AT5 - MNC

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Aluno: Jaime M Della Corte(2017102687)
Turma: 832
Atividade 5
Exercício 1 : Considere a função f(x) = x3 + 3x – 1 e encontre a derivada f’(x) desta função e escreva a expressão φ(X[k-1]) = X) = X[k-1]) = X -f(X[k-1]) = X)/f’(X[k-1]) = X) substituindo f e f’ pela função e a derivada encontrada 
R: x - (x³ + 3x- 1)/(3x+1)
Exercício 2 : Tomando a função f(x) = x3 + 3x – 1 e a função derivada f’(x) encontrada no exercício 1 altere o programa Programa de Método Newton Raphson colocando as funções f(x) e f’(x) citadas tomando o ponto inicial x0=0 com precisão 0.000001 (0.000001 para o compilador C On Line e 0,000001 no Dev C++).
 |f(x[0])|=|f(0,000000)|=1,000000
x[1]=x[0] - f(x[0])/f'(x[0])=1,000000 e |x[1]-x[0]|=1,000000
x[2]=x[1] - f(x[1])/f'(x[1])=0,250000 e |x[2]-x[1]|=0,750000
x[3]=x[2] - f(x[2])/f'(x[2])=0,383929 e |x[3]-x[2]|=0,133929
x[4]=x[3] - f(x[3])/f'(x[3])=0,287089 e |x[4]-x[3]|=0,096839
x[5]=x[4] - f(x[4])/f'(x[4])=0,348913 e |x[5]-x[4]|=0,061823
x[6]=x[5] - f(x[5])/f'(x[5])=0,305324 e |x[6]-x[5]|=0,043589
x[7]=x[6] - f(x[6])/f'(x[6])=0,334325 e |x[7]-x[6]|=0,029001
x[8]=x[7] - f(x[7])/f'(x[7])=0,314183 e |x[8]-x[7]|=0,020142
x[9]=x[8] - f(x[8])/f'(x[8])=0,327793 e |x[9]-x[8]|=0,013609
x[10]=x[9] - f(x[9])/f'(x[9])=0,318415 e |x[10]-x[9]|=0,009377
x[11]=x[10] - f(x[10])/f'(x[10])=0,324793 e |x[11]-x[10]|=0,006378
x[12]=x[11] - f(x[11])/f'(x[11])=0,320416 e |x[12]-x[11]|=0,004377
x[13]=x[12] - f(x[12])/f'(x[12])=0,323402 e |x[13]-x[12]|=0,002986
x[14]=x[13] - f(x[13])/f'(x[13])=0,321357 e |x[14]-x[13]|=0,002045
x[15]=x[14] - f(x[14])/f'(x[14])=0,322754 e |x[15]-x[14]|=0,001397
x[16]=x[15] - f(x[15])/f'(x[15])=0,321797 e |x[16]-x[15]|=0,000956
x[17]=x[16] - f(x[16])/f'(x[16])=0,322451 e |x[17]-x[16]|=0,000654
x[18]=x[17] - f(x[17])/f'(x[17])=0,322004 e |x[18]-x[17]|=0,000447
x[19]=x[18] - f(x[18])/f'(x[18])=0,322310 e |x[19]-x[18]|=0,000306
x[20]=x[19] - f(x[19])/f'(x[19])=0,322100 e |x[20]-x[19]|=0,000209
x[21]=x[20] - f(x[20])/f'(x[20])=0,322243 e |x[21]-x[20]|=0,000143
x[22]=x[21] - f(x[21])/f'(x[21])=0,322146 e |x[22]-x[21]|=0,000098
x[23]=x[22] - f(x[22])/f'(x[22])=0,322213 e |x[23]-x[22]|=0,000067
x[24]=x[23] - f(x[23])/f'(x[23])=0,322167 e |x[24]-x[23]|=0,000046
x[25]=x[24] - f(x[24])/f'(x[24])=0,322198 e |x[25]-x[24]|=0,000031
x[26]=x[25] - f(x[25])/f'(x[25])=0,322177 e |x[26]-x[25]|=0,000021
x[27]=x[26] - f(x[26])/f'(x[26])=0,322191 e |x[27]-x[26]|=0,000015
x[28]=x[27] - f(x[27])/f'(x[27])=0,322181 e |x[28]-x[27]|=0,000010
x[29]=x[28] - f(x[28])/f'(x[28])=0,322188 e |x[29]-x[28]|=0,000007
x[30]=x[29] - f(x[29])/f'(x[29])=0,322183 e |x[30]-x[29]|=0,000005
x[31]=x[30] - f(x[30])/f'(x[30])=0,322187 e |x[31]-x[30]|=0,000003
x[32]=x[31] - f(x[31])/f'(x[31])=0,322184 e |x[32]-x[31]|=0,000002
x[33]=x[32] - f(x[32])/f'(x[32])=0,322186 e |x[33]-x[32]|=0,000002
x[34]=x[33] - f(x[33])/f'(x[33])=0,322185 e |x[34]-x[33]|=0,000001
x[35]=x[34] - f(x[34])/f'(x[34])=0,322186 e |x[35]-x[34]|=0,000001
A raiz aproximada é x[35]=0,322186 e f(0,322186)=0,000001
Maior que o Metodo iterativo.