For correction: Newton's Divided Difference method polynomial (nested form)

I'm taking a MSc course in Applied Numerical Analysis and the programming language/software for the class is Matlab which is a fairly unfamiliar territory to me and I have limited time to master all of its syntax and semantics.

We were asked to derive a 6th order polynomial p(x) (where n =6) that is approximately equal to the function f(x) = log10(x) and subsequently solve for f(x) when the value of x = 1.43 using the Newton's Divided difference as follows:

p(x) = a0 + (x-x0)[a1 + (x-x1)[a2 + (x-x2)[a3 + (x-x3)[a4 + % (x-x4)[a5]]]]]

Here's the best I could come up with for now:

NewtonsDiffMtd_150517_.m (Assignment)
    clear
    clc
    close all

      % Given Data points are:
    X = [1.2 1.3 1.4 1.5 1.6 1.7];          % Values for X
    Y = log10(X);                           % Y = f(X) = log(X)
    x_i = 1.43;                                % value of interest to be 
                                           %interpolated

    % First, we find the Newton Coefficient, a. That is,
    %function a = newtonCoeff(X,Y)
    n = length(X(:,1));                     % no. of X data points to be read
    a = Y;                                  % Stores the value of Y data point
                                            % Y data points in the array a
    for k = 2:n                             % For loop iterates over each elmt
       a(k:n) = (a(k:n) - a(k-1))./(X(k:n)- X(k-1));
    end

    % Next, we program Newton Method Polynomial: 
    % Given by: p(x) = a0 + (x-x0)[a1 + (x-x1)[a2 + (x-x2)[a3 + (x-x3)[a4 + 
    % (x-x4)[a5]]]]]
    % NOTE: This is achieved by deploying a backward recursion:

    p = a(n);               
    for k = 1:n-1;
        p = a(n-k) + (x_i - X(n-k))*p;
    end

    % Display the results:
    disp(['Values of p:                               ' num2str(p)]);

I used an algorithm provided in a book: "Numerical Methods for Engineering with MATLAB" by Jaan Kiusalaas.

Please, I would be glad if you could help me correct this so that the each values of x is displayed and that the value of interest (x_i) is used in the final result of p(x).

Thank you in advance.

Answer 1

[beebee]

Guess what guys, I sent an acquaintance the above code and he was able to spot the error. It's in line 7:

I had written:

X = [1.2 1.3 1.4 1.5 1.6 1.7];

when it should have been:

X = [1.2; 1.3; 1.4 1.5 1.6 1.7];

What was missing was the semi-colon i.e.: ";". And as you might guess, I tried it out and it worked perfectly fine.

Thank you.