can anyone help me prepare the code

2021 4 19
1 个回答
更新时间：2021 4 26
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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]'

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回答（1 个）

per isakson
per isakson 2021-4-21
Ran in:

0 个投票

Ran in:
Problems with your script
  • The long lines make the code hard to read.
  • Numerous statement separators are missing
I added line breaks and semicolons. Now the script runs.
%%
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
C = 
Ua=E*C
Ua = 
Ue=[6*c0 6*c1 6*c2 6*c3]'
Ue = 4×1
0 2 4 6

6 个评论
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Relly Syam
Relly Syam 2021-4-26
编辑：Relly Syam 2021-4-26
thank you very much for the answer, could you help me write it down using the 'for loop' function to make it shorter
Relly Syam
Relly Syam 2021-4-26
I run it on matlab mobile, what do you think about the result, where the approximate solution and the exact solution are the same
Walter Roberson
Walter Roberson 2021-4-26
https://www.mathworks.com/help/symbolic/sym.euler.html
The function accepts vectors for both arguments
Walter Roberson
Walter Roberson 2021-4-26
You are only doing symbolic computation, which should be exact for those formulas. I do not see any approximation calculations?
Relly Syam
Relly Syam 2021-4-26
The above code is part of the process of estimating the integral using the weighted residual collocation method. the example above uses n = 3. what do you think what i did was a little wrong?
Relly Syam
Relly Syam 2021-4-26
编辑：per isakson 2021-4-26
I am trying to approximate this equation

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