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variable store in loop

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yogeshwari patel
yogeshwari patel 2023-1-4
编辑： VBBV 2023-1-5
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.

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回答（1 个）

VBBV
VBBV 2023-1-4
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
VBBV
VBBV 2023-1-5
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

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