MATLAB Answers

Newton Raphson on Mathlab

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Sunny
Sunny on 3 Nov 2014
Answered: Meysam Mahooti on 5 Dec 2019
I was wondering how to solve this equation in matlab? the abc would be replaced by 742. Any help would be deeply appreciated.

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Answers (7)

Torsten
Torsten on 3 Nov 2014
p=[0.5 1 -10 14.742];
r=roots(p);
Best wishes
Torsten.

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Sunny
Sunny on 5 Nov 2014
Yes I have the root now to solve the equation, the root is -6 how do I now plug this into the equation and solve it?
Matt Tearle
Matt Tearle on 5 Nov 2014
What do you mean? The roots in r already solve the equation. The real root near -6 is the first element of r:
>> fprintf('%10.6f\n',r(1))
-6.083918
You can use the polyval function to evaluate a polynomial at a given value of x. For example:
x = linspace(-8,6);
y = polyval(p,x);
plot(x,y)
grid on
If you evaluate at x = r(1) you'll see that it solves the equation (to within ~10^-13):
polyval(p,r(1))

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Matt Tearle
Matt Tearle on 5 Nov 2014
Torsten showed how to find the roots of a polynomial. More generally, to find the zeros of any function:
f = @(x) 0.5*x.^3 + x.^2 - 10*x + 14.742; % define the function
x0 = -4; % initial guess for the solution
xroot = fzero(f,x0) % solve
(If you're interested in the details, according to the doc, "the [fzero] algorithm, created by T. Dekker, uses a combination of bisection, secant, and inverse quadratic interpolation methods")

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Sunny
Sunny on 5 Nov 2014
So to complete this manually x0=-4?
Matt Tearle
Matt Tearle on 5 Nov 2014
I don't understand what you mean by "complete this manually". The assignment problem says to solve the equation. One way is to do N-R by hand. The other is "repeat using MATLAB". So I guess the question is what "repeat" entails. Repeat the solving of the equation? Or repeat solving the equation with N-R?
We've shown two ways you can solve the equation in MATLAB: roots (for solving polynomial equations) and fzero (for solving general nonlinear equations), but neither of these use N-R.
If you want to implement Newton-Raphson in MATLAB then that's a bigger issue. That requires knowing the basics of MATLAB programming. Given that this is a homework problem, you'll need to show what you've attempted and ask for some specific pointers.

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Sunny
Sunny on 13 Nov 2014
I've done the newton raphson manually on paper and I done it on matlab, Can someone help me compare the two on similarities or the differences?

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Torsten
Torsten on 13 Nov 2014
As Matt already mentioned, neither MATLAB's "fzero" nor MATLAB's "roots" uses N-R.
So it's difficult to compare the two methods.
You should ask your supervisor what he/she means by
"Repeat using MATLAB and compare the two methods.".
Best wishes
Torsten.

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Farhad Sedaghati
Farhad Sedaghati on 3 Aug 2015
The following link is the matlab code to perform Newton-Raphson's Method:

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sabik EL YATIM
sabik EL YATIM on 27 Jul 2019
hello, Recently, a part of the Matlab code I found on the resolution system of nonlinear equations using the method of Newton-Raphson with the Jacobian matrix (I also left it in my comments). However, although he provides me with the basic code, I can not make it work, no matter how hard I try. I spent many hours trying to present function func, but to no avail, I often did not understand how I used it if you can help me
function [x,F, niter] = newtonsys(Ffun ,Jfun,x0 ,tol ,...
nmax, varargin )
% NEWTONSYS cherche un zéro d’un système non linéaire
% [ZERO ,F, NITER ]= NEWTONSYS(FFUN, JFUN,X0 ,TOL , NMAX)
% tente de trouver le vecteur ZERO, racine d’ un
% système non linéaire défini dans FFUN et dont
% la matrice jacobienne est définie dans la
% fonction JFUN. La racine est cherchée autour
% du vecteur X0.
% La variable F renvoie le résidu dans ZERO
% NITER renvoie le nombre d’ itérations nécessaires
% pour calculer ZERO. FFUN et JFUN sont des fonctions
% MATLAB définies dans des M- files.
niter = 0; err = tol + 1; x = x0;
while err >= tol & niter < nmax
J = feval(Jfun ,x , varargin {:});
F = feval(Ffun ,x , varargin {:});
delta = - J\F;
x = x + delta;
err = norm( delta );
niter = niter + 1;
end
F = norm( feval( Ffun,x, varargin {:}));
if (niter == nmax & err > tol)
fprintf ([Pas de convergence dans le nombre ,...
d’’ iterations imparti \n ]);
fprintf ([La valeur retournée a un résidu ,...
relatif de %e\n’],F);
else
fprintf ([La méthode a convergé à l’’ itération,...
%i avec un résidu %e\n’],niter,F);
end
return

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