Use \ccSum.

This commit is contained in:
Sylvain Pion 2009-02-02 17:07:31 +00:00
parent 40200e96cf
commit 2c5ebdb4f4
3 changed files with 24 additions and 24 deletions

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@ -7,11 +7,11 @@ re-produces linear functions exactly. The interpolation of
$\Phi(\mathbf{x})$ is given as the linear combination of the neighbors' function $\Phi(\mathbf{x})$ is given as the linear combination of the neighbors' function
values weighted by the coordinates: values weighted by the coordinates:
\begin{displaymath} \begin{displaymath}
Z^0(\mathbf{x}) = \sum_i \lambda_i(\mathbf{x}) z_i. Z^0(\mathbf{x}) = \ccSum{i}{}{ \lambda_i(\mathbf{x}) z_i}.
\end{displaymath} \end{displaymath}
Indeed, if $z_i=a + \mathbf{b}^t \mathbf{p_i}$ for all natural Indeed, if $z_i=a + \mathbf{b}^t \mathbf{p_i}$ for all natural
neighbors of $\mathbf{x}$, we have neighbors of $\mathbf{x}$, we have
\[ Z^0(\mathbf{x}) = \sum_i \lambda_i(\mathbf{x}) (a + \mathbf{b}^t\mathbf{p_i}) = a+\mathbf{b}^t \mathbf{x}\] \[ Z^0(\mathbf{x}) = \ccSum{i}{}{ \lambda_i(\mathbf{x}) (a + \mathbf{b}^t\mathbf{p_i})} = a+\mathbf{b}^t \mathbf{x}\]
by the barycentric coordinate property. The first example in by the barycentric coordinate property. The first example in
Subsection~\ref{subsec:interpol_examples} shows how the function is Subsection~\ref{subsec:interpol_examples} shows how the function is
called. called.
@ -31,20 +31,20 @@ Sibson's $Z^1$ interpolant is a combination of the linear interpolant
$Z^0$ and an interpolant $\xi$ which is the weighted sum of the first $Z^0$ and an interpolant $\xi$ which is the weighted sum of the first
degree functions degree functions
$$\xi_i(\mathbf{x}) = z_i $$\xi_i(\mathbf{x}) = z_i
+\mathbf{g_i}^t(\mathbf{x}-\mathbf{p_i}),\qquad \xi(\mathbf{x})= \frac{\sum_i \frac{\lambda_i(\mathbf{x})} +\mathbf{g_i}^t(\mathbf{x}-\mathbf{p_i}),\qquad \xi(\mathbf{x})= \frac{\ccSum{i}{}{ \frac{\lambda_i(\mathbf{x})}
{\|\mathbf{x}-\mathbf{p_i}\|}\xi_i(\mathbf{x}) }{\sum_i {\|\mathbf{x}-\mathbf{p_i}\|}\xi_i(\mathbf{x}) } }{\ccSum{i}{}{
\frac{\lambda_i(\mathbf{x})}{\|\mathbf{x}-\mathbf{p_i}\|}}.$$ \frac{\lambda_i(\mathbf{x})}{\|\mathbf{x}-\mathbf{p_i}\|}}}.$$
Sibson observed that the combination of $Z^0$ and $\xi$ reconstructs exactly Sibson observed that the combination of $Z^0$ and $\xi$ reconstructs exactly
a spherical quadric if they are mixed as follows: a spherical quadric if they are mixed as follows:
$$ $$
Z^1(\mathbf{x}) = \frac{\alpha(\mathbf{x}) Z^0(\mathbf{x}) + Z^1(\mathbf{x}) = \frac{\alpha(\mathbf{x}) Z^0(\mathbf{x}) +
\beta(\mathbf{x}) \xi(\mathbf{x})}{\alpha(\mathbf{x}) + \beta(\mathbf{x}) \xi(\mathbf{x})}{\alpha(\mathbf{x}) +
\beta(\mathbf{x})} \textrm{ where } \alpha(\mathbf{x}) = \beta(\mathbf{x})} \mbox{ where } \alpha(\mathbf{x}) =
\frac{\sum_i \lambda_i(\mathbf{x}) \frac{\|\mathbf{x} - \frac{\ccSum{i}{}{ \lambda_i(\mathbf{x}) \frac{\|\mathbf{x} -
\mathbf{p_i}\|^2}{f(\|\mathbf{x} - \mathbf{p_i}\|)}}{\sum_i \mathbf{p_i}\|^2}{f(\|\mathbf{x} - \mathbf{p_i}\|)}}}{\ccSum{i}{}{
\frac{\lambda_i(\mathbf{x})} {f(\|\mathbf{x} - \mathbf{p_i}\|)}} \frac{\lambda_i(\mathbf{x})} {f(\|\mathbf{x} - \mathbf{p_i}\|)}}}
\textrm{ and } \beta(\mathbf{x})= \sum_i \lambda_i(\mathbf{x}) \mbox{ and } \beta(\mathbf{x})= \ccSum{i}{}{ \lambda_i(\mathbf{x})
\|\mathbf{x} - \mathbf{p_i}\|^2,$$ \|\mathbf{x} - \mathbf{p_i}\|^2},$$
where in Sibson's original work, where in Sibson's original work,
$f(\|\mathbf{x} - \mathbf{p_i}\|) = \|\mathbf{x} - \mathbf{p_i}\|$. $f(\|\mathbf{x} - \mathbf{p_i}\|) = \|\mathbf{x} - \mathbf{p_i}\|$.
@ -77,8 +77,8 @@ Knowing the gradient $\mathbf{g_i}$ for all $\mathbf{p_i} \in
exactly quadratic functions. This interpolant is not $C^1$ continuous exactly quadratic functions. This interpolant is not $C^1$ continuous
in general. It is defined as follows: in general. It is defined as follows:
\begin{displaymath} \begin{displaymath}
I^1(\mathbf{x}) = \sum_i \lambda_i(\mathbf{x}) I^1(\mathbf{x}) = \ccSum{i}{}{ \lambda_i(\mathbf{x})
(z_i + \frac{1}{2} \mathbf{g_i}^t (\mathbf{x} - \mathbf{p_i})) (z_i + \frac{1}{2} \mathbf{g_i}^t (\mathbf{x} - \mathbf{p_i}))}
\end{displaymath} \end{displaymath}
@ -89,9 +89,9 @@ $f$ from the function values on the data sites. For the data point
$\mathbf{p_i}$, we determine $\mathbf{p_i}$, we determine
$$\mathbf{g_i} $$\mathbf{g_i}
= \min_{\mathbf{g}} = \min_{\mathbf{g}}
\sum_j \ccSum{j}{}{
\frac{\lambda_j(\mathbf{p_i})}{\|\mathbf{p_i} - \mathbf{p_j}\|^2} \frac{\lambda_j(\mathbf{p_i})}{\|\mathbf{p_i} - \mathbf{p_j}\|^2}
\left( z_j - (z_i + \mathbf{g}^t (\mathbf{p_j} -\mathbf{p_i})) \right), \left( z_j - (z_i + \mathbf{g}^t (\mathbf{p_j} -\mathbf{p_i})) \right)},
$$ $$
where $\lambda_j(\mathbf{p_i})$ is the natural neighbor coordinate where $\lambda_j(\mathbf{p_i})$ is the natural neighbor coordinate
of $\mathbf{p_i}$ with respect to $\mathbf{p_i}$ associated to of $\mathbf{p_i}$ with respect to $\mathbf{p_i}$ associated to

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@ -514,7 +514,7 @@ the case if and only if there are real coefficients
$\lambda_1,\ldots,\lambda_n$ such that $p$ is a convex combination of $\lambda_1,\ldots,\lambda_n$ such that $p$ is a convex combination of
$p_1,\ldots,p_n$: $p_1,\ldots,p_n$:
\[ \[
p = \sum_{j=1}^{n}~\lambda_j~p_j, \quad \sum_{j=1}^{n}~\lambda_j = 1, p = \ccSum{j=1}{n}{~\lambda_j~p_j}, \quad \ccSum{j=1}{n}{~\lambda_j} = 1,
\quad \lambda_j \geq 0 \mbox{~for all $j$.} \quad \lambda_j \geq 0 \mbox{~for all $j$.}
\] \]
The problem of testing the existence of such $\lambda_j$ can The problem of testing the existence of such $\lambda_j$ can
@ -526,11 +526,11 @@ $h_1,\ldots,h_n,h$, we have
\[q_j = h_j \cdot (p_j \mid 1) \mbox{~for all $j$, and~} q = h \cdot \[q_j = h_j \cdot (p_j \mid 1) \mbox{~for all $j$, and~} q = h \cdot
(p\mid 1).\] Now, nonnegative $\lambda_1,\ldots,\lambda_n$ are (p\mid 1).\] Now, nonnegative $\lambda_1,\ldots,\lambda_n$ are
suitable coefficients for a convex combination if and only if suitable coefficients for a convex combination if and only if
\[\sum_{j=1}^n~ \lambda_j(p_j \mid 1) = (p\mid 1), \] \[\ccSum{j=1}{n}{~ \lambda_j(p_j \mid 1)} = (p\mid 1), \]
equivalently, if there are $\mu_1,\ldots,\mu_n$ equivalently, if there are $\mu_1,\ldots,\mu_n$
(with $\mu_j = \lambda_j \cdot h/{h_j}$ for all $j$) such that (with $\mu_j = \lambda_j \cdot h/{h_j}$ for all $j$) such that
\[ \[
\sum_{j=1}^n~\mu_j~q_j = q, \quad \mu_j \geq 0\mbox{~for all $j$}. \ccSum{j=1}{n}{~\mu_j~q_j} = q, \quad \mu_j \geq 0\mbox{~for all $j$}.
\] \]
The linear program now tests for the existence of nonnegative $\mu_j$ The linear program now tests for the existence of nonnegative $\mu_j$

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@ -349,8 +349,8 @@ then $\lambda_i\geq 0$ ($\lambda_i\leq 0$, respectively).
&&\leq 0 & \mbox{if $l_j=-\infty$.} &&\leq 0 & \mbox{if $l_j=-\infty$.}
\end{array} \end{array}
\] \]
\item \[\qplambda^T\qpb \quad<\quad \sum_{j: \qplambda^TA_j <0} \qplambda^TA_j u_j \item \[\qplambda^T\qpb \quad<\quad \ccSum{j: \qplambda^TA_j <0}{}{ \qplambda^TA_j u_j }
\quad+\quad \sum_{j: \qplambda^TA_j >0} \qplambda^TA_j l_j.\] \quad+\quad \ccSum{j: \qplambda^TA_j >0}{}{ \qplambda^TA_j l_j}.\]
\end{enumerate} \end{enumerate}
{\bf Proof:} Let us assume for the purpose of obtaining a contradiction {\bf Proof:} Let us assume for the purpose of obtaining a contradiction
@ -358,10 +358,10 @@ that there is a feasible solution $\qpx$. Then we get
\[ \[
\begin{array}{lcll} \begin{array}{lcll}
0 &\geq& \qplambda^T(A\qpx -\qpb) & \mbox{(by $A\qpx\qprel \qpb$ and 1.)} \\ 0 &\geq& \qplambda^T(A\qpx -\qpb) & \mbox{(by $A\qpx\qprel \qpb$ and 1.)} \\
&=& \sum_{j: \qplambda^TA_j <0} \qplambda^TA_j x_j &=& \ccSum{j: \qplambda^TA_j <0}{}{ \qplambda^TA_j x_j }
\quad+\quad \sum_{j: \qplambda^TA_j >0} \qplambda^TA_j x_j - \qplambda^T \qpb \\ \quad+\quad \ccSum{j: \qplambda^TA_j >0}{}{ \qplambda^TA_j x_j} - \qplambda^T \qpb \\
&\geq& \sum_{j: \qplambda^TA_j <0} \qplambda^TA_j u_j &\geq& \ccSum{j: \qplambda^TA_j <0}{}{ \qplambda^TA_j u_j }
\quad+\quad \sum_{j: \qplambda^TA_j >0} \qplambda^TA_j l_j - \qplambda^T \qpb & \quad+\quad \ccSum{j: \qplambda^TA_j >0}{}{ \qplambda^TA_j l_j} - \qplambda^T \qpb &
\mbox{(by $\qpl\leq \qpx \leq \qpu$ and 2.)} \\ \mbox{(by $\qpl\leq \qpx \leq \qpu$ and 2.)} \\
&>& 0 & \mbox{(by 3.)}, &>& 0 & \mbox{(by 3.)},
\end{array} \end{array}