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

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s=3; E=[]; for i=1:s for j=1:s E(i,j)=euler(i,c); end end E
  1 Comment
Relly Syam
Relly Syam on 25 Mar 2021
Edited: KSSV on 25 Mar 2021
s=3; E=[];
for i=1:s
for j=1:s
E(i,j)=euler(i,c);
end
end
E

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Answers (1)

KSSV
KSSV on 25 Mar 2021
c = 0:0.5:1
  9 Comments
KSSV
KSSV on 25 Mar 2021
E is a cell now...you need to access using: E{1}, E{2}
Relly Syam
Relly Syam on 25 Mar 2021
Edited: Relly Syam on 25 Mar 2021
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

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