Plotting a graphical converge inside a While - Newton Raphson numerical method

Hi, I'm trying to get a command inside a while so as the Newton Rapshon code shows a graphical converge. If the values of the aproximations are connected between a line, it shows something like a spider web. Can anyone help me out, please?

Here is my code:

clear 
clc
syms x
f = input('Introducir la función: '); %function
p0 = input ('Introducir valor semilla: '); %first aproximation
TOL = input ('Introducir la tolerancia de error: '); %error
fplot (f)
grid on
hold on
derivada = diff(f);
derivada = inline (derivada);
f = inline (f);
eabs = 100;
i =0;

while eabs>TOL
    p = p0 - (f(p0))/(derivada(p0));
    eabs = abs(((p-p0)/p)*100);
    p0 = p;
    i = i+1;
end
fprintf('\n Valor= %8.3f  ',p0)
plot (p)
fplot (0)
hold off

[Walter Roberson]

Is it strictly necessary that the lines be drawn during the execution of the while loop, or could it be drawn after the while finishes?

[Luis Francisco Sanchez]

It's not necessary. I just need the spider web to be drawn over the original graph.

[Walter Roberson]

allp = p0;
allfp = [];
while eabs>TOL
    fp = f(p0);
    p = p0 - fp/(derivada(p0));
    eabs = abs(((p-p0)/p)*100);
    p0 = p;
    i = i+1;
    allp(end+1) = p;
    allfp(end+1) = fp;
end
plot(allp(1:end-1), allfp);

[Luis Francisco Sanchez]

Thank you very much, Walter. Works perfectly

[Luis Francisco Sanchez]

I have another question. I'm solving an exercise where I need to find the f(x) = 0 of the funciton:

0 = -x +cos(x)+tan(x), since is a periodical graph with multiple y=0, how can I make the code and graph look into a specific range of x? I got {-pi/2,pi/2} as the range of the exercise above

[Walter Roberson]

p = asin(sin(p));

[David Wilson]

Another solution assuming you are looking for an iterative numerial stragegy that stays strictly within the given bounds is to use bisection. Sure it is slow, but it will stay within the -pi/2 to +pi/2 bound.

F = @(x)  -x +cos(x)+tan(x); % function of interest 

eps = 0.0; tol = 1e-6;  % assume some stopping tolerance, set eps to 0.1 if nervous
x = [-pi/2+eps, pi/2-eps]; 

Fx = F(x); % should be vectorised
assert(prod(sign(Fx)) < 0,'x does not bracket root')

while abs(diff(x)) > tol % Now start bisection routine 
  m = mean(x); % mid point, m = (a + b)/2
  fm = F(m); % find f(m)
  if Fx(1)*fm > 0
    x(1)=m; Fx(1)=fm; % discard left
  else
    x(2)=m; % discard right
  end % if
end % while

Either value of x is a reasonable solution.