# variable store in loop

1 view (last 30 days)
yogeshwari patel on 4 Jan 2023
Edited: VBBV on 5 Jan 2023
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.

VBBV on 4 Jan 2023
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
VBBV on 5 Jan 2023
Edited: VBBV on 5 Jan 2023
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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