## Wednesday, February 17, 2016

### HP Prime: Hermite Interpolation – 2 Points Known

HP Prime:  Hermite Interpolation – 2 Points Known

 Hermite Interpolation:  Many calculations of divided differences

Introduction

The program HERMITE2 interpolates two data points to determine the value of an unknown function at a third point.  What is unique about Hermite Interpolation is that that not only the two data points are known, but the slopes (first derivatives) are also given.   Hermite Interpolation will require a lot of calculations, especially when more than two points are known.

The 2-Known Points Case

We have points (x0, y0, y’0) and (x1, y1, y’1) and we want to determine y for a given value of x.  The approximation of y is determined by divided differences.

For two points known:

 z0:  x0, y0 z01 = y’0 z1: x0, y0 z02 = (z12 – z01)/(x1 – x0) z12 = (y1 – y0)/(x1 – x0) z13 = (z13 – z02)/(x1 – x0) z2: x1, y1 z13 = (z23 – z12)/(x1 – x0) z23 = y’1 z3: x1, y1

And y = H3(x) = y0 + y’0 * (x – x0) + z02 * (x – x0)^2 + z13 * (x – x0)^2 * (x – x1)

The arguments of HERMITE2 are as follows:  x0, y0, y’0 (labeled dx0), x1, y1, y’1 (labeled dx1), x.

HP Prime:  Program HERMITE2

EXPORT HERMITE2(x0,y0,dy0,x1,y1,dy1,x)
BEGIN
// Hermite Interpolation
// 2 points
// 2016-02-16
LOCAL z12,dx,z02,z13,z03,y;
dx:=x1-x0;
z12:=(y1-y0)/dx;
z02:=(z12-dy0)/dx;
z13:=(dy1-z12)/dx;
z03:=(z13-z02)/dx;
y:=y0+dy0*(x-x0)+
z02*(x-x0)^2+
z03*(x-x0)^2*(x-x1);
RETURN y;
END;

Example:

Given: x0 = 0, y0 = 0, y’0 = 1; x1 = 1, y1 = 1, y’1 = 0.78
For x = 0.5, result is y = 0.5275
For x = 0.78, result is y = 0.80944656

Source:

Faires, J. Douglas and Burden, Richard.  Numerical Methods – 3rd Edition  Thompson Brooks/Cole: Pacific Grove, CA  2003

Until next time everyone, happy computing!     Eddie

This blog is property of Edward Shore.  2016

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