Introduction
Theory
HOWTO
Error Analysis
Examples
Questions
Applications in Engineering
Matlab
Maple

# Introduction

To this point, we have found interpolating polynomials which
pass through *n* points. Here we will see how we can not
only match points, but also derivatives and higher derivatives
at points.

This technique also generalizes for interpolating non-polynomial
functions or points in more than two dimensions.

# Background

Useful background for this topic includes:

# References

# Theory

# HOWTO

# Problem

Given *n* constraints which match either values (0th derivative), 1^{st} derivatives, 2^{nd} derivatives, *etc*, up to the (*n* − 1)^{st}
derivative such that for any 1 ≤ *k* ≤ *n*, there are *k* constraints which are derivatives of order *k* − 1 or less.

# Assumptions

There are no conflicting or equivalent constraints.

# Process

For each of *n* constraints on the *x*-value *x*_{i} which has the *k*^{th} derivative, define the entry in the Vandermonde matrix **V** = (*v*_{i,j}) as:

where **y** is the set of contrsaints. The solution to this system of linear equations
are the coefficients of the interpolating polynomial.

For example, given five constraints, the rows for constraints at *x*_{i} of 0th, 1st, 2nd, 3rd, and 4th derivatives would be

*x*_{i}^{4} *x*_{i}^{3} *x*_{i}^{2} *x*_{i} 1
- 4
*x*_{i}^{3} 3*x*_{i}^{2} 2*x*_{i} 1 0
- 12
*x*_{i}^{2} 6*x*_{i} 2 0 0
- 24
*x*_{i} 6 0 0 0
- 24 0 0 0 0

respectively.

# Error Analysis

To be completed.

# Examples

1. Find the polynomial which has the value 1, slope 0.5, and concavity 0.25 at the point *x* = 2.

The system of linear equations is

which has a solution **c** = (0.125, 0, 0.5)^{T}. A plot is shown in Figure E1.

Figure E1. The polynomial which has value, slope, and concavity of 1, 0.5, and 0.125 at *x* = 2.

2. Find the polynomial which has value 1 at *x* = 1, slope 2 at *x* = 2, and concavity 3 at *x* = 3.

The system of linear equations is

Notice that the point at which the concavity is fixed is irrelevant.

This has the solution **c** = (1.5, -4, 3.5)^{T}. A plot is shown in Figure E2.

Figure E2. The polynomial which has value, slope, and concavity of 1, 0.5, and 0.125 at *x* = 1, 2, and 3, respectively.

3. Find the polynomial which has value and slope 1 at *x* = 0 and value and slope *e* at *x* = 1. Plot
the solution and plot the error |p_{3}(*x*) − *e*^{x}|.

The system of linear equations is

which has the solution **c** = (3 − *e*, 2*e* - 5, 1, 1)^{T}. The plot of the polynomial p_{3}(*x*) = (3 − *e*)*x*^{3} + (2*e* − 5)*x*^{2} + *x* + 1 and
the error as compared to the exponential function are shown in Figure E3a and E3b, respectively.

Figure E3a. The polynomial with values and slopes 1 at *x* = 0 and *e* at *x* = 1.

Figure E3b. The difference of the approximation with the exponential function.

4. Find the polynomial which matches the value, slope, and concavity of the sine function at *x* = 1 and *x* = 2.

The system of linear equations is

which has the solution **c** = (0.0039113, 0.012864, -0.17864, -0.0011108, 1.0074, -0.0029858)^{T}. The plot and error is shown in Figures E4a and E4b, respectively.

Figure E4a. The polynomial with values, slopes, and concavity matching the sine function at *x* = 1 and *x* = 2.

Figure E4b. The difference of the approximation with the sine function.

# Questions

# Applications to Engineering

This is used for splines but can also be used in interpolating points
found as solutions to initial-value problems.

# Matlab

# Maple