How do I create a for loop for the this example?
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clear
clc
syms E0 E1 E2 E3 c0 c1 c2 c3 t r0 r1 r2 r3 K0 K1 K2 K3 format long
c0=0 c1=1/3; c2=2/3; c3=1; r0=0; r1=1; r2=2; r3=3;
E0=[euler(r0,sym(c0)) euler(r0,c1) euler(r0,c2) euler(r0,c3)]
E1=[euler(r1,c0) euler(r1,c1) euler(r1,c2) euler(r1,c3)]
E2=[euler(r2,c0) euler(r2,c1) euler(r2,c2) euler(r2,c3)]
E3=[euler(r3,c0) euler(r3,c1) euler(r3,c2) euler(r3,c3)]
E=[E0' E1' E2' E3']
K0=[int(euler(r0,t), 0,c0) int(euler(r0,t), 0,c1) int(euler(r0,t), 0,c2) int(euler(r0,t), 0,c3)]
K1=[int(euler(r1,t), 0,c0) int(euler(r1,t), 0,c1) int(euler(r1,t), 0,c2) int(euler(r1,t), 0,c3)]
K2=[int(euler(r2,t), 0,c0) int(euler(r2,t), 0,c1) int(euler(r2,t), 0,c2) int(euler(r2,t), 0,c3)]
K3=[int(euler(r3,t), 0,c0) int(euler(r3,t), 0,c1) int(euler(r3,t), 0,c2) int(euler(r3,t), 0,c3)]
K=[K0' K1' K2' K3']
Ek=E-K
F=[6*c0-3*c0^2;6*c1-3*c1^2;6*c2-3*c2^2;6*c3-3*c3^2]
InvInv_Ek=inv(Ek)
C=inv(Ek)*F
Ua=E*C
Ue=[6*c0 6*c1 6*c2 6*c3]'
1 commentaire
Walter Roberson
le 26 Avr 2021
https://www.mathworks.com/help/symbolic/sym.euler.html
The function accepts vectors for both arguments.
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