% \def\updated{07mar04}%14nov97 \magnification=1200\input carlformat\def\cite#1{[#1]}\parskip10pt % % %piecewise cubic interpolation % %
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Piewise cubic interpolation %

Except, perhaps, for broken line interpolation, piecewise cubic interpolation is the most ancient spline approximation method. This method constructs the interpolant to given data $(x_i,y_i)$ (with $x$ strictly increasing) in the following simple way: On the interval $[x_i .. x_{i+1}]$, the interpolant is the unique cubic polynomial $p_i$ which matches the prescribed values $y_j$ and slopes $s_j$ at $x_j$, $j=i,i+1$. Different variants differ only in just how the slopes $s_i$ are chosen. %

It is clear that the resulting interpolant is in $C^{(1)}$. In (complete) cubic spline interpolation, $s_i = f(x_i)$ for $i=1,n$, while all other $s_i$ are chosen to make the interpolant be even in $C^{(2)}$. All other variants of piecewise cubic interpolation choose the $s_i$ by some local considerations. %

In piecewise cubic Hermite interpolation, $s_i = f(x_i)$, all $i$, with $f$ the function being interpolated. %

In Bessel interpolation, $s_i$ is chosen as the slope at $x_i$ of the parabola which matches the data at $x_{i-1}, x_i, x_{i+1}$, except for $i=1,n$ where the slope of the parabola through the nearest three data points is used. %

In Akima interpolation (see Akima70, Akima72), $s_i$ is chosen by certain geometric considerations as the following average: $$%
s_i = (w_{i+1} dy_{i-1} + w_{i-1} dy_i) / (w_{i+1} + w_{i-1}), $$%
with $dy_j := \dvd{x_j,x_{j+1}}y := (y_{j+1}-y_j)/(x_{j+1}-x_j)$ the divided difference, and $$%
w_i := |dy_i - dy_{i-1}|. $$%
The additional data $(x_i,y_i)$ for $i=-1,0$ and $i=n+1,n+2$ are obtained by evaluating the parabolic interpolant to the nearest three points at the additional data sites, with $x_i := x_{i+2} + (x_1 - x_3)$ for $i=-1,0$ and a corresponding formula for the additional two sites at the other end. Note that only the sequence $dy$ needs extension and that the choice of the additional data sites is handpicked to give, e.g., $dy_0 = dy_1 + (dy_1-dy_2)$, $dy_{-1} = dy_1 + 2(dy_1-dy_2)$ (consider the construction of the additional $dy_i$ as the job of extending the divided difference table at the points $x_1,x_2,x_3$). %

14nov97 \bigskip\rightline{last updated \updated} % % \bye