Problems with the application of Newton's method

1 Ansicht (letzte 30 Tage)
Aryo Aryanapour
Aryo Aryanapour am 29 Dez. 2021
Kommentiert: Aryo Aryanapour am 16 Jan. 2022
function [] = newton_raphson(func, diff, x0)
%UNTITLED Summary of this function goes here
% Detailed explanation goes here
x = x0;
maxiter = 200;
tol = 10^(-5);
eps = 0.4;
c_s = 5.67*10^(-8);
alpha_k = 4;
s1 = 0.250;
s2 = 0.015;
lamda1 = 0.35;
lamda2 = 22.7;
Tw_1 = 1200;
T_l = 10;
func = @(x) eps * c_s * x^4 + (alpha_k + 1/(s1/lamda1+s2/lamda2)) * x - ((1/(s1/lamda1+s2/lamda2)) * Tw_1 + alpha_k * T_l);
diff = @(x) 4 * eps * c_s * x^3 + (alpha_k + 1/(s1/lamda1+s2/lamda2));
newton_raphson(func, diff, 200)
for i = 1:maxiter
if
diff(x(i)) < tol
fprintf('Pitfall hast occured a better initial guess\n');
return;
end
x(i+1) = x(i) - func(x(i))/diff(x(i));
abs_error(i+1) = abs((x(i+1)-x(i))/x(i+1))*100;
if
abs(x(i+1) - x(ix)) < tol
fprintf('The Root has converged at x = %.10f\n', x(i+1));
else
fprintf('Iteration no: %d,current guess x = %.10f, error = %.5f', i, x(i+1), abs_error(i+1));
end
end
end
Can someone help me please. Unfortunately, I'm not quite fit in Matlab and have recently started working with functions. I don't know what's wrong with this code. Unfortunately I don't get a result. It had to come out with an X value of around 290.
Thanks a lot

Akzeptierte Antwort

Torsten
Torsten am 29 Dez. 2021
Bearbeitet: Torsten am 29 Dez. 2021
function main
%UNTITLED Summary of this function goes here
% Detailed explanation goes here
x0 = 20;
maxiter = 200;
tol = 10^(-5);
eps = 0.4;
c_s = 5.67*10^(-8);
alpha_k = 4;
s1 = 0.250;
s2 = 0.015;
lamda1 = 0.35;
lamda2 = 22.7;
Tw_1 = 1200;
T_l = 10;
func = @(x) eps * c_s * x^4 + (alpha_k + 1/(s1/lamda1+s2/lamda2)) * x - ((1/(s1/lamda1+s2/lamda2)) * Tw_1 + alpha_k * T_l);
diff = @(x) 4 * eps * c_s * x^3 + (alpha_k + 1/(s1/lamda1+s2/lamda2));
xsol = newton_raphson(func, diff, x0)
end
function xsol = newton_raphson(func, diff, x0)
%UNTITLED Summary of this function goes here
% Detailed explanation goes here
x(1) = x0;
maxiter = 200;
tol = 10^(-5);
for i = 1:maxiter
if diff(x(i)) < tol
fprintf('Pitfall hast occured a better initial guess\n');
return;
end
x(i+1) = x(i) - func(x(i))/diff(x(i));
abs_error(i+1) = abs((x(i+1)-x(i))/x(i+1))*100;
if abs(x(i+1) - x(i)) < tol
fprintf('The Root has converged at x = %.10f\n', x(i+1));
else
fprintf('Iteration no: %d,current guess x = %.10f, error = %.5f', i, x(i+1), abs_error(i+1));
end
end
xsol = x(end);
end
  9 Kommentare
Torsten
Torsten am 16 Jan. 2022
Bearbeitet: Torsten am 16 Jan. 2022
Sorry, but I'm no engineer. I can't help you in this respect.
I thought the question was how to obtain deduced quantities from the result xsol in general.
Aryo Aryanapour
Aryo Aryanapour am 16 Jan. 2022
Thorsten
dont be sorry
You are the best. You have been able to help me a lot more than I thought. Thank you again for always replying so quickly.

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