- The function handles have to be defined with the variable that is present in the expression
- While passing a function handle to subs(), you need not place it in quotes as this is a variable
- You may have to use double() to actually substitute the numeric values while referencing f0, f0_prime.
How do I solve "Conversion to logical from sym is not possible" error on line 26
26 vues (au cours des 30 derniers jours)
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clear; clc;
sym h;
f = @(x) 10*pi*h^2-((pi*h^3)/3)-1000;
F = @(x) 20*pi*h-pi*h^2; %calculated by hand
n = 6;
eps = 1*10^-(n+1); %epsilon value
h0 = 6;
for i = 1:20
f0 = vpa(subs('f','h','h0'));
f0_prime = vpa(subs('F','h','h0'));
Y = (h0-f0)/f0_prime; % The Formula
error = abs(Y-h0);
if error < eps %checking the error value after each iteration <===== LINE 26
break
end
h0 = Y;
end
Y = Y - rem(Y,10^-n)
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Réponses (1)
Jemima Pulipati
le 18 Déc 2020
Hello,
From my understanding, you are trying to use a symbolic variable in an 'if' condition. But the if condition checks for a definite value and here since there is a symbolic variable, it throws an error.
Here are few observations from the code:
Here is the modified code
clear; clc;
sym h;
f = @(h) 10*pi*h^2-((pi*h^3)/3)-1000;
F = @(h) 20*pi*h-pi*h^2; %calculated by hand
n = 6;
eps = 1*10^-(n+1); %epsilon value
h0 = 6;
for i = 1:20
f0 = vpa(subs(f,'h','h0'));
f0_prime = vpa(subs(F,'h','h0'));
Y = (h0-double(subs(f0)))/double(subs(f0_prime)); % The Formula
error = abs(Y-h0);
if error < eps %checking the error value after each iteration <===== LINE 26
break
end
h0 = Y;
end
Y = Y - rem(Y,10^-n)
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