Lagrangian interpolation index exceeds matrix dimensions

fid = fopen('force.m','r');
n = fscanf(fid,'%f',1);
a = fscanf(fid,'%f',[2 n]);
x = a(1,:); x = x';
fx = a(2,:); fx = fx';
fclose(fid);
xx = input('\nEnter x value to find f(x):   ');
yy = 0.;
for i = 1:n
    al = 1.;
    for j = 1:n
        if j ~= i
            al = al*(xx-x(j))/(x(i)-x(j));
        end
    end
    yy = yy + al*fx(i);
end
fprintf('\nValue of f(x) at x = %8.6f', xx)
fprintf('is %14.6e\n' , yy)

When I run my code, the command window keep telling me that

>> interpolation

Index exceeds matrix dimensions.

Error in interpolation (line 4)

x = a(1,:); x = x';

and I don't know how to fix this

Answer 1

[Walter Roberson]

>> fid = fopen('force.m','r');

n = fscanf(fid,'%f',1)

n =

[]

>> type force.m

F = [0.107 100;0.172 200;0.238 300;0.351 400;0.388 500;0.417 600;0.432 700;0.441 800]

Your code is expecting force.m to contain a number that defines the number of rows that follow.

a = fscanf(fid,'%f',[2 n]);

and then it is expecting 2 columns (not 2 rows) with the number of rows of data given by the first number in the file.

The code is not expecting something in the form of a MATLAB assignment statement, and is totally unprepared to deal with the variable name, the equal sign, the [ and ], and the semi-colons.

Example of format that would be expected:

8
0.107 100
0.172 200
0.238 300
0.351 400
0.388 500
0.417 600
0.432 700
0.441 800

You need to either change the format of the file, or else you need to change how you read the file.

[Peerapon Amornsuppasil]

Thank you for your answer. I try changing the format like you said and get it now.