How to compute the sum term ?

Hi, I hope you can help me in this problem :

Compute the sum term solution : which .

I don't need

% I have two solution U1 and U2
alpha = 0.6;
p     = 0;
S     = 0;
for n=2:50
    p = p+1;
    qnp=(n-p+1)^(1-alpha)-(n-p)^(1-alpha);
    S = S+qnp*(U1-U2);
    
    % Compute the new solution
    
    U = 0.5*(U2+S); 
    
    % Overwrite the two solution
    U1 = U2;
    U2 = U;
end
% Resultat  
U

The problem is

• when n=2 then p=1, qnp=, and S = S+qnp*(U1-U2)=0+()*(U1-U2)= ()*(U1-U2)

It's Correct , then overwrite U1=U2 (i write it u2) and U2=U ( i write it u3)

• When n=3 then p=2, qnp=, and S = S+qnp*(U1-U2) = ()*(U1-U2) + ()*(u2-u3).

It's not Correct because S = ()*(U1-U2) +()*(u2-u3) .

• When n=4, then p=3 and the sum term is exactly : S = S+qnp*(U1-U2) = +qnp*(u3-u4)

which u3 and u4 has overwrited.

So, How can i compute the sum correctly ?

Thanks.

Answer 1

[David Hill]

alpha = 0.6;
U(1)=1;%you need to specify U(1) and U(2)
U(2)=.5;
S=2^(1-alpha)*(U(1)-U(2));
U(3)=(U(2)+S)/2;
for n=3:50
    S=S+sum(((n-[1:n-1]+1).^(1-alpha)-(n-[2:n]).^(1-alpha)).*(U(1:n-1)-U(2:n)));
    U(n+1) = (U(n)+S)/2;
end

[Yamina chbak]

Thanks @David Hill yes it's working you use index of the solution U(1),U(2),...U(n)

But it's possible to no use index and just overwrite the solution U1=U2, U2=U which U is the new solution ? if you say yes, Can you say me how ?