Decreasing varible into a loop

12 vues (au cours des 30 derniers jours)
Miguel Angel Villegas Hernandez
Commenté : Steven Lord le 20 Juil 2021
Hello
I'm new programming and I have to sove this.
I Have to decrease one varible using i=0 and increasing by 1, i=i+1, but at same time I have another expression that depends of "i" and i need to drecreasing that varible, let you show:
i = 0;
x = b;
Fx = Fb;
Ek = abs(xe-x);
EA = Ek/Ek+1; %Here I've to solbe it
while 1
DeltaX = -Fx/(Fb-Fa)*(b-a);
x = x+DeltaX;
Fx = f(x);
Ek = abs(xe-x);; %Here I've to solbe it
disp ([i a b x Fx DeltaX Ek ]);
if(abs(DeltaX)<Tolerancia && abs(Fx)<Tolerancia)||i >=n
break;
end
a = b;
Fa = Fb;
b = x;
Fb = Fx;
i = i+1;
end
the equation is:
EA = Ek / E_k-1 it must be E_0/E_1 i =0 at the same time k = 2
Thanks
  4 commentaires
Miguel Angel Villegas Hernandez
This is my code
f=@(x)x^3-3*x-2
function [Raiz,Iter,Error] = Secante(f,a,b,Tolerancia,IterMax);
xe = -1; %scalar constant
x = a;
Fa = f(a);
x = b;
Fb = f(b);
if abs(Fa) < abs(Fb)
t = a;
a = b;
b = t;
t = Fa;
Fa = Fb;
Fb = t;
end
printf('Calculo de la raiz por el metodo de la secante\n i a b x Fx DeltaX Ek \n');
i = 0;
x = b;
Fx = Fb;
Ek = abs(xe-x); % Ek=abs(-1- x(i))
%here I want to introduce EA=Ek(i)/Ek(k-1),where "k" is another kund of counter linked with "i" counter
% it must be i=0 at same time k=2 to get Ek(0) / Ek(1), Ek(1) / Ek(2)
while 1
DeltaX = -Fx/(Fb-Fa)*(b-a);
x = x+DeltaX;
Fx = f(x);
Ek = abs(xe-x);
disp ([i a b x Fx DeltaX Ek ]);
if(abs(DeltaX)<Tolerancia && abs(Fx)<Tolerancia)||i >=IterMax
break;
end
a = b;
Fa = Fb;
b = x;
Fb = Fx;
i = i+1;
end
Raiz = x;
if abs(DeltaX)<Tolerancia && abs(Fx)<Tolerancia
Error = 0;
else
Error = 1;
end
end
I hope be explicit.
Thanks
Steven Lord
Steven Lord le 20 Juil 2021
There's no element 0 in an array in MATLAB, so E(0) will error if E is a regular numeric array.

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