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\begin{document}
\title{ \bf The Application of Cayley-Bacharach Theorem to Bivariate Lagrange
Interpolation }
\author{Xue-Zhang Liang , Li-Hong Cui and Jie-Lin Zhang\\
(Institute of Mathematics,Jilin
University,Changchun,130012,China)}
\date{}
\maketitle
\begin{center}\bf Abstract\end{center}
\indent In this paper, we give a new proof of the famous
Cayley-Bacharach theorem by means of interpolation, and deduce a
general method of constructing properly posed set of nodes for
bivariate Lagrange interpolation . As a result, we generalize the
main results in [1], [2] and [3]
to the more extensive situations.\\
{\bf Keywords:} bivariate Lagrange interpolation; properly posed
set of nodes for bivariate Lagrange interpolation; Lagrange
interpolation along a plane algebraic curve.
\begin{center}{\Large \bf 1. Introduction }\end{center}
In this paper what we will discuss is the problem of bivariate Lagrange
interpolation in two-dimensional real plane
${\mathbf{R}}^2$, which is closely related to the problem of Lagrange
interpolation along a plane algebraic curve . Let $n$ be a nonnegative integer and
$e_n=\displaystyle\frac{1}{2}(n+1)(n+2)$. ${\mathbf{P}}_n$ denotes the space
of all bivariate polynomials of total degree $\leq n$, i.e.
$${\mathbf{P}}_n = \{\sum \limits_{0\leq {i+j}\leq n} a_{ij} x^i y^j |a_{ij} \in \mathbf{R}\}.$$
\indent
In 1965, Liang gave the definition about a properly posed
set of nodes for bivariate Lagrange interpolation in [1] as follows .\\
\indent {\bf Definition A ([1])}~~ Let ${ \cal
A}=\{Q_i\}_{i=1}^{e_n}$ be a set of $e_n$ distinct points on
${\mathbf{R}}^2$. Given any set of real numbers
$\{f_i\}_{i=1}^{e_n}$ , we seek a polynomial $g(x,y)\in
{\mathbf{P}}_n$ satisfying
\begin{equation}\label{Eq1}
g(Q_i)=f_i,\hspace{10pt}i=1,\cdots ,e_n.
\end{equation}
If for any given set $\{f_i\}_{i=1}^{e_n}$ of real numbers there
exists a unique solution for the equation system (1), we call the
interpolation problem a properly posed interpolation problem and
call the corresponding set ${\cal A}=\{Q_i\}_{i=1}^{e_n}$ of
nodes a properly posed set of nodes ( or PPSN, for short ) for
${\mathbf{P}}_n$.\\
\indent In the case of bivariate interpolation, different from
the univariate interpolation, the condition
$e_n=\dim{\mathbf{P}}_n=
\displaystyle\frac{1}{2}(n+1)(n+2)$ is only necessary for ${\cal A}=\{Q_i\}_{i=1}^{e_n}$
being a PPSN for
${\mathbf{P}}_n$, but is not sufficient. In view
of this situation, Liang gave two theorems in [1] and [7] as follows:\\
\indent {\bf Theorem A ([1])}~~ A set ${\cal
A}=\{Q_i\}_{i=1}^{e_n}$ of points is a PPSN for ${\mathbf{P}}_n$
if and only if ${\cal A}=\{Q_i\}_{i=1}^{e_n}$ is not contained in
any plane algebraic curve in ${\mathbf{P}}_n$. (We call $g(x,y)=0$
a plane algebraic curve in $\mathbf{P_n}$ if $g(x,y)\in
\mathbf{P_n}$ and $g(x,y) \not\equiv 0$. )\\
\indent{\bf Theorem B ([7])}~~ If ${\cal A}=\{Q_i\}_{i=1}^{e_n}$
is a PPSN for $\mathbf{P_n}$, and none of these points is on an
irreducible curve $q(x,y)=0$ of degree $l$ ($l=1$ or $l=2$, $l=1$
means a straight line; $l=2$ means a conic ), then ${\cal
A}=\{Q_i\}_{i=1}^{e_n}$ together with the $(n+3)l-1$ points being
distinct and selected freely on the irreducible curve $q(x,y)=0$
of degree $l$
must constitute a PPSN for ${\mathbf{P}}_{n+l}$ . \\
\indent Since any point on ${\mathbf{R}}^2$ must constitute a
PPSN for ${\mathbf{P}}_0$, basing on it, we can construct some
PPSNs for a series of interpolation spaces
${\mathbf{P}}_{1},{\mathbf{P}}_{2},{\mathbf{P}}_{3},\dots$ by
repeated using Theorem B. At the same time, we call the
constructive methods given in Theorem B a Line-Superposition
Process ( $l=1$) and a Conic-Superposition Process ($l=2$).\\
\indent In 1998, Liang and l$\ddot{u}$ [2] not only proposed the
following concept of bivariate Lagrange interpolation along an
algebraic curve
without multiple factors ( or ACWMF, for short )but also gave a method
of constructing the PPSN along an ACWMF, and proved two recursive construction theorems. \\
\indent {\bf Definition B ([2])}~~ Let $k$ be a natural number,
\begin{equation}\label{Eq2}
e_n(k)=(\begin{array}{c}
n+2 \\
2 \
\end{array})-(\begin{array}{c}
n+2-k \\
2 \
\end{array})=\left\{\begin{array}{ll}
\frac{\displaystyle 1}{\displaystyle 2}(n+1)(n+2), & \hspace{10pt} n<k \\
\frac{\displaystyle 1}{\displaystyle 2}k(2n+3-k), & \hspace{10pt} n \geq
k,
\end{array}\right.
\end{equation}
and
$q(x,y)=0$ be an ACWMF of degree $k$.
Also suppose that ${\cal B}=\{Q_i\}_{i=1}^{e_n(k)}$ is a set of
$e_n(k)$ distinct points on $q(x,y)=0$.
Given any set $\{f_i\}_{i=1}^{e_n(k)}$ of real numbers, we seek a
polynomial $g(x,y)\in {\mathbf{P}}_n$ satisfying
\begin{equation}\label{Eq3}
g(Q_i)=f_i,\hspace{10pt}i=1,\cdots ,e_n(k).
\end{equation}
We call the set ${\cal B}=\{Q_i\}_{i=1}^{e_n(k)}$ of nodes a PPSN
for polynomial interpolation of degree $n$ along the ACWMF
$q(x,y)=0$ of degree $k$ and write ${\cal{B}}\in I_n(q)$ ( where
$I_n(q)$ denotes the set of all the PPSN for polynomial
interpolation of degree $n$ along the ACWMF $q(x,y)=0$ ) , if for
each given set $\{f_i\}_{i=1}^{e_n(k)}$ of real numbers there
always exists a
solution for the equation system (3).\\
\indent {\bf Theorem C ([2])}~~ Suppose an ACWMF $q(x,y)=0$ of
degree $k$ and a straight line $l(x,y)=0$ meet exactly at $k$
distinct points ${\cal{C}}={\{Q_i\}}_{i=1}^k$, ${\cal{B}}\in
I_n(q)$ $ (n\geq k-2)$, and ${\cal{B}}\bigcap{\cal{C}}=\O$.
Then we have $${\cal{B}}\bigcup{\cal{C}}\in I_{n+1}(q).$$
\indent {\bf Theorem D ([2])}~~ Let the set ${\cal
A}=\{Q_i\}^{e_n}_{i=1}$ of points be a PPSN for ${\bf P}_n$. If
none of these points is on an ACWMF $q(x,y)=0$ of degree $k$, then
for any ${\cal D}\in I_{n+k}(q)$, ${\cal D}\cup{\cal A}$ must be a
PPSN for ${\bf P}_{n+k}$.\\
\indent We call the above two theorems Recursive Construction
Theorems.\\
\indent Theorem C gives a method
of constructing PPSN for polynomial interpolation along an ACWMF
by using the intersection of
a straight line with a curve of
degree $k$. According to Theorem C and Theorem D,
we know that Theorem C is a generalization of Theorem B.
In 2000, Liang and Cui further generalized the method given in
Theorem C to the case that a conic intersects with a curve of degree $k$. \\
\indent {\bf Theorem E ([3])}~~ Suppose an ACWMF $q(x,y)=0$ of
degree $k$ and a conic $c(x,y)=0$ meet
exactly at $2k$ distinct points
${\cal{D}}={\{Q_i\}}_{i=1}^{2k}$, ${\cal{B}}\in I_n(q)$ $(n\geq
k-2)$, and ${\cal{D}}\bigcap{\cal{B}}=\O$. Then we have
$${\cal{D}}\bigcup{\cal{B}}\in I_{n+2}(q).$$
\indent The following theorem will play a very important role in the proof of our main
results( see section 2) .\\
\indent {\bf Theorem F ([5])}~~ Let $e_n(k)$ be defined as (2),
$\{Q_i\}_{i=1}^{e_n(k)}$ a set of $e_n(k)$ distinct points on an ACWMF
$q(x,y)=0$ of degree $k$. Then $\{Q_i\}_{i=1}^{e_n(k)}\in
I_n(q)$, if and only if, for any polynomial $g(x,y)\in
{\mathbf{P}}_n$ which satisfies the following zero-interpolation condition
$$g(Q_i)=0,\hspace{10pt}i=1,\cdots ,e_n(k),$$
there always exists a polynomial $r(x,y) \in
{\mathbf{P}}_{n-k}$ ($n\geq k$) such that
$$g(x,y)=q(x,y)r(x,y),$$
( when $n<k$, it means $r(x,y)\equiv
0$ ).\\
\indent {\bf Remark 1 ([3])}~~ In Definition B the condition `` If for each given
set $\{f_i\}_{i=1}^{e_n(k)}$ of real numbers, there always exists a polynomial
$g(x,y)\in {\mathbf{P}}_n$ satisfying
$g(Q_i)=f_i,i=1,\cdots,e_n(k)$." can be replaced by the condition `` If the relations $g(x,y)\in {\mathbf{P}}_n$ and
$g(Q_i)=0,i=1,\cdots,e_n(k)$ , imply $g(x,y)\equiv 0 $
along the curve $q(x,y)=0$ of degree $k$".\\
\indent{\bf Remark 2}~~ Let the set ${\cal
A}=\{Q_i\}^{e_n}_{i=1}$ be a PPSN for ${\bf P}_n$, then ${\cal A}$
can be produced by the Recursive Construction Theorem D.\\
\indent{\bf Proof of Remark 2:} Suppose that
$l_1(x,y),\cdots,l_{e_n}(x,y)$
are the basic polynomials of Lagrange interpolation for ${\cal A}$. Then\\
$~~~~~~~~~~~~~~~~~~~~~~~~~~~l_i(Q_j)=\left\{\begin{array}{ll}
1, & j=i \\
0, & j\neq i
\end{array}\right. \hspace {20pt} i, j =1, \cdots ,e_n.
$\\ Choose a point in ${\cal A}$ freely, we might as well suppose
it is $Q_1$. It is obvious that $Q_1$ is a PPSN for ${\bf P}_{0}$.
Then by using Theorem A and Theorem F, we know that
$\cal B $$=\{Q_i\}^{e_n}_{i=2}$ must be a PPSN for polynomial
interpolation of degree $n$ along the ACWMF
$l_1(x,y)=0$ of degree $n$.\\
\indent In fact, suppose that there exists a polynomial $p(x,y)\in
{\bf P}_{n}$ ( $p(x,y) \not\equiv 0$) satisfying $p(Q_i)=0,$
$\forall Q_i \in \cal B$. By Theorem A, we have $p(Q_1)\neq 0$.\\
\indent Next set $\tilde{p}(x,y)=p(x,y)/p(Q_1)$, then we have
$\tilde{p}(Q_1)=1$ ,
$\tilde{p}(Q_i)=0,$ $i=2 ,\cdots , e_n$.\\
\indent That is to say, $ \tilde{p}(x,y)$ is the basic polynomial
corresponding to the point $Q_1$, i.e. $l_1(x,y)$. Therefore
$p(x,y)=p(Q_1)l_1(x,y)$. By Theorem F, $\{Q_i\}^{e_n}_{i=2}$ is
a PPSN for polynomial interpolation of degree $n$ along the curve
$l_1(x,y)=0$ of degree $n$ .\\
\indent From Remark $2$, we know
that the key of constructing the PPSN for ${\bf P}_{n}$ is the
construction of
PPSN for interpolation polynomial of degree $n$ along a curve of degree $n$.\\
\indent A very interesting problem is whether the conclusions given
in Theorem C and Theorem E
can be generalized to the general case that a curve of
degree $m$ (here $m$ is allowed that $m > 2$) intersects with
another one of degree $n$. In this paper, we give an affirmative
answer on the problem and deduce a general method of constructing
PPSN for Lagrange interpolation along an algebraic curve by using
Cayley-Bacharach
Theorem [6] in algebraic geometry.\\
\indent Cayley-Bacharach Theorem which we will use in this paper is as follows:\\
\indent {\bf Theorem G ([6])}~~ Let $m,n$ and $r$ be natural
numbers, and $3\leq r\leq \min {\{m,n\}}+2$. Suppose that the two
curves $p(x,y)=0$ of degree $m$ and $q(x,y)=0$ of degree $n$ meet
exactly at $mn$ distinct points. If $f(x,y)\in
{\mathbf{P}}_{m+n-r}$ and the curve $f(x,y)=0$ passes through
$mn-\displaystyle\frac{1}{2}(r-1)(r-2)$ points of those $mn$
points, then it must pass through the
$\displaystyle\frac{1}{2}(r-1)(r-2)$ remainder points, unless
these $\displaystyle\frac{1}{2}(r-1)(r-2)$ remainder points lie on
one
curve of degree $r-3$.\\
\indent In this paper, we give a new proof of
Cayley-Bacharach theorem by means of interpolation and acquire the following main result by
using Theorem G. \\
\indent {\bf Theorem $\bf 1$ } Let
$\{0\}=\mathbf{P}_{-1}=\mathbf{P}_{-2}=...$ denote the space of
zero polynomial, and under these circumstances we regard the
corresponding PPSN as the empty set. Suppose $m$ and $n$ are
natural numbers, $m\leq n$, and $\sigma$ is an integer number
satisfying $\sigma \geq 1-m$. Also suppose that the two
curves $p(x,y)=0$ of degree $m$ and $q(x,y)=0$ of degree $n$
meet exactly at $mn$ distinct points ${\cal
A}=\{Q_i\}_{i=1}^{mn},$ and $ {\cal B}\subseteq {\cal A}$ is a PPSN for
$\mathbf{P}_{-\sigma}$. Then we have\\
(1) If ${\cal C}_1$ is a PPSN for polynomial interpolation of degree $m+\sigma
-3$ along the curve $p(x,y)=0$ of degree $m$, and ${\cal C}_1\bigcap
{\cal A}=\O$, then ${\cal C}_1\bigcup ({\cal A}\backslash {\cal B})$ must be a PPSN
for polynomial interpolation of degree $m+n+\sigma-3$ along the
curve $p(x,y)=0$ of degree $m$ ;\\
(2) If ${\cal C}_2$ is a PPSN for polynomial interpolation of
degree $n+\sigma -3$ along the curve $q(x,y)=0$ of degree $n$,
and ${\cal C}_2\bigcap {\cal A}=\O$, then ${\cal C}_2\bigcup ({\cal
A}\backslash {\cal B})$ must be a PPSN for polynomial
interpolation of degree $m+n+\sigma-3$ along the
curve $q(x,y)=0$ of degree $n$.\\
\indent {\bf Remark 3} In Theorem 1, if $m+\sigma-3<0$ the empty
set ( and only the empty set ) is regarded as a PPSN for
polynomial interpolation
of degree $m+\sigma-3$ along a given curve .
\begin{center}{\Large \bf 2. The Proof of Theorems }\end{center}
\indent In order to prove Theorem 1 and Cayley-Bacharach
Theorem (i.e. Theorem G), we need the following lemma .\\
\indent {\bf Lemma 1}~~ Let $m,n$ and $r$ be natural numbers. If
the two
curves $p(x,y)=0$ of degree $m$ and $q(x,y)=0$ of degree
$n$ meet exactly at $mn$ distinct points, and the curve $f(x,y)=0$
of degree $r$ passes through these $mn$ distinct points, then
there must exist polynomials $\alpha(x,y)\in {\mathbf{P}}_{r-m}$
and
$\beta(x,y)\in {\mathbf{P}}_{r-n}$ such that
$$f(x,y)=\alpha(x,y)p(x,y)+\beta(x,y)q(x,y).$$
\indent {\bf Proof :} The proof can be realized by
proving the following three propositions.\\
\indent {\bf Proposition 1} The Lemma holds when $r\geq mn-1$.\\
\indent {\bf Proposition 2} The Lemma holds when $mn-2\geq r \geq
m+n-1$.\\
\indent {\bf Proposition 3} The Lemma holds when $m+n-2\geq r
\geq1$.\\
\indent Now we prove them respectively.\\
\indent {\bf Proof of Proposition 1 :} Let ${\cal{A}}={\{Q_i\}}_{i=1}^{mn}$
denote the set of these $mn$ points of intersection, then there exists a
set ${\{l_i(x,y)\in {\mathbf{P}}_r\}}_{i=1}^{mn} $ of
polynomials which constitute the basic polynomials of Lagrange
interpolation for $\cal{A}$. In fact it can be constructed as
follows: For each point $Q_i$ $\in \cal A$ ($i=1,\cdots , mn$), we
make a straight line $L_i(x,y)=0$ which passes through $Q_i$ $\in
\cal A$ and doesn't pass through any other point in $\cal A$. Thus
we obtain a set $\{L_i(x,y)=0\}_{i=1}^{mn}$ of straight lines. Let
$$\displaystyle l_i(x,y)=\displaystyle\frac{L_1(x,y)L_2(x,y)\cdots L_{i-1}(x,y)L_{i+1}(x,y)
\cdots L_{mn}(x,y)}{L_1(Q_i)L_2(Q_i)\cdots
L_{i-1}(Q_i)L_{i+1}(Q_i)\cdots L_{mn}(Q_i)},$$ then it is obvious
that $l_i(Q_j)=\left\{\begin{array}{ll}
1, & j=i \\
0, & j\neq i
\end{array}\right. $ \hspace {20pt}$i, j =1,\cdots,mn.$\\
\indent We select $d=\displaystyle\frac{1}{2}m(2r-2n+3-m)$
distinct points ${\{Q_i\}}_{i=1+mn}^{d+mn}$ on the curve
$p(x,y)=0$ , such that ${\cal{B}}={\{Q_i\}}_{i=1+mn}^{d+mn}$
constitutes a PPSN for polynomial interpolation of degree $r-n$
along the curve $p(x,y)=0$ (it is realizable according to Theorem
C) and $\cal B \bigcap \cal A =\O $. For any given set
${\{{\widetilde{f}_i}\}}_{i=1}^{d+mn}$ of real numbers, we
construct a polynomial $\widetilde{g}(x,y)\in {\mathbf{P}}_r$ as
follows:
\begin{equation}\label{Eq4}
\widetilde{g}(x,y)=\sum\limits_{j=1}^{mn}\widetilde{f}_jl_j(x,y) +
\widetilde{\beta} (x,y)q(x,y).
\end{equation}
\indent Where $\widetilde{\beta}(x,y)\in {\mathbf{P}}_{n-r}$ and
satisfies
\begin{equation}\label{Eq5}
\widetilde{\beta}(Q_i)=(\widetilde{g}(Q_i)-\sum\limits_{j=1}^{mn}\widetilde{f}_jl_j(Q_i))/
q(Q_i), \hspace{10pt}i=1+mn,\cdots, d+mn.
\end{equation}
\indent Since ${\cal{B}}={\{Q_i\}}_{i=1+mn}^{d+mn}\in I_{r-n}(p)$
, there must exist a polynomial $\widetilde{\beta}(x,y)\in
{\mathbf{P}}_{r-n}$ satisfying (5). From (4) and (5), we know that there
exists the polynomial $\widetilde{g}(x,y)\in {\mathbf{P}}_r$ such
that
$$\widetilde{g}(Q_i)=\widetilde{f}_i,\hspace{20pt} i=1,\cdots,d+mn.$$
The number of points in $\cal{B}\bigcup\cal{A}$ is
$$\frac{1}{2}m(2r-2n+3-m)+mn=\frac{1}{2}m(2r+3-m),$$
which is exactly equal to the number of points in a PPSN for
polynomial interpolation of degree $r$ along the curve $p(x,y)=0$
of degree $m$.\\
\indent So it follows from Definition B that
$\cal{B}\bigcup\cal{A}$ constitutes a PPSN for polynomial
interpolation of degree $r$ along the curve $p(x,y)=0$.\\
\indent Also since ${\cal{B}}={\{Q_i\}}_{i=1+mn}^{d+mn}\in
I_{r-n}(p)$, there must exist a polynomial $\beta(x,y)\in
{\mathbf{P}}_{r-n}$ satisfying
$$\beta(Q_i)=\frac{f(Q_i)}{q(Q_i)} \hspace{20pt}, \forall \hspace{5pt} Q_i\in \cal{B}.$$
Let \begin{equation} \label{6} g(x,y)=f(x,y)-\beta(x,y)q(x,y)
~~(~g(x,y)\in {\mathbf{P}}_r ~).
\end{equation}
Then $g(x,y)$ satisfies
$$g(Q_i)=f(Q_i)-\beta(Q_i)q(Q_i)=0,\hspace{20pt}\forall \hspace{5pt} Q_i\in \cal B \bigcup \cal A.$$
Because $\cal B \bigcup \cal A $$\in I_r(p)$, from Remark 1 we
know that $g(x,y)\equiv 0$ along the curve $p(x,y)=0$. Hence based
on Bezout Theorem [4] there exists $\alpha(x,y)\in
{\mathbf{P}}_{r-m}$ such that
\begin{equation}\label{Eq7}
g(x,y)=\alpha(x,y)p(x,y).
\end{equation}
Combining (6) and (7), we have
$$f(x,y)=\alpha(x,y)p(x,y)+\beta(x,y)q(x,y),$$
where $\alpha(x,y)\in {\mathbf{P}}_{r-m}$ and $\beta(x,y)\in
{\mathbf{P}}_{r-n}$. Thus we complete the proof of Proposition 1 .\\
\indent {\bf Proof of Proposition 2 :} We use the descending
induction for $r$ to prove this proposition, i.e. by
hypothesizing its truth for the polynomial of degree $r+1$ we are
to prove Lemma 1 is true for the polynomial of degree $r$.
Suppose $f(x,y)\in {\mathbf{P}}_r$, and the curve $f(x,y)=0$
passes through these $mn$ distinct points of intersection (denoted
by the set $\cal A$$ =\{Q_i\}_{i=1}^{mn}$ ). We make a straight
line $l(x,y)=0$ such that it meets with the curve $q(x,y)=0$
exactly at $n$ distinct points ${\{Q_i\}}_{i=1}^n$ and
$\{Q_i\}_{i=1}^{n}$ $\bigcap \cal A =\O$. Since $l(x,y)f(x,y)\in
{\mathbf{P}}_{r+1}$, according to the assumption of induction,
there must exist polynomials $\alpha_1(x,y)\in
{\mathbf{P}}_{r+1-m}$ and $\beta_1(x,y)\in {\mathbf{P}}_{r+1-n}$
such that
$$l(x,y)f(x,y)\equiv \alpha_1(x,y)p(x,y)+\beta_1(x,y)q(x,y)
\ \equiv (\alpha_1(x,y)-\lambda(x,y)q(x,y))p(x,y)$$
\begin{equation}\label{Eq8}
+(\beta_1(x,y)+\lambda(x,y)p(x,y))q(x,y),
\end{equation}
where $\lambda(x,y)\in {\mathbf{P}}_{r+1-m-n}$ is an arbitrary
polynomial of degree $r+1-m-n$. From (8) we know
$$\alpha_1(Q_i)=0 ,~~
\alpha_1(Q_i)-\lambda(Q_i)q(Q_i)=0 , ~~\forall \hspace{5pt}
{Q_i}\in {\{Q_i\}}_{i=1}^n.$$
Now we freely choose $r+2-m-n$
distinct points ${\{Q_i\}}_{i=n+1}^{r+2-m}$ on the straight line again such that
$ {\{Q_i\}}_{i=1}^n \bigcap
{\{Q_i\}}_{i=n+1}^{r+2-m}=\O$. Then by using the
Line-Superposition Process in Theorem A, we can construct a PPSN
$\widetilde{\cal A}$ for ${\mathbf{P}}_{r+1-m-n}$ such that ${\{Q_i\}}_{i=n+1}^{r+2-m}\subset
\widetilde{\cal A}$ and no point in $\widetilde{\cal A}$ lies on the curve $q(x,y)=0$.
Therefore, by the interpolation condition
$$\alpha_1(Q_i)-\lambda(Q_i)q(Q_i)=0, \hspace{10pt} \forall \hspace{5pt} Q_i \in \widetilde{\cal A},$$
we can obtain a unique polynomial $\lambda(x,y)\in
{\mathbf{P}}_{r+1-m-n}$ . At the same time, we have
$$\alpha_1(Q_i)-\lambda(Q_i)q(Q_i)=0, \hspace{10pt} \forall \hspace{5pt} Q_i \in
{\{Q_i\}}_{i=1}^{n}\bigcup {\{Q_i\}}_{i=n+1}^{r+2-m} .$$
Since $\alpha_1(x,y)-\lambda(x,y)q(x,y) \in {\mathbf{P}}_{r+1-m}$,
$l(x,y)$ is a factor of
$\alpha_1(x,y)-\lambda(x,y)q(x,y)$. From (8) we know that $l(x,y)$
must be a factor of $\beta_1(x,y)+\lambda(x,y)p(x,y)$. After
reducing $l(x,y)$ from both sides of the equation (8), we know
that there exist $\alpha(x,y)\in {\mathbf{P}}_{r-m}$ and
$\beta(x,y)\in {\mathbf{P}}_{r-n}$ such that
$$f(x,y)=\alpha(x,y)p(x,y)+\beta(x,y)q(x,y).$$
\indent The proof of Proposition 2 is completed.\\
\indent {\bf Proof of Proposition 3 :} The method of proving this proposition is similar
to that of Proposition 2, so we omit it here.\\
\indent Up to now we finish the proof of Lemma 1.\\
\indent Next we give a new proof of Cayley-Bacharach
Theorem (i.e. Theorem G )by interpolation .\\
\indent {\bf Proof of Cayley-Bacharach Theorem:}
We consider the cases of $r=3$ and $r>3$ respectively.\\
\indent ( i ) For $r=3$, let $Q_i\in \cal A$ $=\{Q_i\}_{i=1}^{mn}$
be the remainder point. We make a straight line $l(x,y)=0$ which
passes through the point $Q_i$ and doesn't pass through any other
point in $\cal A$, such that it meets the curve $p(x,y)=0$ at
$m$ distinct points and the curve $q(x,y)=0$ at $n$ distinct
points respectively. Then $l(x,y)f(x,y)\in {\mathbf{P}}_{m+n-2}$
and the curve $l(x,y)f(x,y)=0$ passes through all the $mn$ points
in $\cal A$. It follows from Lemma 1 that there exist
$\widetilde{\alpha}(x,y)\in {\mathbf{P}}_{n-2}$ and
$\widetilde{\beta}(x,y)\in {\mathbf{P}}_{m-2}$ such that
\begin{equation}\label{Eq9}
l(x,y)f(x,y)=\widetilde{\alpha}(x,y)p(x,y)+\widetilde{\beta}(x,y)q(x,y).
\end{equation}
From (9) we know that $l(x,y)=0$ intersects with
$\widetilde{\alpha}(x,y)=0$
at not less than $n-1$ points and
intersects with $\widetilde{\beta}(x,y)$ at not less
than $m-1$ points. So it follows from Bezout Theorem [4]that
\begin{equation}\label{Eq10}
\widetilde{\alpha}(x,y)=\alpha(x,y)l(x,y),
\end{equation}
\begin{equation}\label{Eq11}
\widetilde{\beta}(x,y)=\beta(x,y)l(x,y),
\end{equation}
where $\alpha(x,y)\in {\mathbf{P}}_{n-3}$ and $\beta(x,y)\in
{\mathbf{P}}_{m-3} $ . Substituting (10) , (11 ) into (9) and
reducing $l(x,y)$ from both sides of the equation (9) we have
$$f(x,y)=\alpha(x,y)p(x,y)+\beta(x,y)q(x,y).$$
\indent From the above equation we have $f(Q_i)=0$.\\
\indent ( ii ) For $r>3$, let
$d=\displaystyle\frac{1}{2}(r-1)(r-2)$ and $\hat{\cal
A}=$${\{Q_i\}}_{i=1}^d$ be the set of the remainder points.
Since $\hat{ \cal A}$ does not lie on any curve of degree $r-3$, it
follows from Theorem B that $\hat{\cal A}$ must constitute a PPSN
for ${\mathbf{P}}_{r-3}$. Let ${\{l_i(x,y)\in
{\mathbf{P}}_{r-3}\}}_{i=1}^d$ denote the corresponding basic
polynomials of Lagrange interpolation and $\cal A$ denote the set
of $mn$ points of intersection, then $l_i(x,y)f(x,y)\in
{\mathbf{P}}_{m+n-3}$ and $l_i(x,y)f(x,y)=0$ passes through all
points in $\cal A$ but $Q_i$ . By the conclusion from the case
of $r=3$, we know that $l_i(x,y)f(x,y)=0$ must pass through all
$mn$ points of intersection. It follows from Lemma 1 that there
exist $\alpha(x,y)\in {\mathbf{P}}_{n-3}$ and $\beta(x,y)\in
{\mathbf{P}}_{m-3} $ such that
$$l_i(x,y)f(x,y)=\alpha(x,y)p(x,y)+\beta(x,y)q(x,y).$$
Hence for any ${Q_i}\in \cal{A}$, we have $l_i(Q_i)f(Q_i)=0 $. But
$l_i(Q_i)=1\neq 0$, so $f(Q_i)=0$. Taking
$i=1,\cdots,d$ , we get the conclusion .\\
\indent This completes the proof of Cayley-Bacharach Theorem.\\
\indent {\bf Proof of Theorem 1 :} The proofs of two conclusions
in Theorem 1 are similar, so we only prove the first conclusion
here. We consider three cases : $\sigma\geq3,~\sigma<0$ and
$\sigma=0,1,2.$\\
\indent ( i ) For $\sigma\geq3$, in this case we have
$\cal{B}=\O$. By Definition A and Definition B, we know that the
number of points in ${{\cal{C}}_1}\bigcup(\cal{A}\backslash
\cal{B})=$${\cal{C}}_1\bigcup\cal{A}$ is
$$\frac{1}{2}m(2(m+\sigma-3)+3-m)+mn=\frac{1}{2}m(2(m+n+\sigma-3)+3-m),$$
which is exactly equal to the number of the points contained in a
PPSN for polynomial interpolation of degree
$m+n+\sigma-3$ along the curve $p(x,y)=0$ of degree $m$.\\
\indent Let $g(x,y)\in {\mathbf{P}}_{m+n+\sigma-3}$ which
satisfies
$$g(Q_i)=0,\hspace{10pt}\forall\hspace{7pt}Q_i\in
{\cal{C}}_1\bigcup\cal{A}.$$
Since $g(Q_i)=0$ for every $Q_i\in\cal{A}$, it
follows from Lemma 1 that there exist $\alpha(x,y)\in
{\mathbf{P}}_{n+\sigma-3}$ and $\beta(x,y)\in
{\mathbf{P}}_{m+\sigma-3} $ such that
\begin{equation}\label{Eq12}
g(x,y)=\alpha(x,y)p(x,y)+\beta(x,y)q(x,y).
\end{equation}
Also since
\begin{equation}\label{Eq13}
g(Q_i)=0,\hspace{20pt}\forall\hspace{7pt}Q_i\in{\cal{C}}_1,
\end{equation}
combining (12) and (13), we have
$$\beta(Q_i)q(Q_i)=0,\hspace{20pt}\forall\hspace{7pt}Q_i\in{\cal{C}}_1.$$
But for every $Q_i\in{\cal{C}}_1$ we have $q(Q_i)\neq 0$, hence
${\cal{B}}(Q_i)=0$. Because $\beta(x,y)\in
{\mathbf{P}}_{m+\sigma-3}$ and ${\cal{C}}_1\in I_{m+\sigma-3}(p)$,
from Theorem F we know that there exists $\widetilde{r}(x,y)\in
{\mathbf{P}}_{\sigma-3}$ such that
\begin{equation}\label{Eq14}
\beta(x,y)=p(x,y)\widetilde{r}(x,y).
\end{equation}
Substituting (14) into (12) , we have
$$g(x,y)=p(x,y)r(x,y),$$
where $r(x,y)=\alpha(x,y)+\widetilde{r}(x,y)q(x,y)\in
{\mathbf{P}}_{n+\sigma-3}$. Then it follows from Theorem F that
\begin{center}
${\cal{C}}_1\bigcup\cal{A}= $${\cal C}_1 \bigcup ({\cal A}
\backslash {\cal B})$ $ \in I_{m+n+\sigma-3}(p).$
\end{center}
\indent ( ii ) For $\sigma<0$, according Definition A and Definition B and by a simple calculation,
we firstly get that the number
of points in ${\cal C}_1\bigcup ({\cal A}\backslash {\cal
B})$ is exactly equal to the number of points which should be
contained in a PPSN for polynomial interpolation of degree
$m+n+\sigma-3$
along the curve $p(x,y)=0$ .\\
\indent Suppose that $g(x,y)\in {\mathbf{P}}_{m+n+\sigma-3}$ and
$g(Q_i)=0$ for every $Q_i\in {\cal C}_1\bigcup ({\cal A}\backslash
{\cal B})$. Since $\cal B$ is a PPSN for ${\mathbf{P}}_{-\sigma}$,
from Theorem B , $\cal B$ is not contained in any curve of degree
$-\sigma$. Also because $g(Q_i)=0$ for every $Q_i\in {\cal
A}\backslash {\cal B}$, from Cayley-Bacharach Theorem, we have
$g(Q_i)=0$ for each $Q_i\in \cal A$. At the same time, by using
Lemma 1 we have
\begin{equation}\label{Eq15}
g(x,y)=\alpha(x,y)p(x,y)+\beta(x,y)q(x,y),
\end{equation}
where $\alpha(x,y)\in {\mathbf{P}}_{n+\sigma-3}$ and
$\beta(x,y)\in
{\mathbf{P}}_{m+\sigma-3} $.\\
\indent Since $g(Q_i)=0$ for every $Q_i\in {\cal C}_1$, combining
with (15) we have $\beta(Q_i)q(Q_i)=0$. By using Bezout Theorem we
can prove that $q(Q_i)\neq 0$. So $\beta(Q_i)=0$. Also because
${\cal C}_1 \in I_{m+\sigma-3}(p)$ and $\beta(x,y)\in
{\mathbf{P}}_{m+\sigma-3}$, therefore from Remark 1 we have
$\beta(x,y)\equiv 0$ along the curve $p(x,y)=0$. So $g(x,y)\equiv
0$ along the curve $p(x,y)=0$. Then it follows from Remark 1 that
${\cal C}_1\bigcup ({\cal A}\backslash {\cal B})\in
I_{m+n+\sigma-3}(p)$.\\
\indent ( iii ) For $\sigma=0,1,2$, we can complete the proof just
like that in case (ii). So we omit it here.\\
\indent This finishes the proof of Theorem 1.
\begin{center}{\Large \bf 3. Extension, Corollaries and Example of Theorem }\end{center}
We have an extension of Cayley-Bacharach Theorem as follows:\\
\indent {\bf Theorem 2 } Let $m, n $ and $r$ be natural numbers,
$m\leq n$, and $3\leq r\leq m+2$. Suppose that the two curves
$p(x,y)=0$ of degree $m$ and $q(x,y)=0$ of degree $n$ meet
exactly at $mn$ distinct points $\cal A =$${\{Q_i\}}_{i=1}^{mn}$,
and $e_n(k)$ is defined just like (2). Also suppose that
$\widetilde{\cal A}=$${\{Q_i\}}_{i=1}^{e_{r-3}(m)}$ $\subset \cal
A$ is a PPSN for polynomial interpolation of degree $r-3$ along
the curve $p(x,y)=0$ of degree $m$. Then we have : If $f(x,y)\in
{\mathbf{P}}_{m+n-r}$ and the curve $f(x,y)=0$ passes through
$mn-e_{r-3}(m)$ points in $\cal B =$ $\cal A \backslash \widetilde{\cal A}$,
then it must pass through the $e_{r-3}(m)$ remainder points in $\cal A$.\\
\indent For Theorem 1, we deduce the following three corollaries which are
convenient to use.\\
\indent Let $\sigma\leq 0$ and $r=3-\sigma$
in Theorem 1, then we
have\\
\indent {\bf Corollary 1}~~ Suppose that $m$ and $n$ are natural
numbers, $m\leq n$, $r$ is a nonnegative integer, and $3\leq r
\leq m+2 $. If the two curves $p(x,y)=0$ of degree $m$
and $q(x,y)=0$ of degree $n$ meet exactly at
$mn$ distinct points ${\cal A} ={\{Q_i\}}_{i=1}^{mn}$, and the
set $\cal B \subset \cal A$ is a PPSN for
${\mathbf{P}}_{r-3}$, then we have \\
\indent (1) If ${\cal D}_1$ is a PPSN for polynomial
interpolation of degree $m-r$ along the curve $p(x,y)=0$ of
degree $m$ , ${\cal D}_1\bigcap
{\cal A}=\O,$ then ${\cal D}_1\bigcup ({\cal A}\backslash {\cal B})$ must constitute a PPSN
for polynomial interpolation of degree $m+n-r$ along the
curve $p(x,y)=0$ ;\\
\indent(2) If ${\cal D}_2$ is a PPSN for polynomial interpolation
of degree $n-r$ along the curve $q(x,y)=0$ of degree $n$ , ${\cal
D}_2\bigcap {\cal A}=\O,$ then ${\cal D}_2\bigcup ({\cal
A}\backslash {\cal B})$ must constitute a PPSN for polynomial
interpolation of degree $m+n-r$ along the curve
$q(x,y)=0$ .\\
\indent Let $\sigma\geq 1$ and $r=\sigma-1$ in Theorem 1, then we
have\\
\indent {\bf Corollary 2}~~ Suppose that $m$ and $n$ are natural
numbers, and $r$ is a nonnegative integer. If the two curves
$p(x,y)=0$ of degree $m$ and $q(x,y)=0$ of degree $n$ meet
exactly at
$mn$ distinct points ${\cal A} ={\{Q_i\}}_{i=1}^{mn}$, then we have \\
\indent (1) If ${\cal E}_1$ is a PPSN for polynomial interpolation
of degree $m+r-2$ along the curve $p(x,y)=0$ of degree $m$,
${\cal E}_1\bigcap
{\cal A}=\O,$ then ${\cal E}_1\bigcup {\cal A}$ must constitute a PPSN
for polynomial interpolation of degree $m+n+r-2$ along the
curve $p(x,y)=0$ ;\\
\indent (2) If ${\cal E}_2$ is a PPSN for polynomial interpolation
of degree $n+r-2$ along the curve $q(x,y)=0$ of degree $n$ ,
${\cal E}_2\bigcap {\cal A}=\O,$ then ${\cal E}_2\bigcup {\cal A}$
must constitute a PPSN for polynomial interpolation of degree
$m+n+r-2$ along the curve
$q(x,y)=0$ .\\
\indent Furthermore let $\sigma\geq 1$ and $r=m+\sigma-3$ or
$r=n+\sigma-3$ in
Theorem 1, then we have\\
\indent {\bf Corollary 3}~~ Suppose that $m$ and $n$ are natural
numbers, and $r$ is a nonnegative integer. If the two curves
$p(x,y)=0$ of degree $m$ and $q(x,y)=0$ of degree $n$ meet
exactly at $mn$ distinct points ${\cal A}
={\{Q_i\}}_{i=1}^{mn}$, then we have \\
\indent (1) If $r\geq m-2$ , ${\cal B}_1 \in I_r(p)$ and
${\cal B}_1\bigcap
{\cal A}=\O,$ then ${\cal B}_1\bigcup {\cal A}\in I_{r+n}(p)$;\\
\indent (2) If $ r\geq n-2$ , ${\cal B}_2 \in I_r(q)$ and
${\cal B}_2\bigcap
{\cal A}=\O,$ then ${\cal B}_2\bigcup {\cal A}\in I_{r+m}(q)$.\\
\indent {\bf Remark 4 }~~If we take $n=1$ and $n=2$ in case (1)
(or take $m=1$ and $m=2$ in case(2) ) in Corollary 3 , then
Theorem
C and Theorem E can be acquired respectively.\\
\indent Finally we give a concrete example. Suppose that an
elliptic circumference adding its long axis (say, its equation is
$p(x,y)=0, p(x,y)\in {\mathbf{P}}_3$ ) and another elliptic
circumference adding its long axis (say, its equation is
$q(x,y)=0, q(x,y)\in {\mathbf{P}}_3$ ) meet exactly at nine
distinct points $\{1,2,\dots,9\}$(See Figure 1 ), then these nine
points do not constitute a PPSN for any subspace of
${\mathbf{P}}_3$. But if we move any one point of them along the
curve $p(x,y)=0$ to a new position(e.g. move 1 to $1'$), then
these current nine points$\{1',2,\dots 9\}$ must constitute a PPSN
for polynomial interpolation of degree $3$ along the curve
$p(x,y)=0$. Moreover, if we add an arbitrary point (say, node 10)
out of the two curves $p(x,y)=0$ and $q(x,y)=0$, then we get a
PPSN $\{1',2,\dots 9,10\}$ for ${\mathbf{P}}_3$.
\begin{center}
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\vspace {3pt}
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\end{document}