newton

1)
a) f=log(x)-x+2;

newton(2,'f','f1',0.001,10)
=====================================iter x error f(x)
===================================== 02.0000000000 0.0000000000
0.6931471806
1 3.3862943611 1.3862943611
0.16655814642 3.1499383938 0.2363559673
0.0025554987
3 3.1461942570 0.0037441368
0.0000007070
43.1461932206 0.0000010364
0.0000000000 =====================================La raiz aproximada es: 3.146193220620662 El valor de f(p) es:0.000000000000054
b) f=x-(3*cos(x));
>> newton(1,'f','f1',0.001,10)
=====================================iter x error f(x)===================================== 0 1.0000000000 0.0000000000 -0.6209069176
1 1.1761731459 0.1761731459 0.0227920696
2 1.1701265811 0.0060465648 0.0000211863
3 1.1701209500 0.0000056311 0.0000000000=====================================La raiz aproximada es: 1.170120950007557 El valor de f(p) es: 0.000000000018552
c) f=x*(exp(x))-2

>> newton(1,'f','f1',0.001,10)=====================================iter x error f(x)
===================================== 0 1.0000000000 0.0000000000 0.7182818285
1 0.8678794412 0.13212055880.0671626666
2 0.8527833734 0.0150960678 0.0007730906
3 0.8526055264 0.0001778470 0.0000001058 =====================================La raiz aproximada es: 0.852605526368922
Elvalor de f(p) es: 0.000000105841616
d) f=(x^3)-5*x+1;

>> newton(2,'f','f1',0.001,10)
=====================================
iter x error f(x)...