1, \ldots, \mu+\nu,$$

$$a_0+a_1 x_i+\cdots+a_\mu x_i^\mu-f_i\left(b_0+b_1 x_i+\cdots+b_\nu x_i^\nu\right)=0, \quad i=0,1, \ldots, \mu+\nu$$

## 数学代写|数值分析代写Numerical analysis代考|Inverse and Reciprocal

Differences. Thiele’s Continued Fraction

$$\begin{gathered} \varphi\left(x_i, x_j\right)=\frac{x_i-x_j}{f_i-f_j}, \ \varphi\left(x_i, x_j, x_k\right)=\frac{x_j-x_k}{\varphi\left(x_i, x_j\right)-\varphi\left(x_i, x_k\right)}, \quad \varphi\left(x_i, \ldots, x_l, x_m, x_n\right)=\frac{x_m-x_n}{\varphi\left(x_i, \ldots, x_l, x_m\right)-\varphi\left(x_i, \ldots, x_l, x_n\right)} . \end{gathered}$$

$$\Phi^{n, n}(x)=\frac{P^n(x)}{Q^n(x)}$$

$$\Phi^{n, n}\left(x_i\right)=f_i \quad \text { for } i=0,1, \ldots, 2 n$$

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