How can I use for loop to generate the data.

I am trying to generate data using for loop. I have included all of the data below:

p1 is the values or array that will be varied, and fun equation and result will give different values. Also, p4 is fixed.

I sort of know that

for 1:length(p1) ... 
end ... 

but, I am not sure..

p1 = [6.78,2.79,9.8,11.93,7.81,8.83,5.81,12.85,10.94,7.84,5.29,4.83,9.77];
p4 = 29.85;
 
fun = @(p2) p2/p4*(1 -((gamma-1)*(p2/p1-1))./(2*gamma*(2*gamma+(gamma+1)*(p2/p1-1))).^.5).^(-2.*gamma./(gamma-1))-1;
result = fzero(fun,10)/p1;
end

[BobH]

'for' needs a variable and then you can access each element of p1 using that variable in parens, as shown in the examples.

p1(1)
ans =
    6.7800

[Sun Kyoo Choi]

Could you provide an example code please?

For example,

p1 = [6.78,2.79,9.8,11.93,7.81,8.83,5.81,12.85,10.94,7.84,5.29,4.83,9.77];
for 1:length(p1)
p4 = 29.85;
fun = @(p2) p2/p4*(1 -((gamma-1)*(p2/p1-1))./(2*gamma*(2*gamma+(gamma+1)*(p2/p1-1))).^.5).^(-2.*gamma./(gamma-1))-1;
result = fzero(fun,10)/p1;
...
end

In this way..so that I can obtain different result values with respect to p1. 

[Walter Roberson]

for K= 1:length(p1)
p4 = 29.85;
fun = @(p2) p2/p4*(1 -((gamma-1)*(p2/p1(K)-1))./(2*gamma*(2*gamma+(gamma+1)*(p2/p1(K)-1))).^.5).^(-2.*gamma./(gamma-1))-1;
result(K) = fzero(fun,10)/p1(K) ;
...
end

[Ameer Hamza]

You don't need a for loop to get results for different values of p1. You can simply use vector operations

p1 = [6.78,2.79,9.8,11.93,7.81,8.83,5.81,12.85,10.94,7.84,5.29,4.83,9.77];
p4 = 29.85;
fun = @(p2) p2/p4*(1 -((gamma-1)*(p2/p1-1))./(2*gamma*(2*gamma+(gamma+1)*(p2/p1-1))).^.5).^(-2.*gamma./(gamma-1))-1;
result = fzero(fun,10)./p1;

[Walter Roberson]

Ameer, that will not work.

p1 is a row vector.

fzero passes scalars into the function, so inside fun, p2 will be a scalar.

p2/p1 is then scalar / row vector . That will give you an error about incompatible dimensions. At the very least you need to change the p2/p1 to p2./p1 in all places

However, p2./p1 will give a vector result, so the value of fun would be a vector. fzero() can never find the zero of a vector: fzero() is restricted to processing functions that produce scalars.

fsolve() can handle vectors of values, but when you pass the scalar 10 as the initial guess, that would tell fsolve() to find a single scalar value that makes all of the expressions 0, which is not something that can be done. So you would have to change the initial guess of 10 into 10*ones(size(p1)) . But that be inefficient as it would try to adjust the entries of a vector of length 13 against each other to try to reduce the calculation error, which is something that doesn't need to be done as we know the expressions are independent.

In other posts by the same user, I show how to use arrayfun() to deal with this situation more efficiently.

[Ameer Hamza]

Walter, correct. Thanks for pointing out. I overlooked this issue.