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

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Chao Wang
Chao Wang 2019-2-23
评论： Walter Roberson 2019-2-24
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.
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Chao Wang
Chao Wang 2019-2-23
编辑：Chao Wang 2019-2-23
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
Walter Roberson 2019-2-24
x = sym('x', [94, 1]);
myfunc = parameter*[0;x;12]==b;
sol = solve(myfunc, x);
X = vpa( struct2cell(sol) );

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