variable store in loop

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.

Answer 1

[VBBV]

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]

Use str2sym

[yogeshwari patel]

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]

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]

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]

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]

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