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# 数学代写|现代代数代考Modern Algebra代写|MATH342

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## 数学代写|现代代数代考Modern Algebra代写|Cauchy interpolation

The polynomial interpolation problem is, given a collection of sample values $v_i=$ $f\left(u_i\right) \in F$ for $0 \leq i<n$ of an unknown function $f: F \longrightarrow F$ at distinct points $u_0, \ldots, u_{n-1}$ of a field $F$, to compute a polynomial $g \in F[x]$ of degree less than $n$ that interpolates $g$ at those points, so that $g\left(u_i\right)=v_i$ for all $i$. We saw in Section 5.2 that such a polynomial always exists uniquely and learned how to compute it using the Lagrange interpolation formula.

A more general problem is Cauchy interpolation or rational interpolation, where furthermore $k \in{0, \ldots, n}$ is given and we are looking for a rational function $r / t \in F(x)$, with $r, t \in F[x]$, such that
$$t\left(u_i\right) \neq 0 \text { and } \frac{r\left(u_i\right)}{t\left(u_i\right)}=v_i \text { for } 0 \leq i<n, \quad \operatorname{deg} r<k, \quad \operatorname{deg} t \leq n-k .$$
Like polynomial interpolation, Cauchy interpolation can be used to approximate real-valued functions given only by their values at a finite set of points. Empirically, it is often the case that the approximation error is smaller for rational functions than for polynomials, in particular, when the function to be approximated has singularities; we will see an example below.

Obviously $t=1$ and $r=g$, where $g$ is an interpolating polynomial as above, is a solution to (20) for $k=n$, but it is not clear whether solutions for other values of $k$ exist. Multiplying (20) by $t\left(u_i\right)$ and dropping the requirement that it be nonzero, we obtain the weaker condition
$$r\left(u_i\right)=t\left(u_i\right) v_i \text { for } 0 \leq i<n, \quad \operatorname{deg} r<k, \quad \operatorname{deg} t \leq n-k .$$
Now for any $i, r\left(u_i\right)=t\left(u_i\right) v_i=t\left(u_i\right) g\left(u_i\right)$ if and only if $r \equiv \operatorname{tg} \bmod \left(x-u_i\right)$, and by the Chinese Remainder Theorem, Corollary 5.3, (21) is in turn equivalent to (17) with $m=\left(x-u_0\right) \cdots\left(x-u_{n-1}\right)$. The following consequence of Theorem 5.16 on rational function reconstruction gives a complete answer on existence and uniqueness of a solution to (20).

Let $F$ be a field and $g=\sum_{i \geq 0} g_i x^i \in F[[x]]$ with all $g_i \in F$ be a formal power series (Section 25.3). A Padé approximant to $g$ is a rational function $\rho=r / t \in F(x)$, with $r, t \in F[x]$ and $x \nmid t$, that “approximates” $g$ to a sufficiently high power of $x$. More precisely, $r / t$ is a $(k, n-k)$-Padé approximant to $g$ if
$$x \nmid t \text { and } \frac{r}{t} \equiv g \bmod x^n, \quad \operatorname{deg} r<k, \quad \operatorname{deg} t \leq n-k ;$$
the congruence is equivalent to $r \equiv \operatorname{tg} \bmod x^n$. Obviously $r=\sum_{0 \leq i<n} g_i x^i$, the Taylor expansion of order $n$ of $g$ around 0 , and $t=1$ is an (n,0)-Padé approximant for each $n \in \mathbb{N}$, but it is not clear whether approximants for $k<n$ exist. A more general question is to ask for Padé approximants around $u$ of a formal power series in $x-u$ for an arbitrary $u \in F$. This may be reduced to (22) by performing the shift of variable $x \longmapsto x+u$.

In numerical analysis, one is interested in approximating arbitrary (sufficiently smooth) real-valued functions by “simple” functions such as polynomials or rational functions. Taylor expansions and Padé approximants provide such approximations in the vicinity of the origin (or any other point, after an appropriate change of variable). As in the case of interpolation, it was observed empirically that sometimes rational functions yield a much smaller approximation error, in particular when the function to be approximated has singularities; see Example 5.23 below.
The similarity with Cauchy interpolation is clear: instead of prescribing the values of $\rho$ at $n$ distinct points $u_0, \ldots, u_{n-1}$, we have $u_0=\cdots=u_{n-1}=0$ and prescribe an initial segment of the Taylor expansion of $\rho$ at $u_0$. Indeed the statements of the previous section carry over almost literally if we replace $m=\left(x-u_0\right) \cdots\left(x-u_{n-1}\right)$ by $m=x^n$. The following is a consequence of Theorem 5.16.