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Refuerzo
 

Refuerzo

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    Refuerzo Refuerzo Document Transcript

    • EJERCICIO 2 METODO DE SECANTE Hi-1 Hi F(Hi) F(Hi-1) 1 0.08 1 -4.1803 3.762612 2 1 0.515810322 1.562394045 -4.1803 3 0.51581032 0.647542061 0.317424845 1.56239404 4 0.64754206 0.681129178 -0.039731616 0.31742484 5 0.68112918 0.677392803 0.000780059 -0.03973162 6 0.6773928 0.677464748 1.83596E-06 0.00078006 0.67746475 0.677464917 -8.52838E-11 1.836E-06 7 0.67746492 0.677464917 0 -8.5284E-11 EJERCICIO 3 METODO PUNTO FIJO G1(x) N° ITERACIONES X F(x)=-x^3-4*x^2+10 G(x)=-x^3-4*x^2+10+x error 1 1 5 6 83.3333333 2 6 -350 -344 101.744186 G1(x) N° ITERACIONES X G(x)=(10/4+x)^1/2 error 1 1 1.414213562 29.28932188 2 1.41421356 1.359040217 4.059728641 3 1.35904022 1.36601822 0.510827894 4 1.36601822 1.365129741 0.065083782 5 1.36512974 1.365242771 0.008279089 6 1.36524277 1.36522839 0.001053368 7 1.36522839 1.36523022 0.000134019 8 1.36523022 1.365229987 1.70512E-05 9 1.36522999 1.365230017 2.16941E-06 10 1.36523002 1.365230013 2.76012E-07 11 1.36523001 1.365230013 3.51169E-08 12 1.36523001 1.365230013 4.46791E-09 13 1.36523001 1.365230013 5.68452E-10 14 1.36523001 1.365230013 7.23272E-11 15 1.36523001 1.365230013 9.20557E-12 16 1.36523001 1.365230013 1.17103E-12 17 1.36523001 1.365230013 1.46378E-13 18 1.36523001 1.365230013 1.62643E-14 19 1.36523001 1.365230013 0
    • Hi+1 ERROR x F(x) 20 0.51581032 93.8697146 0 3.8197 0.64754206 20.3433485 1 -4.1803 0 0.68112918 4.9310936 2 -24.1803 0 1 0.6773928 0.55158173 3 -50.1803 -20 0.67746475 0.01061966 4 -76.1803 -40 0.67746492 2.5054E-05 5 -96.1803 0.67746492 1.1637E-09 6 -104.1803 -60 0.67746492 0 7 -94.1803 -80 -100 -120
    • TaNqUe 2 3 4 5 6 7 8 F(x)
    • CALCULATION OF THE ROOT we first make the graph of the function to get an inkling of what the value of the root and so allocate an appropriate x f(X) 0 1 GRAPH OF TH 0.1 0.71873075 1.2 0.2 0.47032005 1 0.3 0.24881164 0.8 0.4 0.04932896 0.6 0.5 -0.13212056 0.4 0.6 -0.29880579 0.2 0.7 -0.45340304 0 0.8 -0.59810348 -0.2 0 0.2 0.9 -0.73470111 -0.4 -0.6 -0.8 -1 we now make the evaluation of each of the methods BISECTION METHOD N° ITERATION Xi Xs Xr=Xi+Xs/2 F(Xi) F(Xr) 0 0 2 1 1 -0.86466472 1 0 1 0.5 1 -0.13212056 2 0 0.5 0.25 1 0.35653066 3 0.25 0.5 0.375 0.35653066 0.09736655 4 0.375 0.5 0.4375 0.09736655 -0.02063798 5 0.375 0.4375 0.40625 0.09736655 0.03749731 6 0.40625 0.4375 0.421875 0.03749731 0.00821964 7 0.421875 0.4375 0.4296875 0.00821964 -0.00626086 8 0.421875 0.4296875 0.42578125 0.00821964 0.00096637 9 0.42578125 0.4296875 0.427734375 0.00096637 -0.00265049 10 0.42578125 0.42773438 0.426757813 0.00096637 -0.00084287 11 0.42578125 0.42675781 0.426269531 0.00096637 6.1544E-05 FALSE POSITION METHOD N° ITERACION Xi Xs Xr=Xs-F(Xs)(Xi-Xs)/F(Xi)-F(Xs) F(Xi) F(Xs) 0 0 2 0.67076181 1 -1.98168436 1 0 0.67076181 0.47594889 1 -0.40931479 2 0 0.47594889 0.436673946 1 -0.08994112 3 0 0.43667395 0.428480265 1 -0.01912266
    • 4 0 0.42848027 0.426760405 1 -0.00403004 5 0 0.4267604 0.426398958 1 -0.00084767 6 0 0.42639896 0.426322976 1 -0.00017823 SECANTE METHOD Xi-1 xi F(xi) F(xi-1) xi+1 0 0 2 -1.981684361 1 0.67076181 1 2 0.67076181 -0.409314792 -1.98168436 0.32473829 2 0.67076181 0.32473829 0.197580806 -0.40931479 0.43738964 3 0.32473829 0.43738964 -0.020435594 0.19758081 0.42683035 4 0.43738964 0.42683035 -0.0009772 -0.02043559 0.42630007 NEWTON METHOD xi F(x) F´(x) ERROR 0 1 -0.86466472 -1.270670566 1 0.31952094 0.20827694 -2.055595758 212.968537 2 0.42084287 0.01014051 -1.861966764 24.0759539 3 0.426289 2.5474E-05 -1.852628949 1.27756709 4 0.42630275 1.612E-10 -1.852605502 0.0032254 5 0.42630275 0 -1.852605502 2.0411E-08 FIXED POINT METHOD G1(x) N° ITERACIONESX F(x) G(x) 1 0 1 1 2 1 -0.86466472 0.135335283 3 0.13533528 0.62753249 0.762867769 4 0.76286777 -0.54540672 0.217461047 5 0.21746105 0.42985405 0.647315095 6 0.64731509 -0.37331592 0.273999173 7 0.27399917 0.30410665 0.57810582 8 0.57810582 -0.26342979 0.314676031 9 0.31467603 0.21826097 0.532936999 10 0.532937 -0.1885103 0.344426695 11 0.3444267 0.15772482 0.502151511 12 0.50215151 -0.13585166 0.366299849 13 0.36629985 0.11435795 0.480657799 14 0.4806578 -0.09826832 0.382389484 15 0.38238948 0.08304731 0.465436796 16 0.4654368 -0.07122761 0.394209182 17 0.39420918 0.060354 0.454563181 18 0.45456318 -0.05168714 0.402876038 19 0.40287604 0.04387577 0.446751809
    • 20 0.44675181 -0.03753232 0.40921949 21 0.40921949 0.03190022 0.441119714 22 0.44111971 -0.02726464 0.413855074 23 0.41385507 0.02319384 0.437048918
    • ROOT e root and so allocate an appropriate interval to assess GRAPH OF THE FUNCTION f(X) 0.4 0.6 0.8 1 F(Xi)*F(Xr) Ea -0.86466472 -0.13212056 100 0.35653066 100 0.03471416 33.3333333 -0.00200945 14.2857143 0.00365098 7.69230769 0.00030821 3.7037037 -5.1462E-05 1.81818182 7.9432E-06 0.91743119 -2.5613E-06 0.456621 -8.1453E-07 0.22883295 5.9474E-08 0.11454754 THOD F(Xr) F(Xi)*F(Xr) Ea -0.40931479 -0.40931479 -0.08994112 -0.08994112 40.93147915 -0.01912266 -0.01912266 8.994111954 -0.00403004 -0.00403004 1.912265637
    • -0.00084767 -0.00084767 0.403003816 -0.00017823 -0.00017823 0.084767304 -3.7469E-05 -3.7469E-05 0.017822515 ERROR 198.168436 250 106.554579 25.7553761 200 2.47388323 0.12439283 150 100 50 0 0 2 -50
    • METODO SECANTE METODO NEWTON % ERROR RELATIVO N° ITERACIONES % ERROR RELATIVO N° ITERACIONES 198.1684361 0 212.9685366 1 106.5545793 1 24.07595393 2 25.75537607 2 1.277567089 3 2.473883231 3 0.003225405 4 0.124392826 4 2.04108E-08 5
    • ERROR Vs ITERACIONES METODO BISECCION METODO SECANTE METODO NEWTON 2 4 6 8 10 12
    • METODO BISECCION N° ITERACIONES % ERROR RELATIVO N° ITERACIONES 100 0 100 1 33.3333333 2 14.2857143 3 7.69230769 4 3.7037037 5
    • 1.81818182 6 0.91743119 7 0.456621 8 0.22883295 9 0.11454754 10
    • r t 0 0 1 Tanque 2 44.6042958 3 46.6094872 4 47.16 50 5 47.5164445 40 30 20 Tiempo 4-3 44.6042958 10 0 0 1 2 3
    • Tanque t 4 5 6