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symbolic calculation error in code

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YOGESHWARI PATEL
YOGESHWARI PATEL el 22 de Mzo. de 2021
Comentada: YOGESHWARI PATEL el 24 de Mzo. de 2021
syms x n
syms t
% syms m
% m=0.7;
U=zeros(1,2,'sym');
A=zeros(1,2,'sym');
B=zeros(1,2,'sym');
U(1)=exp((sqrt(2)*x)/4)/(exp((sqrt(2)*x)/4)+exp(-(sqrt(2)*x)/4));
for k=1:10
A(1)=0;
B(1)=0;
for i=1:k
A(1)=A(1)+U(i)*U(k-i+1) ;
end
for j=1:k
for r=1:j
B(1)=U(r)*U(j-r+1)*U(k-r+1);
end
end
U(k+1)=((diff(U(k),x,2)-B(1)+2*A(1)-U(k)))/k;
end
disp (U)
for k=1:10
series(x,t)=series(x,t)+U(k)*(power(t,k-1));
end
series
C=zeros(5,5);
for x=1:5
e=(x-1)/1;
for t=1:5
f=(t-1)/10;
C(x,t)=series(e,f)
end
end
vpa(C,15)
This code is not showing the answer.BUSY is also shown in the workspace.

Respuesta aceptada

Walter Roberson
Walter Roberson el 22 de Mzo. de 2021
tic
syms x n
syms t
% syms m
% m = 0.7;
U = zeros(1,2,'sym');
A = zeros(1,2,'sym');
B = zeros(1,2,'sym');
U(1) = simplify(exp((sqrt(2)*x)/4)/(exp((sqrt(2)*x)/4)+exp(-(sqrt(2)*x)/4)));
for k = 1:10
A(1) = 0;
B(1) = 0;
for i = 1:k
A(1) = simplify(A(1)+U(i)*U(k-i+1)) ;
end
for j = 1:k
for r = 1:j
B(1) = simplify(U(r)*U(j-r+1)*U(k-r+1));
end
end
U(k+1) = simplify(((diff(U(k),x,2)-B(1)+2*A(1)-U(k)))/k);
%disp(U(k+1))
end
disp (U)
for k = 1:10
series(x,t) = simplify(series(x,t)+U(k)*(power(t,k-1)));
end
series
series(x, t) = 
C = zeros(5,5);
for x = 1:5
e = (x-1)/1;
for t = 1:5
f = (t-1)/10;
C(x,t) = series(e,f);
end
end
vpa(C,15)
ans = 
toc
Elapsed time is 47.994421 seconds.
  7 comentarios
Walter Roberson
Walter Roberson el 24 de Mzo. de 2021
k = 20 took about 600 seconds for me; it looks like the time increase might be cubic in k (which does not surprise me.)
So if you were planning to take this to k = 1000 like in your earlier question, you should expect on the order of 295 days to execute, if your system has enough memory.
You could attempt to rewrite the whole calculation as numeric work, in hope that would speed it up -- which would be entirely possible. However, beyond k = 245 or so, the denominator of the expression would overflow to infinity, so that is the approximate upper limit on numeric work before the calculation becomes completely useles. The calculation is likely to become mostly useless for numeric work long before that.
YOGESHWARI PATEL
YOGESHWARI PATEL el 24 de Mzo. de 2021
Thank you

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