Evidencia de aprendizaje. Fundamentos de análisis numérico

a) desarrolla el polinomio de Taylor para la siguiente función: f ( x )=ex

Alrededor del punto x0=0

>>> x = 0x = 0>>> x = 0; z = exp(x)z = 1

b) Encuentra n tal que la cantidad de cifras significativas del resultado sean 5

>>> x =1.63; z = exp(x)z = 5.1039>>> x =1.625; z = exp(x)z = 5.0784>>> x =1.621; z = exp(x)z = 5.0581>>> x =1.619; z = exp(x)z = 5.0480>>> x =1.612; z = exp(x)z = 5.0128>>> x =1.61; z = exp(x)z = 5.0028>>> x =1.6; z = exp(x)z = 4.9530>>> x =1.611; z = exp(x)z = 5.0078>>> x =1.609; z = exp(x)z = 4.9978>>> x =1.6012; z = exp(x)z = 4.9590>>> x =1.62; z = exp(x)z = 5.0531>>> x =1.602; z = exp(x)z = 4.9629>>> x =1.605; z = exp(x)z = 4.9779>>> x =1.607; z = exp(x)z = 4.9878>>> x =1.604; z = exp(x)z = 4.9729>>> x =1.603; z = exp(x)z = 4.9679>>> x =1.60; z = exp(x)z = 4.9530>>> x =1.61; z = exp(x)z = 5.0028>>> x =1.608; z = exp(x)

z = 4.9928>>> x =1.609; z = exp(x)z = 4.9978>>> x =1.607; z = exp(x)z = 4.9878>>> x =1.6095; z = exp(x)z = 5.0003>>> x =1.6094; z = exp(x)z = 4.9998>>> x =1.60945; z = exp(x)z = 5.0001>>> x =1.6044; z = exp(x)z = 4.9749>>> x =1.60944; z = exp(x)z = 5.0000

>>> cd 'C:/Actividades'miFuncion3>>>Introduce el num. de aproximaciones en el calculo: 2Introduce el valor de x: 3y =20.0855 7.3891 20.0855 54.5982

ALUMNO: ALQUICIRA MEJIA MARCO ANTONIO

MATRICULA: AL10505211