Problems with the application of Newton's method

Question

Aryo Aryanapour 2021-12-29

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1619370-problems-with-the-application-of-newton-s-method

评论： Aryo Aryanapour 2022-1-16

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

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Answer 1

Torsten 2021-12-29

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1619370-problems-with-the-application-of-newton-s-method#answer_864625

编辑：Torsten 2021-12-29

在 MATLAB Online 中打开

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

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Aryo Aryanapour 2021-12-30

在 MATLAB Online 中打开

clear 
clc
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here
x0 = 280;
eps = 0.4;
c_s = 5.67*10^(-8);
s1 = 0.250;
s2 = 0.015;
lamda1 = 0.35;
lamda2 = 22.7;
Tw_1 = 1200;
T_l = 10;
Tw_1 = 1200+273.15;
T_l = 10+273.15;
lambda_L = 25.12;
betha = 3.543;
v = 144.0;
L = 7;
g = 9.81;
Pr = 0.7095;
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)  
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);
lambda_L = 25.12;
betha = 3.543;
v = 144.0;
L = 7;
g = 9.81;
Pr = 0.7095;
T_l = 10;
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;
    
    Gr = (g*L.^3*betha*(x(i+1)-T_l))/v;
    Ra = Gr*Pr;
    alpha_k = ((0.825+(0.387*Ra^1/6)/(1+(0.492/Pr)^9/16)^8/27)^2)*lambda_L/L;
    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);
alpha_k
end

Hey Torsten, Can you help me again? With the same code, alpha_k is no longer constant, but changes with temperature. I have to calculate Gr as a function of the temperature and then re-calculate alpha_k with Gr. At the end, the x value or xsol must be calculated using the alpha_k. Unfortunately the alpha_k cannot be calculated for me. Can you see why Thanks so much

Torsten 2021-12-30

编辑：Torsten 2021-12-30

在 MATLAB Online 中打开

function main
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here
x0 = 80;
eps = 0.4;
c_s = 5.67*10^(-8);
s1 = 0.250;
s2 = 0.015;
lamda1 = 0.35;
lamda2 = 22.7;
Tw_1 = 1200;
T_l = 10;
Tw_1 = 1200+273.15;
T_l = 10+273.15;
lambda_L = 25.12;
betha = 3.543;
v = 144.0;
L = 7;
g = 9.81;
Pr = 0.7095;
Ra = @(x)Pr*(g*L^3*betha*(x-T_l))/v;
alpha_k = @(x)((0.825+(0.387*Ra(x)^1/6)/(1+(0.492/Pr)^9/16)^8/27)^2)*lambda_L/L;
func = @(x) eps * c_s * x^4 + (alpha_k(x) + 1/(s1/lamda1+s2/lamda2)) * x - ((1/(s1/lamda1+s2/lamda2)) * Tw_1 + alpha_k(x) * T_l);
xsol = newton_raphson(func,x0)  
end
function xsol = newton_raphson(func,x0)
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here
x(1) = x0;
maxiter = 200;
tol = 10^(-5);
for i = 1:maxiter
    
    diff_xi = (func(x(i)+1e-6)-func(x(i)))*1e6;
    if  diff_xi < tol
        fprintf('Pitfall hast occured a better initial guess\n');
        return;
        
    end
    
    x(i+1) = x(i) - func(x(i))/diff_xi;
    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

If you are willing to spend the effort to get an analytical expression for the derivative of "func" with respect to x, you can do so. But I don't have the time.

And it works with a finite-difference approximation for the derivative, as you can see.

Aryo Aryanapour 2022-1-15

在 MATLAB Online 中打开

function main
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here
x0 = 80;
eps = 0.4;
c_s = 5.67*10^(-8);
s1 = 0.250;
s2 = 0.015;
lamda1 = 0.35;
lamda2 = 22.7;
Tw_1 = 1200;
T_l = 10;
Tw_1 = 1200+273.15;
T_l = 10+273.15;
lambda_L = 25.12;
betha = 3.543;
v = 144.0;
L = 7;
g = 9.81;
Pr = 0.7095;
Ra = @(x)Pr*(g*L^3*betha*(x-T_l))/v;
alpha_k = @(x)((0.825+(0.387*Ra(x)^1/6)/(1+(0.492/Pr)^9/16)^8/27)^2)*lambda_L/L;
func = @(x) eps * c_s * x^4 + (alpha_k(x) + 1/(s1/lamda1+s2/lamda2)) * x - ((1/(s1/lamda1+s2/lamda2)) * Tw_1 + alpha_k(x) * T_l);
xsol = newton_raphson(func,x0)  
end
function xsol = newton_raphson(func,x0)
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here
x(1) = x0;
maxiter = 200;
tol = 10^(-5);
for i = 1:maxiter
    
    diff_xi = (func(x(i)+1e-6)-func(x(i)))*1e6;
    if  diff_xi < tol
        fprintf('Pitfall hast occured a better initial guess\n');
        return;
        
    end
    
    x(i+1) = x(i) - func(x(i))/diff_xi;
    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

Hello Torsten

sorry if i have to annoy you again. can you help me again I'm calculating the heat flow in watts here, but matlab won't give me the value. I also need the value for Alpha end printed out so I know how big alpha is. Can you help me to determine these two values. Oh god I need a good read on Matlab, especially functions. There are many youtube videos. But somehow they couldn't help me. I thank you in advance.

by the way this is not new code. This is the one from the previous comment. I just re-pasted it here.

Torsten 2022-1-16

编辑：Torsten 2022-1-16

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 2022-1-16

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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