Azzera filtri
Azzera filtri

can someone help me run this

3 visualizzazioni (ultimi 30 giorni)
Relly Syam
Relly Syam il 7 Mag 2021
Commentato: Relly Syam il 7 Mag 2021
%
clear;
clc;
format long g
syms c0 c1 c2 c3 t r0 r1 r2 r3
c0=0; c1=1/3; c2=2/3; c3=1; r0=0; r1=1; r2=2; r3=3;
EvalAt = [c0, c1, c2, c3];
ktemp = arrayfun(@(EA) euler([r0, r1, r2, r3], EA).', EvalAt, 'uniform', 0);
ptemp=arrayfun(@(EA) int(euler([r0, r1, r2, r3],t), 0,EA).', EvalAt, 'uniform', 0);
E = horzcat(ktemp{:}).'
K = horzcat(ptemp{:}).'
Ek=E-K
Inv_Ek= inv(Ek)
F=[6*c0-3*c0^2;6*c1-3*c1^2;6*c2-3*c2^2;6*c3-3*c3^2]
C=inv(Ek)*F
%solusi aproximasinya
Ua=@(x)(C(1)*euler(0,x)+C(2)*euler(1,x)+C(3)*euler(2,x)+C(4)*euler(3,x))
uaa=[];
xx=[];
k=0;
for i=1:4
uaa(i)=Ua(k);
xx(i)=k;
k=k+.1;
end
uaa'
  2 Commenti
KSSV
KSSV il 7 Mag 2021
Whjat help you want? Code runs without any errors.
Relly Syam
Relly Syam il 7 Mag 2021
I ran the code above on matlab mobile, and the result was not error. I am a little confused, I have been trying to approximate an integral with a certain method but have found the above results. Is that done?

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