How to solve for a series of variables that intersect with each other?

Hi folks, recently I got a problem about solving a series of variables which have the property like:

x1 x2 x3 x4 x5 x6 x7 x8 x9 x10, where x1 and x10 are known, and the relationships are as following:

x2=x1^2+x3^3;
x3=x2^2+x4^2;
x4=x3^2+x5^2;
...

Intuitively there are in total 8 unkown variables and 8 equations so we can get value of each of them, but how can I solve these functions in matlab? What if there are more than just 10 variables, say 100, from x1 to x100, how can I solve the functions for x2 to x99? I tried to solve this as solving a system of linear equation but failed.

[Walter Roberson]

Are the variables positive? Are they real values ? Do you need the complete set of solutions or just one? Do you need closed form solutions or are numeric approximations okay ? Do you have the Symbolic Toolbox ?

[Chao Wang]

It might improper to set function as x2=x1^2+x3^2... what I want to say is that every function has the same dynamic, and these functions are actually linear functions and have more parameters, I just want to simplify it, you can change it into whatever you want.For example, x2=(x1+x3)2, x3=(x2+x4)/2, etc. All of them are real values, and if you change the functions into linear functions, I think there will only one set of solution. Also I have symbolic toolbox.

[Walter Roberson]

You should be able to solve that using \

1/2 -1 1/2 0 0 0 0 .......

0 1/2 -1 1/2 0 0 0 .....

for the A matrix and the b would be all 0

[Chao Wang]

I tried as what you said, and this is what I wrote:

M=95;
r=0.02;
q=0;
dt=0.02;
sigma=0.18;
a=zeros(M-1,1);
b=zeros(M-1,1);
c=zeros(M-1,1);
a(1)=M-1;
b(1)=M-1;
c(1)=M-1;
for j=2:M-1
    a(j)=a(j-1)-1;
    b(j)=b(j-1)-1;
    c(j)=c(j-1)-1;
end
for j=1:M-1
    a(j)=(r-q)*(M+1-j)*dt/2-sigma^2*(M+1-j)^2*dt/2;
    b(j)=1+sigma^2*(M+1-j)^2*dt+r*dt;
    c(j)=-(r-q)*(M+1-j)*dt/2-sigma^2*(M+1-j)^2*dt/2;
end
parameter=zeros(M-1,M+1);
for j=1:M-1
        parameter(j,j)=c(j);
        parameter(j,j+1)=b(j);
        parameter(j,j+2)=a(j);
end
b=zeros(94,1);
myfunc=@(x) parameter*[0;x;12]==b;
x=solve(myfunc,x);

but how to claim the dimension of x? If I don't do anything, the system just thinks [0;x;12] is a 3*1 vector rather than 94*1 vector.

[Walter Roberson]

x = sym('x', [94, 1]);
myfunc = parameter*[0;x;12]==b;
sol = solve(myfunc, x);
X = vpa( struct2cell(sol) );