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Why is my newton method giving me a huge answer? What am i doing wrong?pls help

Asked by Erik Ramirez on 5 Sep 2019
Latest activity Edited by Rena Berman on 19 Sep 2019
x= sym('x');
p = 0.6;
F = x^5+4*x^2-1;
F_prime = diff(F);
N=0;
q = p - subs(F,x,p)/subs(F_prime,x,p);
while(N < 100 && not(abs(p-q) <= .00001 ))
N=N+1;
p = q;
q = p - ((subs(F,x,p))./(subs(F_prime,x,p)));
end
if(N == 100)
print("Too many iterations");
else
display(q)
end

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2 Answers

Answer by John D'Errico
on 5 Sep 2019
 Accepted Answer

When you have a problem like this, plot the function. In fact, plot the function anyway!
Next, it is a bad idea to use variables like p and q, then swaping them back and forth. Use variable names that make sense!!!!!
Next, don't keep things symbolic. That causes problems, because you don't know how to work with symbolic variables that well. Use functions instead.
I'll start with a symbolic function, then differentiate it in symbolic form. But then alllow matlabFunction to convert it to a function handle.
Finally, I cleaned up the logic in your code.
x = sym('x');
xcurrent = 0.5; % start point
xold = inf; % the first time through, you need to pass the while condition
F = x^3+4*x^2-10;
F_fun = matlabFunction(F);
fplot(F_fun)
hold on
plot(xcurrent,F_fun(xcurrent),'o')
F_prime = matlabFunction(diff(F));
N=0;
tolerance = 1.e-6;
while (N < 30) && (abs(xcurrent-xold) > tolerance )
N=N+1;
xold = xcurrent;
xcurrent = xold - F_fun(xold)./F_prime(xold);
plot(xcurrent,F_fun(xcurrent),'o')
end
if(N == 30)
print("Too many iterations");
else
display(xcurrent)
end
That should get the point across, that there is no need to use symbolic tools as heavily as you are.

  1 Comment

this looks way better!! could i show you my secant method im having trouble with it

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Answer by Walter Roberson
on 5 Sep 2019

Your code is not giving you a huge answer: it is giving you a rational answer. You will want to use double() to convert it to floating point.

  3 Comments

would i use double on just when i printed out?

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