different answers for implementing summation

im trying to implement summation in the following 2 ways:

1.

f1=[10 20 30 40 50]
x1=[1 2 3 4 5]

J=0
for i=1:5
J=J+((f1(i)-a*exp(-(x1(i)-mu)^2/sigma))^2)
end

and 2.

f1=[10 20 30 40 50]
x1=[1 2 3 4 5]

J=0

J=@(f,x) ((f-a*exp(-(x-mu)^2/sigma))^2)

for i=1:5
J(f1(i),x1(i))
end

and im getting different final answers for each.

can anyone tell why?

Answer 1

[Guillaume]

Really, the best way of implementing your summation is option 3 which uses vectorised operations:

f1=[10 20 30 40 50]
x1=[1 2 3 4 5]

J = sum((f1 - a*exp(-(x1 - mu).^2 / sigma)) .^ 2)

Your option 2 looks like it wants to use J to store the result as in option 1 (since it has the line J = 0), but then put a function in J on the following line. In the loop you invoke the function but never assign the result to anything. I'm not sure what you expected to happen with that code. If you want to use an anonymous function, you could write your option 2 as:

func = @(f,x) (f-a*exp(-(x-mu)^2/sigma))^2;
J = 0;
for idx = 1 : numel(f1)  %don't hardcode bounds, use numel to get the number of elements
   J = J + func(f1(idx), x1(idx));
end

But again, vectorised code is better:

func = @(f,x) (f-a*exp(-(x-mu).^2/sigma)).^2; %note the use of .^ instead of ^
J = sum(func(f1, x1))

[Terry McGinnis]

thanks.the vectorised code looks the most viable

Answer 2

[Andrei Bobrov]

J = sum(f1-a*exp(-(x1-mu).^2/sigma)).^2)

for 2 variant:

f1=[10 20 30 40 50]
x1=[1 2 3 4 5]

J1=0

J=@(f,x) ((f-a*exp(-(x-mu)^2/sigma))^2)

for ii=1:5
    J1 = J1 + J(f1(ii),x1(ii))
end

[Terry McGinnis]

thanks