how can I use summation and index for the series

how can I use summation and index for the following series by using matlab

Edit:

the pervious Equation solution for

My trying to write the following code, but i don't know if it is right ot no

u0=0;
alpha=0.5;%[0.1 0.3 0.5 0.6 0.66 0.9 1]; matrix 
a=0; b=1;n=100;
h=(b-a)/n;
t=a:h:b;
f=@(t,u) (-u.^4) +  ( gamma(2*alpha+1)./gamma(alpha+1) ).*(t.^alpha) - ( 2./gamma(3-alpha) ).*(t.^(2-alpha)) + (t.^(2*alpha)-t.^2).^4;
exact=@(t) t.^(2*alpha)-t.^2;
%-----a'ks---------------------------------------
a_kth=@(k) (k+1).^(1-alpha)-(k).^(1-alpha);
u_p=zeros(1,n); z_p=zeros(1,n); u=zeros(1,n);
s=0;
for i=1
    u_p(i)=a_kth(i-1)*u0;    %s=s+q^keval(subs(fun,xq^k));
    z_p(i)=gamma(2-alpha)*( h.^alpha ) * f(t(i),u_p(i));
    u(i)=u_p(i)+gamma(2-alpha)*( h.^alpha ) * f(t(i),u_p(i)+z_p(i));
end
%--------------------------------------
for i=2:n
for k=1:i-1
    s=s+(a_kth(i-1-k)-a_kth(i-k))*u(k);
    u_p(i)=a_kth(i-1)*u0+s;    %s=s+q^keval(subs(fun,xq^k));
    z_p(i)=gamma(2-alpha)*( h.^alpha ) * f(t(i),u_p(i));
    u(i)=u_p(i)+gamma(2-alpha)*( h.^alpha ) * f(t(i),u_p(i)+z_p(i));
end
end
u1=[ u0 u];
plot(t,u1)
exact1=exact(t);
error=exact1-u1;

[Walter Roberson]

I suspect that in practice you are going to need to use a loop or arrayfun()

[work wolf]

@Walter Roberson thank you for your comment, please can you guid me about code?

[Torsten]

@work wolf

Be careful with the 0-indices for u and a, but check whether

sum(-diff(a).*u(n:-1:1))

works for

sum_{k=1}^(n-1)(a_{n-1-k}-a_{n-k})*u_{k}

[work wolf]

@Torsten Thank you for your comment. please help me to find full code for that summation.

[Torsten]

Then supply a, n alpha, h, f and t.

Is it correct that f depends on u_n and thus the equation above must be solved for u_n by a root finder ?

[work wolf]

@Torsten

u0=0;
alpha=[0.1 0.3 0.5 0.6 0.66 0.9 1]; %multi values (matrix)
a=0; b=1;n=100;
h=(b-a)/n;
t=a:h:b;
f=@(t,u) (-u.^4) +  ( gamma(2*alpha+1)./gamma(alpha+1) ).*(t.^alpha) - ( 2./gamma(3-alpha) ).*(t.^(2-alpha)) + (t.^(2*alpha)-t.^2).^4;
exact=@(t) t.^(2*alpha)-t.^2;
%-----a'ks---------------------------------------
a_kth=@(k) (k+1).^(1-alpha)-(k).^(1-alpha);
%----------------------------------------------

[Torsten]

In f, how do you want to evaluate t^alpha ? t and alpha have a different number of elements.

You should explain the background of your question so that we are able to understand what you are trying to do in your code.

[work wolf]

the code for each alpha, i.e the code run for o.1 then repeate all steps for another value as 0.3. in same time, can you evaluate for one element as alpha= 0.1 then change alpha to 0.3, manually. i hope understand me :)

[Torsten]

Again my question:

Is it correct that f depends on u_n and thus the equation above must be solved for u_n by a root finder ?

[work wolf]

@Torsten yes. f depends on u_n , where u_n denotes the approximate value of u(t) at t= t_n .

Note: i am trying to write the folowing code , but i don't know if it right or wrong?!

u0=0;
alpha=0.5;%[0.1 0.3 0.5 0.6 0.66 0.9 1]; matrix 
a=0; b=1;n=100;
h=(b-a)/n;
t=a:h:b;
f=@(t,u) (-u.^4) +  ( gamma(2*alpha+1)./gamma(alpha+1) ).*(t.^alpha) - ( 2./gamma(3-alpha) ).*(t.^(2-alpha)) + (t.^(2*alpha)-t.^2).^4;
exact=@(t) t.^(2*alpha)-t.^2;
%-----a'ks---------------------------------------
a_kth=@(k) (k+1).^(1-alpha)-(k).^(1-alpha);
u_p=zeros(1,n); z_p=zeros(1,n); u=zeros(1,n);
s=0;
for i=1
    u_p(i)=a_kth(i-1)*u0;    %s=s+q^keval(subs(fun,xq^k));
    z_p(i)=gamma(2-alpha)*( h.^alpha ) * f(t(i),u_p(i));
    u(i)=u_p(i)+gamma(2-alpha)*( h.^alpha ) * f(t(i),u_p(i)+z_p(i));
end
%--------------------------------------
for i=2:n
    for k=1:i-1
        s=s+(a_kth(i-1-k)-a_kth(i-k))*u(k);
        u_p(i)=a_kth(i-1)*u0+s;    %s=s+q^keval(subs(fun,xq^k));
        z_p(i)=gamma(2-alpha)*( h.^alpha ) * f(t(i),u_p(i));
        u(i)=u_p(i)+gamma(2-alpha)*( h.^alpha ) * f(t(i),u_p(i)+z_p(i));
    end
end
u1=[ u0 u];
plot(t,u1)
exact1=exact(t);
error=exact1-u1;

[Torsten]

But then - starting from u(1) - you will have to solve a nonlinear equation for u(i). Is that true ?

[work wolf]

yeah u(t_o) is known as following

[Torsten]

I have no experience with fractional order differential equations, but since your computations diverge, something must be wrong with the implementation.

[work wolf]

@Torsten thank you so much for your comments. i will editing my question to add more infromation by your converstion.

[Torsten]

You should include a link where the forum can find the discretization used in your code.

[work wolf]

@Torsten How is that?

[Torsten]

A link where the solution scheme for your equation is derived and explained.