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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.

David Hill
on 22 Jan 2021

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

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