ejemplo unidad 2metodods
DESCRIPTION
Ejemplo Unidad 2metododsTRANSCRIPT
xi+1 Ea%xi= 0xi+1= 0.5 100
0.566311 11.7092910.56714317 0.146728710.56714329 2.21064E-050.56714329 5.08968E-130.56714329 00.56714329 0 xi+1=0.56714329 con Ea%=00.56714329 0 el metodo de newton-raphson converge
paso 1: f´(xi)=e^ (-x )-1paso 2: xi+1=xi-(e^(-xi)-xi)/(-e^(-xi)-1)
xi+1 xi F(xi+1) f(xi) xi+1(Aprox.) Ea%0 1 1 -0.6321205588 0.61269983681 0.61269984 -0.63212056 -0.0708139479 0.5638383892 8.66586039
0.61269984 0.56383839 -0.07081395 0.0051823545 0.5671703584 0.587472390.56383839 0.56717036 0.00518235 -4.241924E-05 0.5671433066 0.004769840.56717036 0.56714331 -4.2419E-05 -2.538017E-08 0.5671432904 2.85558E-060.56714331 0.56714329 -2.538E-08 1.2423396E-13 0.5671432904 1.39771E-110.56714329 0.56714329 1.24234E-13 0 0.5671432904 0
xi+1=0.56714329 con Ea%=0. el metodo de la secante converge
xi xi+d*xi d*xi f(xi) f(xi+d*xi) xi+1(Aprox.) Ea%1 1.01 0.01 -0.6321205588 -0.64578102 0.5372626655
0.53726267 0.54263529 0.00537263 0.047082949 0.03857927 0.5670096854 5.24629836070.56700969 0.57267978 0.0056701 0.0002093833 -0.0086678 0.5671434241 0.0235811220.56714342 0.57281486 0.00567143 -2.095861E-07 -0.008879055 0.5671432903 2.360512E-050.56714329 0.57281472 0.00567143 2.1490398E-10 -0.008878844 0.5671432904 2.420405E-080.56714329 0.57281472 0.00567143 -2.203793E-13 -0.008878844 0.5671432904 2.4822E-110.56714329 0.57281472 0.00567143 0 -0.008878844 0.5671432904 00.56714329 0.57281472 0.00567143 0 -0.008878844 0.5671432904 00.56714329 0.57281472 0.00567143 0 -0.008878844 0.5671432904 0
XI+1=0.56714329 CON Ea%=0el metodo de la secante modificada converge
