Please explain for me this code , and what does each function do in detail ?

function s=lagrangeP(x,X,Y)

format long

n=size(X);

L=ones(n);

for i=1:n

for j=1:n

if (i~=j)

L(i)=L(i).*(x-X(j))./(X(i)-X(j));

end

end

end

s=0;

for i=1:n

s=s+Y(i)*L(i);

end

Answer 1

[Kevin Holly]

This was answered here:

https://www.mathworks.com/matlabcentral/answers/3338-lagrange-interpolation-m

[Kevin Holly]

Here is a more detail explanation.

function s=lagrangeP(x,X,Y) %This creates the function lagrangeP. s is the output. x, X, and Y are the inputs.
format long %This allows the display of numbers in the command window to appear long (many decimal places)

%This section preallocates the variable L before it goes through the for loop. This reduces computation time, so you don't an array changing sizes
%after every loop. Now, you will just be replacing values in an array of a set size.
n=size(X);%This finds the size of X and saves the dimensions as a vector array with the variable name n.
L=ones(n);%This creates a n by n matrix of ones equal to the dimensions of X. 

%This section calculates values within the matrix L
for i=1:n %This creates a for loop by giving values for i that starts with 1 and then increases by 1 until it reaches an element size of n
    for j=1:n %This does the same think but for variable j
        if (i~=j) %This if condition states that the line below only applies when the value of i does not equal the value of j
            L(i)=L(i).*(x-X(j))./(X(i)-X(j));%These calculations can be found in the equation below. L(i) goes through each element of the matrix L when in the for loop from L(1) to L(n).
        end
    end
end
s=0;
for i=1:n
    s=s+Y(i)*L(i);
end

You can find more details about the equation in the following:

https://www.mathworks.com/content/dam/mathworks/mathworks-dot-com/moler/interp.pdf

The above link was provided in a comment by Walter in the previously linked MATLAB Answers post.

[Alhussain AlAbbas]

Thank you!

[Kevin Holly]

@Alhussain AlAbbas

No problem. If this answers your question, I would appreciate it if you accepted the answer.