programacion

Para la función: function ans = eficiencia(lamba)
ans= ((tan(lamba)-0.08*(tan(lamba))^(2))/(tan(lamba)+0.08)) -0.8


raiz=secante(‘eficiecia’, 0, 1)
iter =
0
x1 =
0
x2 =
1ans =
-0.8000
y1 =
-0.8000
ans =
0.0326
y2 =
0.0326
convergio =
0
iter =
1
x3 =
0.9608
ans =
0.0386
y3 =
0.0386
x1 =
1
y1 =
0.0326
x2=
0.9608
y2 =
0.0386
iter =
2
x3 =
1.2129
ans =
-0.0368
y3 =
-0.0368
x1 =
0.9608
y1 =
0.0386
x2 =
1.2129
y2 =
-0.0368
iter =
3
x3 =1.0900
ans =
0.0127
y3 =
0.0127
x1 =
1.2129
y1 =
-0.0368
x2 =
1.0900
y2 =
0.0127
iter =
4
x3 =
1.1216
ans =
0.0031
y3 =
0.0031
x1 =1.0900
y1 =
0.0127
x2 =
1.1216
y2 =
0.0031
iter =
5
x3 =
1.1317
ans =
-3.6827e-004
y3 =
-3.6827e-004
x1 =
1.1216
y1 =
0.0031
x2 =
1.1317
y2 =-3.6827e-004
iter =
6
x3 =
1.1306
ans =
9.2459e-006
y3 =
9.2459e-006
convergio =
1
raiz =
1.1306
raiz =
1.1306

>> raiz=Newton_Rap('eficiencia', 0, 1)Aplicacion del METODO DE NEWTON-RAPHSON

iter xsup xcalc (xcalc-xsup)/xsup
----- ------------ ------------ -----------------

ans =

-0.7877


ans=

-0.8000


ans =

-0.8000

1 0 0.0648052 Inf

ans =

-0.3507


ans =

-0.3544


ans =

-0.3544

20.0648052 0.159465 1.46069

ans =

-0.1394


ans =

-0.1408


ans =

-0.1408

3 0.159465 0.264824 0.660697

ans =

-0.0439ans =

-0.0445


ans =

-0.0445

4 0.264824 0.337503 0.274445

ans =

-0.0081


ans =

-0.0085


ans =

-0.0085

5...