Error using double (subs())

2022 8 3
1 个回答
更新时间：2022 8 3
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0 个投票

Hi,
I want to run newton's method and I am not able to convert the syms variable into double even though I am using subs.
Here is my code.
syms initial_R
u = -100;
v = @(initial_R) initial_R.*exp ((2*(1-alpha))/(2+(1-alpha)*u.*initial_R)./(2-alpha*u.*initial_R).^((alpha)^2) * (2+(1-alpha)*u.*initial_R).^(1-alpha^2));
vpr = diff (v,initial_R);
funcr = @(initial_R) vpr;
kmax = 10;
initial_R(1) = 0;
initial_R(2) = 0.0070;
y(2) = v(initial_R(2));
ypr(2) = funcr (initial_R(2));
tol = 0.000001
for k = 2:0.01:kmax
initial_R(k+1) = initial_R(k) - y(k)/ypr(k);
y(k+1) = feval (v,initial_R(k+1));
c = initial_R(k+1) - initial_R(k);
d = subs(c, initial_R(k+1), initial_R(k));
if (abs(double(d))) < tol
disp('The Numerical has ended')
end
ypr(k+1) = feval(funcr, initial_R(k+1));
end
I would like to know how else can I check the tolerance level because everytime I run this code with no matter how many variations, the if line throws an error.
Error using symengine
Unable to convert expression containing symbolic variables into double array. Apply 'subs' function first to substitute values for
variables.
Error in sym/double (line 868)
Xstr = mupadmex('symobj::double', S.s, 0);
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回答（1 个）

Torsten
Torsten 2022-8-3
Ran in:

0 个投票

Ran in:
syms initial_R
u = -100;
alpha = 2;
v = initial_R*exp((2*(1-alpha))/(2+(1-alpha)*u*initial_R)/(2-alpha*u*initial_R)^(alpha)^2*(2+(1-alpha)*u*initial_R)^(1-alpha^2))
v = 
vpr = diff(v,initial_R);
y = matlabFunction(v)
y = function_handle with value:
@(initial_R)initial_R.*exp(1.0./(initial_R.*1.0e+2+2.0).^4.*1.0./(initial_R.*2.0e+2+2.0).^4.*-2.0)
ypr = matlabFunction(vpr)
ypr = function_handle with value:
@(initial_R)exp(1.0./(initial_R.*1.0e+2+2.0).^4.*1.0./(initial_R.*2.0e+2+2.0).^4.*-2.0)+initial_R.*exp(1.0./(initial_R.*1.0e+2+2.0).^4.*1.0./(initial_R.*2.0e+2+2.0).^4.*-2.0).*(1.0./(initial_R.*1.0e+2+2.0).^4.*1.0./(initial_R.*2.0e+2+2.0).^5.*1.6e+3+1.0./(initial_R.*1.0e+2+2.0).^5.*1.0./(initial_R.*2.0e+2+2.0).^4.*8.0e+2)
kmax = 50;
initial_R_old = 1.0;
tol = 0.000001;
k = 0;
dx = 1.0;
while dx > tol && k < kmax
k = k+1;
initial_R_new = initial_R_old - y(initial_R_old)/ypr(initial_R_old);
dx = abs(initial_R_old - initial_R_new);
initial_R_old = initial_R_new;
end
initial_R_new
initial_R_new = 0
y(initial_R_new)
ans = 0

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回答：

Torsten
Torsten
2022-8-3

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