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\renewcommand{\refname}{Bibliography}

\title{Splines and Simplexes}
\author{V.A.Malyshev}
\date{}

\begin{document}
\maketitle

\begin{abstract}
ø{\small\it  $\!\!\!\!\!\!\!\!\!\!\!\!$ Let $V\subset \mathbb{R}^{n}$ be a simplex. For
any nonzero $w\in\mathbb{R}^{n}$ the function
$$
øB(t)=\frac{d}{{\rm vol}V}\frac{d}{dt}{\rm vol}\{x\in V:\; (w,x)\le t\}
$$
is a basis spline.}
\end{abstract}



\noindent {Introduction}
\medskip

\noindent {1. Hermite polynomials}

\noindent {2. Basis splines}

\noindent {3. Splines and volumes}

\medskip\noindent {Bibliography}


%%%%%%%%%%%%%%%%%%%
\section*{Introduction}


In the paper we consider connections between polynomial splines and Euclidean simplexes.
In the first and the second section we collect together necessary definitions and
statements about interpolation polynomials and polynomial splines~\cite{Mape1}. Geometric
properties of basis splines are considered in the third section.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{1. Hermite polynomials}

Let $f(t)$ be a sufficiently smooth function, and let $t_{0}<t_{1}<\ldots<t_{n}$ be
points of interpolation. The polynomial $L(t)$ of degree $\le n$ is called the Lagrange
polynomial if
$$
L(t_{i})=f(t_{i})
$$
for all $0\le i\le n$.




\bigskip
\bigskip
\bigskip

\begin{center}
 \begin{picture}(400,20)
 \put(-40,30){\line(1,0){70}}
 \put(-25,15){{\footnotesize{\it 2000 Mathematics Subject Classification.} Primary 41-01, 65-01,
  41A15, 65D07.}}
 \put(-25,0){{\footnotesize{\it Key words and phrases.} Polynomial splines, Euclidean simplexes.}}
\end{picture}
\end{center}

\newpage


Let
$$
\omega_i(t)=\prod_{\stackrel{\scriptstyle{j=0}}{j\ne i}}^n(t-t_j)
$$
and
$$
L_i(t)=\frac{\omega_{i}(t)}{\omega_{i}(t_{i})}.
$$
Then
$$
L(t)=\sum_{i=0}^{n}f(t_{i})L_{i}(t).
$$
In particular,
$$
F(t)=\sum_{i=0}^{n}F(t_{i})L_{i}(t)
$$
for any polynomial $F(t)$ of degree $\le n$.

The function
$$
\Delta_{n}f(t_{0},\ldots,t_{n})=\sum_{i=0}^{n}\frac{f(t_{i})}{\omega_{i}(t_{i})}.
$$
is called the divided difference of order $n$. We see that
$$
ø\Delta_{n}f(t_{0},\ldots,t_{n})=\frac{1}{n!}L^{(n)}(t).
$$
We recall the well-known Newton formula
$$
L(t)=\Delta_{0}f(t_{0})+\Delta_{1}f(t_{0},t_{1})(t-t_{0})+\cdots+\Delta_{n}f(t_{0},\ldots,t_{n})
(t-t_{0})\ldots(t-t_{n-1}).
$$
By the Newton formula, there exists a point $t_{0}<\xi<t_{n}$ such that
$$
ø\Delta_{n}f(t_{0},\ldots,t_{n})=\frac{1}{n!}f^{(n)}(\xi).
$$
Hence
$$
\lim_{\varepsilon\to0}\Delta_{n}f(t_{0},t_{0}+\varepsilon,\ldots,t_{0}+n\varepsilon)=
\frac{1}{n!}f^{(n)}(t_{0}).
$$
Since
$$
ø\Delta_{n}f(t_{0},t_{0}+\varepsilon,\ldots,t_{0}+n\varepsilon)=
\frac{1}{\varepsilon^{n}}\frac{1}{n!}\sum_{i=0}^{n}(-1)^{i+n}C_{n}^{i}f(t_{0}+i\varepsilon),
$$
we see that
$$
\lim_{\varepsilon\to0}\frac{1}{\varepsilon^{n}}
\sum_{i=0}^{n}(-1)^{i+n}C_{n}^{i}f(t_{0}+i\varepsilon)=f^{(n)}(t_{0}).
$$



Let $t_{0}<t_{1}<\ldots<t_{m}$ be points of interpolation with multiplicities
$r_{0}\ge1,\ldots,r_{m}\ge1$. Let us put
$$
n=r_{0}+\ldots+r_{m}-1.
$$
Then we see $n+1$ points in the sequence
$$
ø\underbrace{t_{0},\ldots,t_{0}}_{r_{0}},\;\underbrace{t_{1},\ldots,t_{1}}_{r_{1}},\;
\cdots,\;\underbrace{t_{m},\ldots,t_{m}}_{r_{m}}.
$$
The polynomial $H(t)$ of degree $\le n$ is called the Hermite polynomial if
$$
H^{(p)}(t_{i})=f^{(p)}(t_{i})
$$
for all $0\le i\le m$ and $0\le p\le r_{i}$.


Let
$$
\Omega_i(t)=\prod_{\stackrel{\scriptstyle{j=0}}{j\ne i}}^m(t-t_j)^{r_{j}}
$$
and
$$
øH_{ip}(t)=\frac{(t-t_{i})^{p}}{p!}\Omega_i(t)\sum_{k=0}^{r_{i}-1-p}
\left.\left(\frac{1}{\Omega_{i}(t)}\right)^{(k)}\right|_{t=t_{i}}\frac{(t-t_{i})^{k}}{k!},
$$
where $0\le i\le m$ and $0\le p\le r_{i}-1$.


\begin{theorem}
We have
$$
H_{ip}^{(q)}(t_{j})=\left\{\begin{array}{ccl}
  1 & ø{\rm if} & øj=i\;{\rm and}\;q=p, \\
  0 & ø{\rm if} & øj=i\;{\rm and}\;q\ne p,\\
  0 & ø{\rm if} & øj\ne i,
\end{array}\right.
$$
where $0\le i,j\le m$, $0\le p \le r_{i}-1$ and $0\le q \le r_{j}-1$.
\end{theorem}
\noindent{\bf Proof.} The identities
$$
H_{ip}^{(q)}(t_{j})=\left\{\begin{array}{ccl}
  1 & ø{\rm if} & øj=i\;{\rm and}\;q=p, \\
  0 & ø{\rm if} & øj=i\;{\rm and}\;q\le p-1,\\
  0 & ø{\rm if} & øj\ne i
\end{array}\right.
$$
are obvious.

Let $u(t)$ be a differentiable functions such that $u(0)\ne 0$. Let us put
$$
øv(t)=\sum_{k=0}^{r}\left.\left(\frac{1}{u(t)}\right)^{(k)}\right|_{t=0}\frac{t^{k}}{k!}.
$$
We claim that
$$
(u(t)v(t))^{(q)}|_{t=0}=0
$$
for all $1\le q\le r$. Indeed,
$$
ø(u(t)v(t))^{(q)}=\sum_{k=0}^{q}C_{q}^{k}u^{(q-k)}(t)v^{(k)}(t).
$$
Hence
$$
(u(t)v(t))^{(q)}|_{t=0}=\sum_{k=0}^{q}C_{q}^{k}u^{(q-k)}(0)
\left.\left(\frac{1}{u(t)}\right)^{(k)}\right|_{t=0}=
\left.\left(u(t)\frac{1}{u(t)}\right)^{(q)}\right|_{t=0}=1^{(q)}=0.
$$
By the statement, we have
$$
øH_{ip}^{(q)}(t_{i})=C_{q}^{p}\left.\left(\Omega_i(t)\sum_{k=0}^{r_{i}-1-p}
\left.\left(\frac{1}{\Omega_{i}(t)}\right)^{(k)}\right|_{t=t_{i}}
\frac{(t-t_{i})^{k}}{k!}\right)^{(q-p)}\right|_{t=t_{i}}=0.
$$
The theorem is proved. $\bullet$
\medskip



By Theorem 1, the Hermite polynomial has the form
$$
H(t)=\sum_{i=0}^{m}\sum_{p=0}^{r_{i}}f^{(p)}(t_{i})H_{ip}(t).
$$
In particular,
$$
F(t)=\sum_{i=0}^{m}\sum_{p=0}^{r_{i}}F^{(p)}(t_{i})H_{ip}(t)
$$
for any polynomial $F(t)$ of degree $\le n$.

The function
$$
ø\Delta_{n}f(\underbrace{t_{0},\ldots,t_{0}}_{r_{0}},\;
\cdots,\;\underbrace{t_{m},\ldots,t_{m}}_{r_{m}})=\frac{1}{n!}H^{(n)}(t)
$$
is called the divided difference of order $n$ with multiple arguments. By definition, we
put
$$
\Delta_{n}f(t_{0}^{r_{0}},\ldots,t_{m}^{r_{m}})=
\Delta_{n}f(\underbrace{t_{0},\ldots,t_{0}}_{r_{0}},\;
\cdots,\;\underbrace{t_{m},\ldots,t_{m}}_{r_{m}}).
$$
It is clear that
$$
\Delta_{n}f(t_{0}^{r_{0}},\ldots,t_{m}^{r_{m}})=\sum_{i=0}^{m}\frac{1}{(r_{i}-1)!}
\sum_{p=0}^{r_{i}-1}C_{r_{i-1}}^{p}f^{(p)}(t_{i})
\left.\left(\frac{1}{\Omega_{i}(t)}\right)^{(r_{i}-1-p)}\right|_{t=t_{i}}.
$$
The right-hand side may be written in the form
$$
\Delta_{n}f(t_{0}^{r_{0}},\ldots,t_{m}^{r_{m}})=\sum_{i=0}^{n}\frac{1}{(r_{i}-1)!}
\left.\left(\frac{f(t)}{\Omega_{i}(t)}\right)^{(r_{i}-1)}\right|_{t=t_{i}}.
$$

Let the Lagrange polynomial $L^{\varepsilon}(t)$ be defined for the interpolation points
$$
øt_{0},t_{0}+\varepsilon,\ldots,t_{0}+(r_{0}-1)\varepsilon,\;\ldots\ldots,\;
 t_{m},t_{m}+\varepsilon,\ldots,t_{m}+(r_{m}-1)\varepsilon.
$$
We put
$$
t_{ip}^{\varepsilon}=t_{i}+p\varepsilon,
$$
where $0\le i\le m$ and $0\le p\le r_{i}-1$. Then
$$
L^{\varepsilon}(t)=\sum_{i=0}^{m}\sum_{p=0}^{r_{i}-1}
f(t_{ip}^{\varepsilon})L_{ip}^{\varepsilon}(t)
$$
with some polynomials $L_{ip}^{\varepsilon}(t)$.

\begin{theorem}
We have
$$
\lim_{\varepsilon\to0}L^{\varepsilon}(t)=H(t)
$$
for any $t$.
\end{theorem}
{\bf Proof.} By the identity
$$
\sum_{p=0}^{r-1}x_{p}y_{p}=\sum_{p=0}^{r-1}
\left(\sum_{k=0}^{p}(-1)^{p+k}C_{p}^{k}x_{k}\right)
\left(\sum_{k=p}^{r-1}C_{k}^{p}y_{k}\right),
$$
we write $L^{\varepsilon}(t)$ in the form
$$
L^{\varepsilon}(t)=\sum_{i=0}^{m}\sum_{p=0}^{r_{i}-1}
\left(\frac{1}{\varepsilon^{p}}\sum_{k=0}^{p}(-1)^{p+k}C_{p}^{k}f(t_{ik}^{\varepsilon})\right)
\left(\varepsilon^{p}\sum_{k=p}^{r_{i}-1}C_{k}^{p}L_{ik}^{\varepsilon}(t)\right).
$$
Let
$$
H^{\varepsilon}_{ip}(t)=\varepsilon^{p}\sum_{k=p}^{r_{i}-1}C_{k}^{p}L_{ik}^{\varepsilon}(t).
$$
We claim that
$$
\lim_{\varepsilon\to0}H^{\varepsilon}_{ip}(t)=H_{ip}(t).
$$
Indeed, we put $F(t)=H_{ip}(t)$ and obtain
$$
H_{ip}(t)=\sum_{j=0}^{n}\sum_{q=0}^{r_{j}-1}
\left(\frac{1}{\varepsilon^{q}}\sum_{k=0}^{q}(-1)^{q+k}C_{q}^{k}H_{ip}(t_{jk}^{\varepsilon})\right)
H^{\varepsilon}_{jq}(t),
$$
where $0\le i\le m$ and $0\le p\le r_{i}-1$. Consider the vector $h$ with elements
$$
h_{ip}=H_{ip}(t),
$$
the vector $h^{\varepsilon}$ with elements
$$
h^{\varepsilon}_{jq}=H^{\varepsilon}_{jq}(t),
$$
and the matrix $A^{\varepsilon}$ with elements
$$
øA^{\varepsilon}_{(ip),(jq)}=
\frac{1}{\varepsilon^{q}}\sum_{k=0}^{q}(-1)^{q+k}C_{q}^{k}H_{ip}(t_{jk}^{\varepsilon}).
$$
Then
$$
h=A^{\varepsilon}h^{\varepsilon}.
$$
We see that
$$
\lim_{\varepsilon\to0}A^{\varepsilon}=E,
$$
where $E$ is the identity matrix. Therefore,
$$
\lim_{\varepsilon\to0}h^{\varepsilon}=h.
$$
Hence
$$
\lim_{\varepsilon\to0}H^{\varepsilon}_{ip}(t)=H_{ip}(t).
$$
Since
$$
ø\lim_{\varepsilon\to0}
\frac{1}{\varepsilon^{p}}\sum_{k=0}^{p}(-1)^{p+k}C_{p}^{k}f(t_{ik}^{\varepsilon})=
f^{(p)}(t_{i}),
$$
we have
$$
\lim_{\varepsilon\to0}L^{\varepsilon}(t)=H(t).
$$
The theorem is proved. $\bullet$
\bigskip

\noindent {\bf Corollary.} We have
$$
ø\Delta_{n}f(t_{0}^{r_{0}},\ldots,t_{m}^{r_{m}})=\lim_{\varepsilon\to0}
\Delta_{n}f(t_{0},t_{0}+\varepsilon,\ldots,t_{0}+(r_{0}-1)\varepsilon,\ldots,
t_{m},t_{m}+\varepsilon,\ldots,t_{m}+(r_{m}-1)\varepsilon).
$$

%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{2. Basis splines}

Let
$$
t^k_{+}=\left\{\begin{array}{lll}
 t^k&{\rm if}&t\ge0,\\
 0  &{\rm if}&t<  0.\\
\end{array}\right.
$$
A function
$$
øf(t)=\sum_{i=0}^{m}c_{i}(t-t_{i})^{k_{i}}_{+}
$$
is called a polynomial spline.

\begin{theorem}
Let $a_1<\ldots<a_m$, $k_1\ge1,\ldots,k_m\ge1$, and $n=k_1+\cdots+k_m$. The family
$$
(t+a_{1})^{n-1},\ldots,(t+a_{1})^{n-k_{1}},\;\ldots\ldots,\;
(t+a_{m})^{n-1},\ldots,(t+a_{m})^{n-k_{m}},
$$
is a basis in the space of polynomials of degree $\le n-1$.
\end{theorem}
\noindent{\bf Proof.} Let $k_{0}=0$, and let
$$
f_{k_{0}+\cdots+k_{i-1}+j}(t)=(t+a_{i})^{n-j},
$$
where $1\le i\le m$ and $1\le j\le k_{i}$. It suffices to check that
$$
\det\{f_{j}(x_{i})\}_{i,j=1}^{n}\ne 0,
$$
where $x_{1}<x_{2}<\ldots<x_{n}$. Let
$$
y_{k_{0}+\cdots+k_{i-1}+j}=a_{i}+\varepsilon(j-1),
$$
where $1\le i\le m$ and $1\le j\le k_{i}$. We put
$$
D(\varepsilon)=\det\{(x_{i}+y_{j})^{n-1}\}_{i,j=1}^{n}.
$$
It is easy to see that
$$
D(\varepsilon)=\prod_{i=0}^{n-1}C_{n-1}^{i}\prod_{0\le i<j\le n-1}(x_{i}-x_{j})
\prod_{0\le i<j\le n-1}(y_{j}-y_{i}).
$$
Let
$$
N=\frac{(k_{1}-1)k_{1}}{2}+\cdots+\frac{(k_{m}-1)k_{m}}{2}.
$$
On the one hand,
$$
øD(\varepsilon)=\varepsilon^{N}
 \prod_{i=0}^{n-1}C_{n-1}^{i}
 \prod_{i=1}^{m}\prod_{j=1}^{k_{i}-1}\frac{(k_{i}-1)!}{j!}
 \prod_{0\le i<j\le n-1}(x_{i}-x_{j})
 \prod_{1\le i<j\le m}(a_{j}-a_{i})^{k_{i}k_{j}}+o(\varepsilon^{N}).
$$
On the other hand,
$$
øD(\varepsilon)=\varepsilon^{N}\prod_{i=1}^{m}\prod_{j=1}^{k_{i}-1}\frac{(n-1)!}{(n-1-j)!}
\det\{f_{j}(x_{i})\}_{i,j=1}^{n}+o(\varepsilon^{N}).
$$
This means that
$$
ø\det\{f_{j}(x_{i})\}_{i,j=1}^{n}= C\prod_{0\le i<j\le n-1}(x_{i}-x_{j})\prod_{1\le
i<j\le m}(a_{j}-a_{i})^{k_{i}k_{j}},
$$
where
$$
C=\prod_{i=0}^{n-1}C_{n-1}^{i}
 \prod_{i=1}^{m}\prod_{j=1}^{k_{i}-1}\frac{(n-1-j)!(k_{i}-1)!}{(n-1)!j!}.
$$
Therefore,
$$
\det\{f_{j}(x_{i})\}_{i,j=1}^{n}\ne 0.
$$
This proves the theorem. $\bullet$
\bigskip




\begin{theorem}
\label{bslpa} Let $t_0<t_1<\ldots<t_m$, $r_0\ge1,r_1\ge1,\ldots,r_m\ge1$, and
$n=r_0+r_1+\cdots+r_m-1$. There exists a unique spline
$$
B(t)=\sum_{i=0}^m\sum_{p=0}^{r_{i}-1}b_{ip}(t-t_i)^{n-1-p}_+
$$
such that
$$
\int\limits_{-\infty}^\infty B(t)\,dt=1
$$
and
$$
\begin{array}{lll}
 B(t)=0&{\rm if}&t\le t_{0},\\
 B(t)>0&{\rm if}&t_{0}< t< t_{m},\\
 B(t)=0&{\rm if}&t\ge t_{m}.\\
\end{array}
$$
\end{theorem}
{\bf Proof.} We claim that
$$
B(t)=ø(-1)^{n}n\Delta_{n}(t-\tau)^{n-1}_{+}(t_{0}^{r_{0}},\ldots,t_{n}^{r_{n}}),
$$
where the divided difference is taken with respect to $\tau$. In particular, we have
$$
b_{ip}=(-1)^{n+p}nC_{n-1}^{p}\frac{1}{(r_{i}-1-p)!}
  \left.\left(\frac{1}{\omega_{i}(\tau)}\right)^{(r_{i}-1-p)}\right|_{\tau=t_{i}},
$$
where $0\le i\le m$ and $0\le p\le r_{i}-1$. Here
$$
\omega_{i}(\tau)=\prod_{\stackrel{\scriptstyle{j=0}}{j\ne i}}^{m}(\tau-t_{j})^{k_{j}},
$$
where $0\le i\le m$.


Let $r_0=\ldots=r_m=1$. In this case $m=n$ and we have to prove that
$$
B(t)=\sum_{i=0}^nb_{i}(t-t_i)^{n-1}_+,
$$
where
$$
b_{i}=n\prod_{\stackrel{\scriptstyle{j=0}}{j\ne i}}^{n}\frac{1}{t_j-t_i}.
$$
Obviously
$$
B(t)=0
$$
for $t\le t_{0}$. Differentiating $n$ times
$$
\sum_{i=0}^n\prod_{\stackrel{\scriptstyle{j=0}}{j\ne
i}}^{n}\frac{t_{j}-\tau}{t_j-t_i}(t-t_{i})^{n-1}=(t-\tau)^{n-1}
$$
with respect to $\tau$, we get
$$
B(t)=0
$$
for $t\ge t_{n}$. Differentiating $n$ times
$$
(t_{n}-\tau)^{n}=\sum_{i=0}^n\prod_{\stackrel{\scriptstyle{j=0}}{j\ne
i}}^{n}\frac{t_{j}-\tau}{t_j-t_i}(t_{n}-t_{i})^{n},
$$
with respect to $\tau$, we get
$$
\sum_{i=0}^n\prod_{\stackrel{\scriptstyle{j=0}}{j\ne
i}}^{n}\frac{1}{t_j-t_i}(t_{n}-t_i)^{n}=1.
$$
Hence
$$
\int\limits_{-\infty}^\infty
B(t)\,dt=\sum_{i=0}^n\prod_{\stackrel{\scriptstyle{j=0}}{j\ne
i}}^{n}\frac{1}{t_j-t_i}(t_{n}-t_i)^{n}=1.
$$

Let us check that
$$
b_{0}+\cdots+b_{k}\ne0
$$
for all $0\le k\le n-1$. Assume that $b_{0}+\cdots+b_{k}=0$ for some $0\le k\le n-1$.
Consider the polynomial
$$
f(\tau)=\sum_{i=0}^{k}\prod_{\stackrel{\scriptstyle{j=0}}{j\ne
i}}^{n}\frac{t_{j}-\tau}{t_j-t_i}
$$
of degree $\le n-1$. We see that
$$
f(t_{0})=\ldots=f(t_{k})=1
$$
and
$$
f(t_{k+1})=\ldots=f(t_{n})=0.
$$
Therefore $f'(\tau)$ has $n-1$ mutually different roots. Hence $f'(\tau)\equiv 0$.
Therefore $f(\tau)\equiv{\rm const}$. The contradiction proves the inequality
$b_{0}+\cdots+b_{k}\ne0$.

Let us check that
$$
B(t)\ne 0
$$
for all $t_{0}<t<t_{n}$. Assume that $B(t_*)=0$ for some $t_0<t_*<t_n$. By the definition
of $B(t)$, we see that
$$
B^{(i)}(t_0)=B^{(i)}(t_n)=0
$$
for all $0\le i\le n-2$. Hence $B^{(n-2)}(t)$ has $n+1$ mutually different roots between
$t_0$ and $t_n$. Since $B^{(n-2)}(t)$ is peace-wise linear with knots
$t_{0}<\ldots<t_{n}$, we have $B^{(n-2)}(t)\equiv 0$ on $[t_{k},t_{k+1}]$ with some $0\le
k\le n-1$. On the interval $[t_{k},t_{k+1}]$ the basis spline $B(t)$ is equal to a
polynomial of degree $\le n-3$. Hence $b_{0}+\ldots+b_{k}=0$. The contradiction proves
that  $B(t)\ne 0$ for all $t_0<t<t_{n}$.

Since
$$
\int\limits_{-\infty}^\infty B(t)\,dt=1,
$$
we have
$$
B(t)> 0
$$
for all $t_{0}<t<t_{n}$. This prove the existence of $B(t)$.

Consider a spline
$$
C(t)=\sum_{i=0}^nc_{i}(t-t_i)^{n-1}_+
$$
such that
$$
\int\limits_{-\infty}^\infty C(t)\,dt=1
$$
and
$$
C(t)=0
$$
for all $t\ge t_{n}$. The system of polynomials
$$
(t-t_{1})^{n-1},\ldots,(t-t_{n})^{n-1}
$$
is a basis in the space of polynomials of degree $\le n-1$. By identities
$$
b_{0}(t-t_{0})^{n-1}=-\sum_{i=1}^nb_{i}(t-t_i)^{n-1}
$$
and
$$
á_{0}(t-t_{0})^{n-1}=-\sum_{i=1}^ná_{i}(t-t_i)^{n-1},
$$
we conclude that
$$
c_{i}=\frac{c_{0}}{b_{0}}b_{i}
$$
for all $i=0,1,\ldots,n$. Hence
$$
C(t)=\frac{c_{0}}{b_{0}}B(t).
$$
By integration, we obtain $c_{0}=b_{0}$. Hence $c_{i}=b_{i}$ for all $i=0,1,\ldots$. This
proves the uniqueness of $B(t)$.

Let $r_0\ge1,r_1\ge1,\ldots,r_m\ge1$ and $n=r_0+r_1+\cdots+r_m-1$.

Let us assume that $n-r_{i}=1$ for some $0\le i\le m$. Then $m=1$. Therefore $r_{0}=1$
and $r_{1}=n$ or $r_{0}=n$ and $r_{1}=1$. If $r_{0}=1$ and $r_{1}=n$, then
$$
øB(t)=\frac{n}{(t_{1}-t_{0})^{n}}(t-t_{0})^{n-1}_{+}-
\sum_{p=0}^{n-1}C_{n-1}^{p}\frac{n}{(t_{1}-t_{0})^{n-p}}(t-t_{1})^{n-1-p}_{+}.
$$
The case $r_{0}=n$ and $r_{1}=1$ is considered analogously.

Let us assume that $n-r_{i}\ge 1$ for all $0\le i\le m$. We put
$$
B(t)=ø(-1)^{n}n\Delta_{n}(t-\tau)^{n-1}_{+}(t_{0}^{r_{0}},\ldots,t_{m}^{r_{m}})
$$
and
$$
øB^{\varepsilon}(t)=(-1)^{n}n\Delta_{n}(t-\tau)^{n-1}_{+}
(t_{0},t_{0}+\varepsilon,\ldots,t_{0}+(r_{0}-1)\varepsilon,\ldots,
t_{m},t_{m}+\varepsilon,\ldots,t_{m}+(r_{m}-1)\varepsilon),
$$
where
$$
t_{ip}^{\varepsilon}=t_{i}+p\varepsilon
$$
for $0\le i\le m$ and $0\le p\le r_{i}-1$. We have
$$
B(t)=\lim_{\varepsilon\to0}B^{\varepsilon}(t).
$$
By continuity the spline $B(t)$ satisfies all the demand conditions. This prove the
existence of $B(t)$.

By Theorem 3, we obtain the following statements. If $r_{0}\ge 2$, then the family
$$
(t-t_{0})^{n-1},\ldots,(t-t_{0})^{n-r_{0}+1}; (t-t_{1})^{n-1},\ldots,(t-t_{1})^{n-r_{1}};
\;\ldots;\; (t-t_{m})^{n-1},\ldots,(t-t_{m})^{n-r_{m}}
$$
is a basis in the space of polynomials of degree $\le n-1$. If $r_{0}=1$, then the family
$$
(t-t_{1})^{n-1},\ldots,(t-t_{1})^{n-r_{1}}; \;\ldots;\;
(t-t_{m})^{n-1},\ldots,(t-t_{m})^{n-r_{m}}
$$
is a basis in the space of polynomials of degree $\le n-1$. This proves the uniqueness of
$B(t)$. $\bullet$
\bigskip

The spline $B(t)$ constructed in Theorem 4 is called the basis spline. For example, if
$m=3$, $t_{0}=0, t_{1}=1, t_{2}=2, t_{3}=3$ and $r_{0}=r_{1}=r_{2}=r_{3}=2$, then $B(t)$
has coefficients
$$
\left[\begin{array}{rr}
  b_{00} & b_{01} \\
  b_{10} & b_{11} \\
  b_{20} & b_{21} \\
  b_{30} & b_{31} \\
\end{array}\right]\;=\;
\left[\begin{array}{rr}
  -77/108 & ø7/6 \\
  -7/4 & ø21/2 \\
   7/4 & ø21/2 \\
   77/108 & ø7/6
\end{array}\right].
$$
The spline
$$
B(t)=b_{00}t^{6}_{+}+b_{01}t^{5}_{+}+b_{10}(t-1)^{6}_{+}+b_{11}(t-1)^{5}_{+}+
b_{20}(t-2)^{6}_{+}+b_{21}(t-2)^{5}_{+}+b_{30}(t-3)^{6}_{+}+b_{31}(t-3)^{5}_{+}
$$
we see on the picture.
\begin{center}
\begin{picture}(400,70)
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%\input p30aug00.tex
\end{picture}
\end{center}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{3. Splines and volumes}



Let $\subset \mathbb{R}^{n}$ be a Euclidean simplex with vertexes  $A_0,A_1,\ldots,A_n$.
Consider a nonzero direction $w\in\mathbb{R}^{n}$. Let us assume that
$$
(w,A_{i})\le (w,A_{j})
$$
for all $0\le i\le j\le n$. If $t_{*}=(w,A_{i})$ for some $0\le i\le n$, then we say that
$t_{*}$ is a critical level of the direction $w$. Let
$$
t_{0}<t_{1}<\ldots<t_{m}
$$
be all critical levels of the direction $w$. By $r_{i}$ we denote the number of vertexes
on the critical level $t_{i}$. We note that $r_{0}+r_{1}+\cdots+r_{m}=n+1$.

Let
$$
\Gamma_{t}=\{x\in V:\;(w,x)=t\}
$$
and
$$
V_{t}=\{x\in V:\;(w,x)\le t\}.
$$
We see that
$$
ø{\rm vol}\Gamma_{t}=\frac{d}{dt}{\rm vol}V_{t}.
$$
Therefore,
$$
F(t)=\frac{{\rm vol}V_{t}}{{\rm vol}V}
$$
is the distribution function and
$$
f(t)=\frac{{\rm vol}\Gamma_{t}}{{\rm vol}V}
$$
is the density of $F(t)$. We claim that $F(t)$ is uniquely defined by
$t_{0},t_{1},\ldots,t_{m}$ and $r_{0},r_{1},\ldots,r_{m}$. In the case  $n=2$ the fact is
obvious. Indeed, the triangles
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\special{em:linewidth 0.5pt}

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 \put(210,30){\circle*{3}}
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 \put(67,27){\small $t_0$}
 \put(67,57){\small $t_1$}
 \put(67,87){\small $t_2$}

%\input piq1.tex
\end{picture}
\end{center}
have equal distribution functions with the density of the form.
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 \put(150,15){\circle*{3}}
 \put(180,45){\circle*{3}}
 \put(210,15){\circle*{3}}

 \put(147,2){\small $t_0$}
 \put(177,2){\small $t_1$}
 \put(207,2){\small $t_2$}

%\input piq2.tex
\end{picture}
\end{center}

\begin{theorem} Let the basis spline $B(t)$ be defined for
$t_{0},t_{1},\ldots,t_{m}$ and $r_{0},r_{1},\ldots,r_{m}$. Then $f(t)=B(t)$.
\end{theorem}
{\bf Proof.}  We first consider the special case $r_0=\ldots=r_m=1$. Note that $m=n$.
Consider cones
$$
K^{i}=A_{i}+{\rm co}\{A_{i}-A_{0},\,A_{i}-A_{1},\,\ldots,\,A_{i}-A_{i-1},\,
A_{i+1}-A_{i},\,A_{i+2}-A_{i},\,\ldots,A_{n}-A_{i}\},
$$
where $i=0,1,\ldots,n$. In the case $n=2$ cones
$$
ø\begin{array}{lll}
 K_{0}&=&A_{0}+{\rm co}\{A_{1}-A_{0},\,A_{2}-A_{0}\}\\
 K_{1}&=&A_{1}+{\rm co}\{A_{1}-A_{0},\,A_{2}-A_{1}\}\\
 K_{2}&=&A_{2}+{\rm co}\{A_{2}-A_{0},\,A_{2}-A_{1}\}\\
\end{array}
$$
are shown on the picture.
\begin{center}
\begin{picture}(400,160)
\special{em:linewidth 0.5pt}

 \put(200,15){\special{em:moveto}}
 \put(280,135){\special{em:lineto}}

 \put(200,15){\special{em:moveto}}
 \put(160,150){\special{em:lineto}}

 \put(220,45){\special{em:moveto}}
 \put(100,145){\special{em:lineto}}


 \put(200,15){\circle*{3}}
 \put(220,45){\circle*{3}}
 \put(181,77){\circle*{3}}

 \put(195,2){\small $A_0$}
 \put(224,40){\small $A_1$}
 \put(167,70){\small $A_2$}
 \put(195,40){\small $K_{0}$}
 \put(208,70){\small $K_{1}$}
 \put(143,120){\small $K_{2}$}

%\input piq3.tex
\end{picture}
\end{center}


Let $x\in K_{i}$. Then for uniquely defined
$$
\lambda_{0}\ge0,\ldots,\lambda_{i-1}\ge0,\lambda_{i+1}\ge0,\ldots,\lambda_{n}\ge0
$$
we have
$$
x=A_{i}+\lambda_{0}(A_{i}-A_{0})+\cdots+\lambda_{i-1}(A_{i}-A_{i-