# Can someone please help me stop "Index exceeds array bounds." in iteration? Any help is highly appreciated.

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Yunusah Abdulai 2021 年 4 月 18 日
コメント済み: Clayton Gotberg 2021 年 4 月 18 日
clc
A = [4 -1 -3;-2 6 1;-1 1 7];
b=[3;9;-6];
D=diag(diag(A))
[L,U]=lu(A)
Mj = D\(U+L)
Mj(:,3)=[];
Mj(:,2)=[];
Vj =D\b
x1=zeros;
x=zeros;
for n = 2:8
x = Mj*('the previous x value')+ Vj;
end
x

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### 採用された回答

Clayton Gotberg 2021 年 4 月 18 日

You're asking for elements in x1 that don't exist. If we look only at the last loop n == 8 and you are asking for x1(7), but x1 is defined just before as being a 3x1 matrix of zeros.
I also want to point out that x1 is always zero in the code you posted here, so x will always equal Vj. Additionally, x is being overwritten on every loop instead of the value in each loop being saved.
##### 2 件のコメント表示非表示 1 件の古いコメント
Clayton Gotberg 2021 年 4 月 18 日
The previous value of x is just x. When programming, you are allowed to change the value of a variable using a reference to itself:
x = 5; % Define an initial value for x
x = x+5: % Define a new value for x: x = x (which we defined as 5) + 5
% Now, x == 10

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### その他の回答 (1 件)

the cyclist 2021 年 4 月 18 日
x1(n-1)
which is
x1(4)
But your vector x1 is length 3, so there is no 4th element to access. That's why you get that error.
##### 1 件のコメント表示非表示 なし
Yunusah Abdulai 2021 年 4 月 18 日
I edited the the code and i have this but i still have a problem with the iteration
clc
A = [4 -1 -3;-2 6 1;-1 1 7];
b=[3;9;-6];
D=diag(diag(A))
[L,U]=lu(A)
Mj = D\(U+L)
Mj(:,3)=[];
Mj(:,2)=[];
Vj =D\b
x1=zeros;
x=zeros;
for n = 2:8
x = Mj*('the previous x value')+ Vj;
end
x

サインインしてコメントする。

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