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\title {More on linear interpolation}
\author{Gerald Farin\\
Computer Science and Engineering\\
Arizona State University\\
Tempe, AZ 85287-5406\\
email {\tt farin@asu.edu}\\ }

\maketitle

\section{Vector arguments}

Since linear interpolation is an affine map, we have the following:
\begin{equation}
{\bf l}((1-s)c + s d) = (1-s){\bf l}(c) + s {\bf l}(d) .
\end{equation}
This simply states that a relationship of points in the domain is left
unchanged for the corresponding points i the range.

We may take this one step further:
if a vector is defined in the domain, then that definition also carries 
over to the range. If $c$ and $d$ are two points on the real line, the
domain, they define a vector $\alpha$ by setting $\alpha = d - c$. The
corresponding vector in the range is then defined as
\begin{equation}
	{\bf l}(\alpha) = {\bf l}(d) - {\bf l}(c).
\end{equation}

\section{Vectors and blossoms}

Menelaos' theorem is an assertion about the commmutativity of supplying two
different (point) arguments to piecewise linear interpolation of a polygon
consisting of three points. Instead of using two points in the real line
as arguments, we may use a point and a vector. If $t$ the point and
$\alpha$ is the vector in the real line, our generalized Menelaos' theorem
states
\beq
	{\bf l}(t,\alpha) = {\bf l}(\alpha, t).
\end{equation}
It is easy to check that both expressions equal $(t-1)\pp_1 + (1-2t)\pp_2
+ t\pp_3.$

\section{Derivatives and blossoms: affine}

First, an identity, closely resembling Leibniz' formula for derivatives.
Let $t$ be a point on the real line and let $\alpha$ be a vector in the
associated 1D linear space. Then
\begin{equation}
	\bb[(t+\alpha)^{<n>}] = \sum_{i=0}^n \binom{n}{i}
	\bb[t^{<n-i>},\alpha^{<i>}].
	\label{eq:sumbloss}
\end{equation}
This is easily proved (check for $n=2$!).

The derivative of a curve $\xx(t)$
with respect to a 1D vector $\alpha$ is defined
as
$$
	D_\alpha \xx(t) = \lim_{h\rightarrow 0} \frac{1}{h}[\xx(t+h\alpha)
	-\xx(t)].
$$
Since we are dealing with curves which are defined via blossoms, we get
$$
	D_\alpha \bb[t^{<n>}]
	 = \lim_{h\rightarrow 0} 
	\frac{1}{h}\big[\bb[(t+h\alpha)^{<n>}]-\bb[t^{<n>}]\big].
$$
Invoking (\ref{eq:sumbloss}), we have
\begin{equation}
	D_\alpha \bb[t^{<n>}]
	 = \lim_{h\rightarrow 0} \big[\sum_{i=0}^n  \binom{n}{i}
	\bb[t^{<n-i>},(h\alpha)^{<i>}] - \bb[t^{<n>}]\big].
\end{equation}
Thus
\begin{equation}
	D_\alpha \bb[t^{<n>}]
	 = \lim_{h\rightarrow 0} \big[\sum_{i=0}^n \binom{n}{i}
	h^i\bb[t^{<n-i>},\alpha^{<i>}] - \bb[t^{<n>}]\big].
\end{equation}
Upon taking the limit and observing all terms  containing
$h$ powers of order two and higher  will vanish, we have
\begin{equation}
	D_\alpha \bb[t^{<n>}]
	 = 
	n\bb[t^{<n-1>},\alpha].
\end{equation}

In most cases, one would choose $\alpha$ to have length one; then the
blossom steps with respect to $\alpha$ are simply difference steps.
and the $D_\alpha$ will become the familiar $\frac{\rm d}{{\rm d}t}$.

\section{Derivatives and blossoms: projective}


The above reasoning has to change if we study derivatives in projective
space. Since there are no vectors, we replace the vecotr $\alpha$ by
a point $s$. Following the same reasoning as above, we obtain
\begin{equation}
	D_s {\underline \bb}[t^{<n>}]
	 = 
	n{\underline \bb}[t^{<n-1>},s].
\end{equation}
Note that the derivative changes as a function of $s$!














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