Summation subroutine giving result as NaN (not a number)!

In the code below I'm trying to produce the following summation notation with the answers expected being 1,-2,1 (second derivative,central finite difference coefficient)

However when I run my code the answer for S = NaN, and I'm a bit lost as to where I've gone wrong.

n = 2;
k = n/2;
 
for j=0:n
    S = 0;
    for i=0:n
        if (i~=j)    
        end
            sum=0;
                for m=0:n
                    if (m~=j) && (m~=i)
                    end
    
                    prod=1.0;
                    for l=0:n
                        if (l~=j) && (l~=i) && (l~=m)
                        prod=prod*(n/2-l)/(j-l);
                        end
                    end
                    prod=prod*1/(j-m);
                    sum=sum+prod;
                end 
                S=S*(1/(j-i))*sum;
    end
end

Any help much appreciated, thanks in advance.

[Sargondjani]

I would recommend to check all your intermediate results. YOu could for example add lines like:

if isnan(prod)

error('variable prod is NaN')

end

Also note that 'sum' and 'prod' are functions, so you shouldnt use them as a variable names.

[Ive J]

Your indices for j and m begin from 0, so you end up with division by zero, so sum is always NaN.

A = NaN;
A + 1
ans =

   NaN

btw, as Sargondjani also mentioned, it's always a bad practice to use MATLAB functions as variables (although MATLAB doesn't throw error).

[Louis Crawford]

I'll give those a go thanks!

Answer 1

[David Goodmanson]

Hi Louis,

the basic problem is that in the product calculation you have

for l=0:n
  if (l~=j) && (l~=i) && (l~=m)
    prod=prod*(n/2-l)/(j-l);
  end
end

which is correct, but in the first sum calculation (similarly for the second) you have

for m=0:n
if (m~=j) && (m~=i)
end
 ...
end

which does not make use of the 'if' check because the 'if' is immediatly followed by an 'end'. The following code changes that. I also used q instead of l (small L) because it is hard to distinguish l from 1, and did not use sum and prod for variables since they are Matlab functions and the possibility of getting mysterioso results is high. I also found that in this instance for whatever reason it was easier to keep track without using indentation, which can always be added.

n = 6;  j = 3;
 
si = 0;                     
for i = 0:n
if (i~=j)    

sm = 0;
for m=0:n
if (m~=j) && (m~=i)

p = 1;
for q=0:n
if (q~=j) && (q~=i) && (q~=m)
p = p*(n/2-1)/(j-q);
end
end

sm = sm + p/(j-m);
end
end

si = si + sm/(j-i);
end
end

si

[Louis Crawford]

Thank you!