profe que pena la demora ese es mi trabajo

1. METODO DE BISECCION
function y=f(x)
y=-12.4+10*(0.5*%pi-asin(x/1)-x*(1-x**2)**0.5);
endfunction
function xw=biseccion(xyi,xzi,aproxi)
i=1;
eabs(1)=100;
if f(xyi)*f(xzi) < 0
xy(1)=xyi;
xz(1)=xzi;
xw (1)=(xy(1)+xz(1))/2;
printf('It.tt Xytt Xztt xw tt f(xw)t Error n');
printf('%2d t %11.7f t %11.7f t %11.7f t %11.7f
n',i,xy(i),xz(i), xw(i),f(xw(i)));
while abs(eabs(i)) >= aproxi
if f(xy(i))*f(xw(i))< 0
xy(i+1)=xy(i);
xz(i+1)=xw(i);
end
if f(xy(i))*f(xw(i))> 0
xy(i+1)= xw(i);
xz(i+1)=xz(i);
end
xw(i+1)=(xy(i+1)+xz(i+1))/2;
eabs(i+1)=abs((xw(i+1)-xw(i))/(xw(i+1)));
printf('%2d t %11.7f t %11.7f t %11.7f t %11.7f t %7.6f
n',i+1,xy(i+1),xz(i+1),xw(i+1),f(xw(i+1)),ea(i+1));
i=i+1;
end
else
printf('No existe una raíz en ese intervalo');
end
endfunction

2. METODO DE N-R
function y=g(x)
y=300-80.425*x+201.0625*(1-2.718281828**(-(0.1)*x/0.25));
endfunction
function pn=puntofijo(pn0,aproxi)
i=1;
eabs(1)=100;
pn(1)=pn0;
while abs(eabs(i))>=aproxi,
pn(i+1) = g(pn(i));
eabs(i+1) = abs((pn(i+1)-pn(i))/pn(i+1));
i=i+1;
end
printf(' i t X(i) Error aprox (i) n');
for j=1:i;
printf('%2d t %11.7f t %7.3f n',j-1,pn(j),eabs(j));
end
endfunction

3. PUNTO FIJO
function y=f(x)
y=2*x**3+x-1;
endfunction
function y=df(x)
y=6*x**2+1;
endfunction
function pn=newtonraphson(pn0,aproxi);
i=1;
eabs(1)=100;
pn(1)=pn0;
while abs(eabs(i))>=tol;
pn(i+1)=pn(i)-f(pn(i))/df(pn(i));
eabs(i+1)=abs((pn(i+1)- pn(i))/pn(i+1));
i=i+1;
end
printf(' i t pn(i) Error aprox (i) n');
for j=1:i;
printf('%2d t %11.7f t %7.6f n',j-1,x(j),eabs(j));
end
endfunction
METODO DE NEWTON
function y=f(x)
y=4*cos(x)-exp(x);
endfunction
function y=df(x)
y=-4*sin(x)-exp(x);
endfunction
function pn=newtonraphson(pn0,aproxi);
i=1;
error(1)=100;
pn(1)=pn0;
while abs(error(i))>=aproxi;
pn(i+1)=pn(i)-f(pn(i))/df(pn(i));
aproxi(i+1)=abs((pn(i+1)-pn(i))/pn(i+1));
i=i+1;
end
printf(' i t pn(i) Error aproxi (i) n');
for j=1:i;
printf('%2d t %11.7f t %7.6f n',j-1,pn(j),eabs(j));
end
endfunction

4. METODO DE LA SECANTE
function y=f(x)
y=4*cos(x)-exp(x);
endfunction
function pn = secante(pn0,pn1,aproxi)
j=2;
i=1;
pn(1)=pn0;
pn(2)=pn1;
eabs(i)=100;
while abs(eabs(i))>=aproxi
pn(j+1)=(pn(j-1)*f(pn(j))-pn(j)*f(pn(j-1)))/(f(pn(j))-f(pn(j-1)));
eabs(i+1)=abs((pn(j+1)-pn(j))/pn(j+1));
j=j+1;
i=i+1;
end
printf(' i tt pn(i) t Error aproxi (i) n');
printf('%2d t %11.7f t n',0,pn(1));
for k=2:j;
printf('%2d t %11.7f t %7.3f n',k-1,pn(k),eabs(k-1));
end
endfunction

5. METODO DE LA SECANTE
function y=f(x)
y=x**2-6;
endfunction
function pn = secante(pn0,pn1,aproxi)
j=2;
i=1;
pn(1)=pn0;
pn(2)=pn1;
eabs(i)=100;
while abs(eabs(i))>=aproxi
pn(j+1)=(pn(j-1)*f(pn(j))-pn(j)*f(pn(j-1)))/(f(pn(j))-f(pn(j-1)));
eabs(i+1)=abs((pn(j+1)-pn(j))/pn(j+1));
j=j+1;
i=i+1;
end
printf(' i tt pn(i) t Error aproxi (i) n');
printf('%2d t %11.7f t n',0,pn(1));
for k=2:j;
printf('%2d t %11.7f t %7.3f n',k-1,pn(k),eabs(k-1));
end
endfunction

6. METODO DE LA FALSA POSICION
function y=f(x)
y=x**2-6;
endfunction
function xw=reglafalsa(xyi,xzi, aproxi)
i=1;
eabs(1)=100;
if f(xyi)*f(xzi) < 0
xy(1)=xyi;
xz(1)=xzi;
xw(1)=xy(1)-f(xy(1))*(xz(1)-xy(1))/(f(xz(1))-f(xy(1)));
printf('It. Xy Xz Xw
f(Xw) Error aprox %n');
printf('%2d t %11.7f t %11.7f t %11.7ft %11.7f
n',i,xy(i),xz(i),xw(i),f(xw(i)));
while abs(ea(i))>=aproxi,
if f(xy(i))*f(xw(i))< 0
xy(i+1)=xy(i);
xz(i+1)=xz(i);
end
if f(xy(i))*f(xw(i))> 0
xy(1)=xw(i);
xz(1)=xz(i);
end
xw(i+1)=xy(i+1)-f(xy(i+1))*(xz(i+1)-xy(i+1))/(f(xz(i+1))-
f(xy(i+1)));
eabs(i+1)=abs((xw(i+1)-xw(i))/(xw(i+1)));
printf('%2d t %11.7f t %11.7f t %11.7f t %11.7ft %7.3f n',
i+1,xy(i+1),xz(i+1),xw(i+1),f(xw(i+1)),eabs(i+1));
i=i+1;
end
else
printf('No existe una raíz en ese intervalo');
end
endfunction

7. c. Como se puede observar, por ambos métodos hay una aproximación a la raíz de 6 ( ), pero por
simple observación notamos claramente que el que más aproximación presenta es el realizado por el
método de la secante.