how to find a function written as a series

i tried to solve the following u(n+1)=-int((x-t)*u(n),t,0,x)

where n from 0 to in but matlab say to not use inf so i want to try it up to100 at least is it possible ?

wha i try asfollows but still have something missed

is any way to use index like sum(ui)=u0+u1+u3+.....

and how to find the answer using taylor seires

(this is to solve an integral equation using Adomiandecomposition method)

syms x t u  
a=0;
b=x;
f=(x-t);
u0=-2+3*x-x^2;

for k=1:inf   %or k=1:100 at least
    u = -1*int(f*u,t,a,b);
end
u=u0+u

Answer 1

[Paul]

Would be nice to see the actual mathematical equation to be solved. Assuming the equation and solution follows the exposition at Adomian decomposition method, the first n terms of the solution would be constructed as

syms x t 
a = 0;
b = x;
f = (x-t);
u0 =-2 + 3*x - x^2;
u(1) = u0;
for k=2:10   %or k=1:100 at least
    u(k) = -1*int(f*subs(u(k-1),x,t),t,a,b);
end
u,disp(char(u))

[3*x - x^2 - 2, (x^2*(x^2 - 6*x + 12))/12, -(x^4*(x^2 - 9*x + 30))/360, (x^6*(x^2 - 12*x + 56))/20160, -(x^8*(x^2 - 15*x + 90))/1814400, (x^10*(x^2 - 18*x + 132))/239500800, -(x^12*(x^2 - 21*x + 182))/43589145600, (x^14*(x^2 - 24*x + 240))/10461394944000, -(x^16*(x^2 - 27*x + 306))/3201186852864000, (x^18*(x^2 - 30*x + 380))/1216451004088320000]

temp = u; % save for later

Summing the u(k) together yields a series-like expression

u = simplify(sum(u)),disp(char(u))

3*x - x^3/2 + x^5/40 - x^7/1680 + x^9/120960 - x^11/13305600 + x^13/2075673600 - x^15/435891456000 + x^17/118562476032000 - x^19/40548366802944000 + x^20/1216451004088320000 - 2

figure
hax = gca;
fplot(u,[0,10])

We can also get a closed form expression for u(x)

syms k n integer
term0(n) = (-1)^(n+1);
term1(n) = x^(2*n);
term2(n) = 3*(n+1);
term3(n) = simplify(symsum(2 + 8*k,k,0,n));
term4(n) = piecewise(n==0,1,symprod(term3(k),k,1,n));
u(n) = term0(n)*term1(n)*(x^2 - term2(n)*x + term3(n))/term4(n);
u(n) = simplify(u(n)),disp(char(u(n)))

(2*(-1)^(n + 1)*x^(2*n)*(6*n - 3*x - 3*n*x + 4*n^2 + x^2 + 2))/factorial(2*n + 2)

% verify against previous solution
double(simplify(temp-u(0:9)))

ans = 1×10

0 0 0 0 0 0 0 0 0 0

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

Closed for expression for u(x)

u(x) = symsum(u(n),n,0,inf),disp(char(u(x)))

2*x^2*(cosh(x*1i)/x^2 - 1/x^2) + (4*cosh(x*1i))/x^2 + (x^2*hypergeom(2, [5/2, 3], -x^2/4))/2 - (x^3*hypergeom(2, [5/2, 3], -x^2/4))/4 + (x^2*hypergeom([2, 2], [1, 5/2, 3], -x^2/4))/3 - 4/x^2 - 6*x*(cosh(x*1i)/x^2 - 1/x^2)

figure
fplot([sum(temp),u(x)],[0,10])

It would be nice to know the actual equation that the OP is trying to solve.

[Aya Alaa eldeen mohamed]

the actual equation

[Torsten]

For comparison: solution is

u(x) = 3*sin(x)-2

syms u(x) v(x) t
eqns = [diff(u,x)==3-2*x-v,diff(v,x)==u];
conds = [u(0)==-2,v(0)==0];
sol = dsolve(eqns,conds);
usol(x) = simplify(sol.u)
ucomp = -2+3*x-x^2-int((x-t)*usol(t),t,0,x)