资源分类:Matlab 工具:MATLAB 7.11 (R2010b)



Piecewise Hermite Cubic Interpolation

Piecewise Hermite cubic interpolation between 2 points knowing derivative values


Syntax: y=p3hermite(x,pointx,pointy,yprime,plt)


pointx = data points of the independent variable

           (The points do not have to be equally spaced)

pointy = data points of the dependent variable. pointy is the value of

           the function at pointx

yprime = data points of the dependent variable's derivative. yprime is the

           derivative of the function at pointx

x = an arbitrary vector that will be interpolated

plt = If plt is a number greater than 0 it will plot the interpolation

      employing the number in plt as handle for the figure


-This function returns the piecewise interpolation "y" of a vector "x".

The algorithm employs two adjacent points (from pointx) and interpolates

with a Hermite cubic polynomial using the function values and the corresponding derivatives.

-pointx, pointy, and yprime must be vectors with the same number of elements.

"x" and "y" have the same number of elements.


Written by Juan Camilo Medina 2011



Suppose you have the values of a function "y(x)" at the points xi={0,4,9},

those are yi={2,-2,sqrt(2)} respectively. You also know the values of the

derivative of y(x) at the same points (pointx) yi'=[0,0,-pi/(2*sqrt(2))] respectively.

You want to interpolate within those values with an arbitrary vector "x"

using piecewise cubic Hermite polynomials




pointy=[2,-2,sqrt(2)]; %function values at pointx

yprime=[0,0,-pi/(2*sqrt(2))]; %derivative of the function at pointx

x=0:0.01:pointx(end); % arbitrary vector to be interpolated


y_ex=2*cos(pi/4*x); % exact value (y corresponds to y=2*cos(pi/4*x))

plot(x,y_ex,'--k'); axis tight; % plots exact solution for comparison

legend('Interpolation Points','Hermite Interpolation','Exact Value','Location','Southeast')


Written by Juan Camilo Medina - The University of Notre Dame

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