Dada la funcin f(x) = 100,1x2 -0.5 - COS (2x) Aplique El mtodo de Newton Rapshon . Para determinar la raz de la funcin dada, tomando como aproximacin inicial, Xo =1 y para un error menor a 0,5%. Tome 5 decimales y demuestre los clculos de la primera iteracin.

f(x) = 100,1x2 -0.5 - COS (2x)

derivando f(x)

f(x) = (100,1x2 )- (1/2)- ( COS (2x) ) f(x) = 2(1.25893)x - [ -2sen (2x) ] f(x) = 2.51785 x + 2 sen (2x)

Formula de NEWTHON RAPHSON f(xi+1) = xi - ( f(x) / f(x)

itXif(x)f (x)Xi+1EP

111,175074,336440,72902////////////

20,729020,056583,822870,714222,07221

30,714220,000333,778080,714140,01221

PRIMERA ITERACION

f(1) = 100,1x2 -0.5 - COS (2x) = (1.25893)(1)2 - 0.5 - cos (2*1) = 1.17507f (1) = 2.51785 x + 2 sen (2x)= 2.51785(1) + 2sen(2*1) = 4.33644f1 (X i+1) = xi - [ f(x) / f(x) ] = 1 [ [ 1.17507 / 4.33644 ] = 0.72902

SEGUNDA ITERACIONf2 (x) =100,1x2-0.5-COS(2x)f (0.72902) =(1.25893)( 0.72902)2 - 0.5 - cos [ 2(0.72902) ] = 0,05658f (0.72902) = 2.51785 x + 2 sen (2x)=(2.51785)(0.72902) + 2 sen [ 2(0.72902) ]= 3,82287f2 (X i+1) = xi - [ f(x) / f(x) ] =0.72902 [ 0,05658 / 3,82287 ] = 0.71422

Ep= | Xr nuevo Xr anterior | * 100 % = | 0.71422 (0.72902) | * 100 = 2.07221 % . | Xr nuevo | | 0.71422 |

TERCERA ITERACION

f3 (x) =100,1x2-0.5-COS(2x)f (0.71422) = (1.25893)( 0.71422 )2- 0.5 - cos [ 2(0.71422) ] = 0,00033f (x) = 2.51785 x - 2 sen (2x)=f (0.71422) = (2.51785)( 0.71422)-2 sen [2(0.71422)])= 3,77808f (X i+1) = xi - [ f(x) / f(x) ] = 0.71422 [ 0,00033 / 3,77808 ] = 0,71414

Ep= | Xr nuevo Xr anterior | * 100 % = | 0,71414 - 0.71422 | x 100% = 0,01221 % | Xr nuevo | | 0,71414 |