MATLAB stores polynomials by storing their coefficients in
a row vector ( 3 x 1 matrix)
Hence the polynomial x2 + 2x +3 would be stored as the vector [1 2 3]
» P = [1 2 3]
1 2 3
MATLAB has a several routines that allows one to manipulate
We can evaluate polynomials at a particular point with the command polyval
The following evaluates the polynomial with coefficients stored in P above at the value x = 2:
3. The conv command can be used to multiply two polynomial. For example, to multiply the above polynomial by (x + 1) -- i.e. to compute (x2 + 2x + 3) (x + 1):
» Q = [1 1]
1 3 5 3
( The result is thus x3+ 3x2 + 5x + 3. )
4. The MATLAB Symbolic Toolbox also has some commands for manipulating polynomials symbolically. The poly2sym command converts a vector of coefficients to a symbolic polynomial.
» syms x
find the polynomial of degree n that passes
through n + 1 data points we can use polyfit.
» help polyfit
To find the
polynomial passing through the points in Example 4.4, page
» X = [1 2 3 5]
» Y = [1.06 1.12 1.34 1.78]
» P = polyfit(X,Y,3)
-0.0200 0.2000 -0.4000 1.2800
Thus the polynomial fitting these points is
-0.02x3 +0.2x2-0.4x +1.28.You can see that MATLAB produces the correct coefficients agreeing with your text.
The function polyfit uses the linear algebra method of solving a system of linear equations to find the coefficients. We are using the Lagrange method which will give the same result, but is much less subject to computer round-off error in the calculations.
6. In order to construct the Lagrange Coefficients for the Lagrange Polynomial in MATLAB, we can use the built-in function poly, which constructs a polynomial with given roots.
Enter the following to construct a polynomial with roots 1 and 2 for example:
» poly([1 2])
1 -3 2
Thus this is the polynomial (x-1)(x-2) = x2 -3x +2 which has roots 1 and 2.
7. Assuming that we want to store the Lagrange coefficient polynomials in the 3x3 array L (with the 1st row being the coefficients for L2,0, the 2nd row being the coefficients for L2,1, and the third row being the coefficients for L2,2 ), we proceed as follows:
» L(1,:)= poly([2 2.5])/((1 - 2)*(1
0.6667 -3.0000 3.3333
» L(2,:)= poly([ 1 2.5])/((2 - 1)*(
2 - 2.5))
0.6667 -3.0000 3.3333
-2.0000 7.0000 -5.0000
» L(3,:)= poly([1 2])/((2.5 - 1)*(2.5
0.6667 -3.0000 3.3333
-2.0000 7.0000 -5.0000
1.3333 -4.0000 2.6667
final Lagrange polynomial is y0*L2,0
Since y0 = f(x0) = 1+2/1 = 3, and similarly y1 = 3, y2 = 3.3, we compute the polynomial P as
» P = 3*L(1,:) + 3*L(2,:) + 3.3*L(3,:)
0.4000 -1.2000 3.8000
9. To see this formatted as a polynomial in x enter:
10. To evaluate the polynomial at 1.5 we use:
Comparing the true value of f(x) = x + 2/x
» 1.5 +2/1.5
for x = 1.2:
» 1.2 + 2/1.5
can plot both of these on a graph to compare the two:
First create the symbolic polynomial from the coefficients in P:
» SP = poly2sym(P)
Plot the function f(x) = x + 2/x on the range x = .5 to 3
» ezplot('x + 2/x', [.5 3])
Plot the polynomial Lagrange interpolating polynomial
SP on the same range on the same graph:
» hold on; ezplot(SP,[0 3])
- X is a vector that contains a list of abscissas
% - Y is a vector that contains a list of ordinates
%Output - C is a matrix that contains the coefficents of
% the Lagrange interpolatory polynomial
% - L is a matrix that contains the Lagrange
% coefficient polynomials
15. Thus to call this function we set up the vectors X and Y with the x and y coefficients of the interpolating points. Then call the function to return the interpolating polynomial in C and the Lagrange coefficients for that polynomial in L. For our above example, it would be:
» X = [1 2 2.5]
» Y = [3 3 3.3]
» [C L] = lagran(X,Y)
We see that the same answer is returned as the one we computed step by step above.
Look at how the coefficients are computed in the body of the function
for k=1:n+1 % Calculate each of n+1 Lagrange coefficientThe final step is to compute the interpolating polynomial:
V=1; % Accumulate computations in V temporarily
% Multiply by (x - X(j))/(X(k) - X(j))
if k~=j % Be sure to skip the k'th one
L(k,:)=V; % Store Lagrange coefficient in kth row of L
C = Y*LThis uses matrix multiplication to compute C = Y * L. It is not difficult to verify using the rules for matrix multiplication that this gives the correct polynomial. For example, the first entry in C will be
16. Exercise: Use lagran to calculate the third degree Lagrange polynomial for cos(x) at the evenly spaced interpolating points for the x-values (abscissas) of 0.0, 0.4, 0.8, and 1.2 (see page 210-211, Example 4.7, part b), text for the computations and page 210 figure 4.12b) comparing the graph of this polynomial to the graph for function cos(x).
» X = [0.0 0.4 0.8 1.2]
» Y = cos(X)
» [C , L ] = lagran(X,Y)
Compare the graph of y = cos(x) to your computed polynomial with coefficients stored in C with the following commands
» hold on
We see they look quite close at least through the interpolating range 0 to 1.2.
error formula of Theorem 4.4 (see page 213 text) to compute
an upper bound of the error for these nodes (interpolating points).
h = distance between evenly spaced nodes = .4 in this case.
M4 is the maximum of the absolute value of the 4th derivative of f(x) = cos(x) on the interpolation range -- we can use 1 for this. Thus our error should be <= (.4)^4/25:
Does this appear to be true? --Calculate some of the errors with the following commands (and compare to table 4.7, page 216)
For part a) Use the lagran function. Set up your X, Y vectors ( X is time values, Y is temp values ).
c) Use the following commands to plot your data points
and the corresponding Lagrange polynomial stored in C
» SP = poly2sym(C)
» clf Clear any previous figures
» plot(X,Y,'r*') Plot the original data points in blue starts
» hold on; Retain the graph for next plot also.
b -- (See Theorem 1.10 -- Mean Value Theorem
for Integrals, page 6). Average temperature is the integral of
function (your approximating polynomial for temperature) over the
limits of the time interval divided by the length of the time
Thus you need to:
1. Write an m-file function called integ.m. This will start with the headng
function D = integral(C)
The body of
the function will create a new polynomial --coefficients stored in D,
whose length will be one more than C (its last entry will be 0 for the
constant term) and whose coefficients are those of the integral of the
stored in C.
After calling the function with the command D = integral(C), you
can use polyval on D to evaluate indefinite integral