Juan Felipe Becerra Casas Código: 3021121892

Taller #2 Métodos Númericos1) usando el método de la bisección resolver con un errror menor al 1%A) F(x) = e^x - 3x = 0 [0,1]

X1 X2 Xr F(x1) F(X2) F(Xr) F(X1) F(Xr)0 1 0.5 1 -0.2817181715 0.14872127 0.14872127

0.5 1 0.75 0.14872127 -0.2817181715 -0.13299998 -0.019779930.5 0.75 0.625 0.14872127 -0.1329999834 -0.00675404 -0.001004470.5 0.625 0.5625 0.14872127 -0.0067540426 0.06755466 0.01004681

0.5625 0.625 0.59375 0.06755466 -0.0067540426 0.02951607 0.001993950.59375 0.625 0.609375 0.02951607 -0.0067540426 0.01115649 0.0003293

0.609375 0.625 0.6171875 0.01115649 -0.0067540426 0.00214465 2.39268E-050.6171875 0.625 0.62109375 0.00214465 -0.0067540426 -0.00231889 -4.9732E-06

B) tg x = x [4,4,5]X1 X2 Xr F(x1) F(X2) F(Xr) F(X1) F(Xr)4 4.5 4.25 1.15782128 4.63733205455 2.00630903 2.32294729

4.25 4.25 4.25 2.00630903 2.00630902786 2.00630903 4.02527592

C) x^ 3 + x = 6X1 X2 Xr F(x1) F(X2) F(Xr) F(X1) F(Xr)

1.55 1.75 1.65 -0.726125 1.109375 0.142125 -0.103200521.55 1.65 1.6 -0.726125 0.142125 -0.304 0.2207421.6 1.65 1.625 -0.304 0.142125 -0.08398438 0.02553125

1.625 1.65 1.6375 -0.08398438 0.142125 0.02830273 -0.00237699

D) x^3 -17 =0X1 X2 Xr F(x1) F(X2) F(Xr) F(X1) F(Xr)0 1 0.5 -17 -16 -16.875 272

0.5 1 0.75 -16.875 -16 -16.578125 2700.75 1 0.875 -16.578125 -16 -16.3300781 265.25

0.875 1 0.9375 -16.3300781 -16 -16.1760254 261.281250.9375 1 0.96875 -16.1760254 -16 -16.0908508 258.816406

0.96875 1 0.984375 -16.0908508 -16 -16.0461464 257.4536130.984375 1 0.9921875 -16.0461464 -16 -16.0232549 256.738342

2) Usando el método Newton resuelva con un error menor al 1 %A) 3x + sen x - e^x = 0

3) Usando el metodo de punto fijo resolver con un error al 1% A) Cos x - x =0 Cos(x) = x

xr F(xr) Error Acción1 0.540302306 no aplica no aplica

0.54030231 0.857553216 85.0815718 CONTINUAR0.85755322 0.65428979 36.9948948 CONTINUAR0.65428979 0.793480359 31.0662688 CONTINUAR0.79348036 0.701368774 17.5417787 CONTINUAR0.70136877 0.763959683 13.1331175 CONTINUAR0.76395968 0.722102425 8.19295974 CONTINUAR0.72210243 0.750417762 5.79658182 CONTINUAR0.75041776 0.731404042 3.77327646 CONTINUAR0.73140404 0.744237355 2.59961912 CONTINUAR0.74423735 0.73560474 1.72435748 CONTINUAR0.73560474 0.741425087 1.17353981 CONTINUAR0.74142509 0.737506891 0.78502148 PARAR

B) x^2 - 2x -3 = 0 x^2 -3 = 2x (x^2 -3)/-2 = xxr F(xr) Error Acción

1 1 no aplica no aplica1 1 0 PARAR

C) e^x - 3x^2 = 0 e^x = 3x^2 e^x = 3x^2 ln(e^x)=3x^2 x=ln(3x^2)

xr F(xr) Error Acción1 1.098612289 no aplica no aplica

1.09861229 1.286707944 8.97607734 CONTINUAR1.28670794 1.602786239 14.6183643 CONTINUAR1.60278624 2.042099317 19.7205521 CONTINUAR2.04209932 2.526569 21.5128165 CONTINUAR

2.526569 2.9523368 19.1750031 CONTINUAR2.9523368 3.263806274 14.4213831 CONTINUAR

3.26380627 3.464400454 9.54313606 CONTINUAR3.46440045 3.583691466 5.79015571 CONTINUAR3.58369147 3.651399098 3.3287188 CONTINUAR

3.6513991 3.688833106 1.85429285 CONTINUAR3.68883311 3.709232642 1.01479267 CONTINUAR3.70923264 3.720262329 0.54996647 PARAR

D) sen ( x) -x = 0 Sen ( x) = xxr F(xr) Error Acción

1 0.841470985 no aplica no aplica0.84147098 0.793973511 18.8395106 CONTINUAR0.79397351 0.777733404 5.98224921 CONTINUAR

0.7777334 0.771942954 2.08813287 CONTINUAR0.77194295 0.769847926 0.75011368 PARAR

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Taller #2 Métodos Númericos1) usando el método de la bisección resolver con un errror menor al 1%

Error DecisiónNo aplica No aplica

33.3333333 continuar20 continuar

11.1111111 continuar5.26315789 continuar2.56410256 continuar1.26582278 continuar0.62893082 detener

Error DecisiónNo aplica No aplica

0 detener

Error Decisión

No aplica No aplica3.125 continuar

1.53846154 continuar0.76335878 detener

Error Decisión

No aplica No aplica33.3333333 continuar14.2857143 continuar6.66666667 continuar3.22580645 continuar1.58730159 continuar0.78740157 detener