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# Soluciones

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### Soluciones

1. 1. Primer punto biseccion -->biseccion(1,0,0.01) It. Xa Xb Xr f(Xr) Error 1 1.0000000 0.0000000 0.5000000 -6.2581515 2 0.5000000 0.0000000 0.2500000 -1.6394539 1.000000 3 0.2500000 0.0000000 0.1250000 0.8144890 1.000000 4 0.2500000 0.1250000 0.1875000 -0.4199467 0.333333 5 0.1875000 0.1250000 0.1562500 0.1957259 0.200000 6 0.1875000 0.1562500 0.1718750 -0.1125364 0.090909 7 0.1718750 0.1562500 0.1640625 0.0414932 0.047619 8 0.1718750 0.1640625 0.1679688 -0.0355476 0.023256 9 0.1679688 0.1640625 0.1660156 0.0029664 0.011765 10 0.1679688 0.1660156 0.1669922 -0.0162922 0.005848 ans = 0.5 0.25 0.125 0.1875 0.15625 0.171875 0.1640625 0.1679688 0.1660156 0.1669922
2. 2. Segundo punto >newtonraphson(1,0.0001); i X(i) Error aprox (i) 0 1.0000000 100.000000 1 0.7142857 0.400000 2 0.6051687 0.180308 3 0.5900220 0.025671 4 0.5897546 0.000453 5 0.5897545 0.000000 Tercer punto -->puntofijo(0,0.01); i X(i) Error aprox (i) 0 0.0000000 100.000 1 300.0000000 1.000 2 -23626.4375000 1.013 3 -1.#INF000 1.#QO Cuarto punto parte a: -->newtonraphson(1,0.0001); i X(i) Error aprox (i) 0 1.0000000 100.000000
3. 3. 1 0.9084390 0.100789 2 0.9047941 0.004028 3 0.9047882 0.000006 Cuarto punto parte b -->secante(%pi/4,%pi/2,0.0001) i x(i) Error aprox (i) 0 0.7853982 1 1.5707963 100.000 2 0.8770026 0.791 3 0.8985446 0.024 4 0.9048658 0.007 5 0.9047880 0.000 ans = 0.7853982 1.5707963 0.8770026 0.8985446 0.9048658 0.904788 Quinto punto parte a: -->secante(3,2,0.0001) i x(i) Error aprox (i)
4. 4. 0 3.0000000 1 2.0000000 100.000 2 2.4000000 0.167 3 2.4545455 0.022 4 2.4494382 0.002 5 2.4494897 0.000 ans = 3. 2. 2.4 2.4545455 2.4494382 2.4494897 Quinto punto parte b: -->reglafalsa(3,2,0.0001) It. Xa Xb Xr f(Xr) Error aprox % 1 3.0000000 2.0000000 2.4000000 -0.2400000 2 3.0000000 2.4000000 2.4444444 -0.0246914 0.018 3 3.0000000 2.4444444 2.4489796 -0.0024990 0.002 4 3.0000000 2.4489796 2.4494382 -0.0002525 0.000 5 3.0000000 2.4494382 2.4494845 -0.0000255 0.000 ans = 2.4
5. 5. 2.4444444 2.4489796 2.4494382 2.4494845