how to run c at intervals [0, 1] with a jump of 0.5

s=3; E=[]; for i=1:s for j=1:s E(i,j)=euler(i,c); end end E

[Relly Syam]

s=3; E=[]; 
for i=1:s 
    for j=1:s 
        E(i,j)=euler(i,c); 
    end 
end 
E

Answer 1

[KSSV]

c = 0:0.5:1

[Relly Syam]

can it work when i and j are running then c is also running

[KSSV]

Are you looking for something like this?

s=3; 
c = 0:0.5:1
E = cell(length(c),1) ; 
for k=1:length(c)
    EE = zeros(s) ; 
for i=1:s 
    for j=1:s 
        EE(i,j)=euler(i,c(k)); 
    end 
end 
E{k} = EE ; 
end

[Relly Syam]

I want to make a matrix [E1(c1) E1(C2) ... E1(Cn); E2(C1) E2(C2) .. E2(Cn); .....; En(C1) En(C2) ... En(Cn)]

[KSSV]

Edited the answer.

[Relly Syam]

it says "Error in euler (line 27) E (k) = EE "

[KSSV]

Yes you are right..it throws error....edited the line now.

[Relly Syam]

The answer is a little different. each row has the same value. looks like this [0.5 0.5 0.5; 0 0 0; -0. 25 -0. 25 -0. 25]

[KSSV]

E is a cell now...you need to access using: E{1}, E{2}

[Relly Syam]

I don't understand the flow. I want to make a matrix [E1(c1) E1(C2) ... E1(Cn); E2(C1) E2(C2) .. E2(Cn); .....; En(C1) En(C2) ... En(Cn)].

so when n = 3 then we get C1 = 0, C2 = 1/2, C3 = 1. and the matrix [E1(c1) E1(C2) E1(C3); E2(C1) E2(C2) E2(C3); E3(C1) E3(C2) E3(C3)] where the result is [-0.5 0 0.5; 0 -0.25 0; 0.25 0 -0.25]. because I want to multiply the results of the matrix above with another matrix.

E=euler(i, j) % i=1:n and j=1:n