Interpolation

Linear Interpolation

A linear function may be written as $f(x) = a_0 + a_1 x.$ Given any two data points, $( x_1, y_1 )$, and $(x_2,y_2)$, we can determine a linear function that exactly passes through these points. We can do this by solving for $a_0$ and $a_1$. Since there are two unknowns we require two equations. They are:

$\begin{align} y_1 &= a_0 + a_1 x_1 \\ y_2 &= a_0 + a_1 x_2 \end{align}$

These may be written in matrix form as

$\left[ \begin{array}{cc} 1 & x_1 \\ 1 & x_2 \end{array} \right] \left( \begin{array}{c} a_0 \\ a_1 \end{array} \right) = \left( \begin{array}{c} y_1 \\ y_2 \end{array} \right)$

Given values for $( x_1, y_1 )$, and $(x_2,y_2)$, these equations may be easily solved in MATLAB. But since they are so simple, we can easily by hand to obtain a general solution,

$\begin{align} a_0 &= \frac{x_1 y_2 -x_2 y_1}{x_1-x_2} \\ a_1 &= \frac{y_1-y_2}{x_1-x_2} \end{align}$

This gives the coefficients, which may then be substituted into the original equation $f(x) = a_0 + a_1 x$ and simplified to obtain

$f(x) = \left( \frac{y_2-y_1}{x_2-x_1} \right) \left( x-x_1 \right) + y_1.$

This is a very convenient equation for performing linear interpolation.

Example

The density of air at various temperatures and atmospheric pressure is given in the following table

Temperature (°F)	1.39	1.16	0.99	0.87	0.78	0.69	0.63	0.58	0.54	0.50	0.46	0.44	0.41
Density (kg/m3)	-10	80	170	260	350	440	530	620	710	800	890	980	1070

Estimate the density of air at 32 °F. Using linear interpolation, we have $\rho(T)=a_0 + a_1 T$. We define $(T_1,\rho_1)=(-9.7,1.39)$ and $(T_2,\rho_2)=(80,1.161)$. Now we can calculate

$\begin{align} \rho(32) &\approx \frac{\rho_2 -\rho_1}{T_2-T_1}\left( T - T_1 \right) + \rho_1 \\ &= \frac{1.16-1.39}{-9.7-80} \left( 32 + 9.7 \right) + 1.39 \\ &= 1.28. \end{align}$

Implementation in Matlab

There are three ways to do interpolation in MATLAB.

1. Solve the linear system.

    T1=-9.7;  T2=80;
    rho1=1.39; rho2=1.16;
    T = 32;
    A = [ 1 T1; 1 T2 ];
    b = [ rho1; rho2 ];
    a = A\b;
    rho = a(1) + a(2)*T;
   

2. Use the general equation we derived above:

    T1=-9.7; T2-80;
    rho1=1.39; rho2=1.16;
    T = 32;
    rho = (rho2-rho1)/(T2-T1) * (T-T1) + rho1;
   

3. Use MATLAB's built-in interpolation tool

    Ti = [-9.7 80];
    rhoi = [1.39 1.16];
    rho = interp1( Ti, rhoi, 32 );
   

All three of these methods provide exactly the same answer. However, using MATLAB's interp1 function allows you to use a vector for the points that you want to interpolate to. For example, if we wanted the density at 30, 50, 70, 10, and 110 °F, we could accomplish this by:

  Ti = [-9.7 80 170 ];
  rhoi = [1.39 1.16 0.995];
  T = 30:20:110;
  rho = interp1( Ti, rhoi, T );

Implementation in Excel

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Polynomial Interpolation

Polynomial interpolation is a simple extension of linear interpolation. In general, an n^th degree polynomial is given as $f(x)=\sum_{i=0}^n a_i x^i.$

If n=1 then we recover a first-degree polynomial, which is linear. The formula gives $f(x)=\sum_{i=0}^1 a_i x^i = a_0 + a_1 x$.

In general, if we want to interpolate a set of data using an n^th degree polynomial, then we must determine n+1 coefficients, $a_0 \ldots a_n$. Therefore, we require n+1 points to interpolate using an n^th degree polynomial.

The equations that must be solved are given as

$\left[\begin{array}{ccccc} 1 & x_{1} & x_{1}^2 & \cdots & x_{1}^{n}\\ 1 & x_{2} & x_{2}^2 & \cdots & x_{2}^{n}\\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 1 & x_{n+1} & x_{n+1}^2 & \cdots & x_{n+1}^{n}\end{array}\right]\left(\begin{array}{c} a_{0}\\ a_{1}\\ \vdots\\ a_{n}\end{array}\right) = \left(\begin{array}{c} y_{1}\\ y_{2}\\ \vdots\\ y_{n+1}\end{array}\right)$

Solving these equations provides the values for the coefficients, $a_0 \ldots a_n$. Once we know these coefficients, we can evaluate the polynomial at any point.

Example

Using a third degree polynomial and the data in the table above, approximate the density of air at 50 °F and 300 °F.

We require 4 points to fit a third degree polynomial, $a_0, \, a_1, \, a_2, \, a_3$. The equations to be solved are:

$\begin{align} \rho_1 &= a_0 + a_1 T_1 + a_2 T_1^2 + a_3 T_1^3 \\ \rho_2 &= a_0 + a_1 T_2 + a_2 T_2^2 + a_3 T_2^3 \\ \rho_3 &= a_0 + a_1 T_3 + a_2 T_3^2 + a_3 T_3^3 \\ \rho_4 &= a_0 + a_1 T_4 + a_2 T_4^2 + a_3 T_4^3 \end{align}$

Once we have solved these equations we have the polynomial coefficients and we can then evaluate the resulting polynomial at any temperature that we wish.

We still must choose what 4 points in the table to use in the above equations. We will choose the 4 points in the table that are closest to the temperature that we wish to interpolate to. Note that if we want to evaluate the density at different temperatures, we may want to use different points in the table to determine the coefficients!

Our procedure to perform the interpolation is:

1. Determine which points we need to use.
2. Set up and solve the linear system to obtain the polynomial coefficients $a_0 \ldots a_3$.
3. Evaluate the polynomial to obtain the temperature.

Implementation in Matlab

The MATLAB code to obtain the density at 50 °F is

% Set the data points from the table.
Ti = [ -9.7   80  170  260  350  440  530 ...
        620  710  800  890  980 1070 ]';
rhoi=[ 1.39 1.16 0.99 0.87 0.78 0.69 0.63 ...
       0.58 0.54 0.50 0.46 0.44 0.41 ]';

% Set the temperature we want to evaluate the density at.
T = 50;

% determine the set of points that we will use to construct
% the interpolant. We want to use points that are closest
% to the temperature that we are interested in.
points = 1:4;

% set up the linear system
A = [ ones(4,1), ones(4,1).*Ti(points), ones(4,1)*Ti(points).^2, ones(4,1)*Ti(points).^3 ];
b = rhoi(points);

% solve for the polynomial coefficients
a = A\b;

% evaluate the density.
rho = a(0) + a(1)*T + a(2)*T^2 + a(3)*T^3;

For T=300 we would probably want to choose a different set of points to determine the polynomial coefficients. Therefore, we would repeat the above with the only change being:

T=300;
points=4:8;

Cubic Spline Interpolation

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Lagrange Polynomial Interpolation

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Given $n_p$ points $(x_k,y_k),$ the $n^\mathrm{th}$ order Lagrange polynomial that interpolates these function values, $f(x)$ are expressed as

$f(x)=\sum_{k=0}^{n} y_{k} L_{k}(x),$

where $L_{k}(x)$ is the Lagrange polynomial given by

$L_{k}(x)=\prod_{\stackrel{i=0}{i\ne k}}^{n}\frac{x-x_{i}}{x_{k}-x_{i}},$

and $n_p = n+1$. In other words, for an $n^\mathrm{th}$ order interpolation, we require $n_p=n+1$ points.

The $\prod$ operator represents the continued product, and is analogous to the $\sum$ operator for summations. For example, $\prod_{i=1}^{3} (x+i) = (x+1)(x+2)(x+3)$.