variable store in loop

2023 1 4
1 个回答
更新时间：2023 1 5
5 次查看（30 天）

0 个投票

syms z x a v
Y(1)=a
T=0
for k=1:2
Y(k+1)=z
v=0
for l=1:k
v=v+kroneckerDelta(sym(l-2))*(k+2-l)*(k+1-l)*Y(k+2-l) % THIS LINE
end
T= v+ 2*(k)*Y(k+1)+(kroneckerDelta(sym(l-2)))==0
a=solve(T,z)
Y(k+1)=a
end
The highlighted part is not getting updated. Why it is so? Y(2)=z is variable and this value is later used to calculate T.

请先登录，再回答此问题。

回答（1 个）

VBBV
VBBV 2023-1-4

1 个投票

syms z x a v
Y(1)=a
T=0
for k=1:2
Y(k+1)=z;
v=0;
for l=1:k
v=v+kroneckerDelta(str2sym('l-2'))*(k+2-l)*(k+1-l)*Y(k+2-l);
end
T= v+ 2*(k)*Y(k+1)+(kroneckerDelta(str2sym('l-2')))==0
a=solve(T,z);
Y(k+1)=a;
end

6 个评论
显示 4更早的评论 隐藏 4更早的评论

VBBV
VBBV 2023-1-4
Use str2sym
yogeshwari patel
yogeshwari patel 2023-1-5
Sir You did get my point .Let me explain it
(kroneckerDelta(('l-2')) part is correct as the value of l varies from 1 to n the answer are 0,1,0,0,0,...... Now in the first for loop when value of l=2 , kroneckerDelta(('l-2') is one and Y(k+2-l)=z .So value that must be stored in variable v= 2z but after each and every calculation it is zero. That is the problem with the code
Image Analyst
Image Analyst 2023-1-5
You're assigning a to Y, not z. What is the point of z? What value does z even have? If you do
whos z
z
as the first lines in your for loop, what does it show?
yogeshwari patel
yogeshwari patel 2023-1-5
I want to generate the linear equation see I have check the calculation but is not correct
syms z x a v
Y(1)=a
T=0
fprintf('&&&&&&&&&')
for k=1:2
Y(k+1)=z
v=0;
for l=1:k
t=kroneckerDelta(sym(l-2))
m=Y(k+1)
y=Y(k+2-l)
n=(k+2-l)*(k+1-l)*Y(k+2-l)
v=v+kroneckerDelta(sym(l-2))*(k+2-l)*(k+1-l)*Y(k+2-l)
end
T= v+ 2*(k)*Y(k+1)+(kroneckerDelta(sym(l-2)))==0
a=solve(T,z);
Y(k+1)=a
fprintf('%%%%%%%%%%%%%%%%%%%%%5')
end
VBBV
VBBV 2023-1-5
Ran in:
Ran in:
syms z x a v
Y(1)=a
Y = 
a
T=0;
for k=1:2
Y(k+1)=z
v=0
for l=1:k
v=v+kroneckerDelta(sym(l-2))*(k+2-l)*(k+1-l)*Y(k+2-l) % THIS LINE
end
T= v+ 2*(k)*Y(k+1)+(kroneckerDelta(sym(l-2)))==0
a=solve(T,z)
Y(k)=a % may be this is change needed
end
Y = 
v = 0
v = 
0
T = 
a = 
0
Y = 
Y = 
v = 0
v = 
0
v = 
T = 
a = 
Y = 
VBBV
VBBV 2023-1-5
编辑：VBBV 2023-1-5
you need to assign kth value to a in the below line
a=solve(T,z)
Y(k)=a % may be this is change needed
end

请先登录，再进行评论。

类别

在 帮助中心 和 File Exchange 中查找有关 Code Performance 的更多信息

产品

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by