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2017-10-30

-- Oct 18 In-Class Exercise
Abhinaya Koduri x0 = 5 f(x) = ((x-3)**2 ) * ((x-2)**2) f'(x) = 2*(x-2) * (x-3) * (2x-5) x1 = 4.4 x2 = 3.958 x3 = 3.636 x4 = 3.407
(Edited: 2017-10-30)
Abhinaya Koduri x0 = 5 f(x) = ((x-3)**2 ) * ((x-2)**2) f'(x) = 2*(x-2) * (x-3) * (2x-5) x1 = 4.4 x2 = 3.958 x3 = 3.636 x4 = 3.407

-- Oct 18 In-Class Exercise
fp(x) = 2(x-3)(x-2)(2x-5) x0 = 5 f(x0) = 36 fp(x0) = 60 x1 = x0 - f(x0)/fp(x0) = 4.4 f(x1) = 11.289 and fp(x1) = 25.536 x2 = x1 - f(x1)/fp(x1) = 3.957 f(x2) = 3.507 and fp(x2) = 10.914 x3 = x2 - f(x2)/fp(x2) = 3.636 x4 = 3.407
(Edited: 2017-10-30)
<nowiki>fp(x) = 2(x-3)(x-2)(2x-5) x0 = 5 f(x0) = 36 fp(x0) = 60 x1 = x0 - f(x0)/fp(x0) = 4.4 f(x1) = 11.289 and fp(x1) = 25.536 x2 = x1 - f(x1)/fp(x1) = 3.957 f(x2) = 3.507 and fp(x2) = 10.914 x3 = x2 - f(x2)/fp(x2) = 3.636 x4 = 3.407</nowiki>
2017-11-01

-- Oct 18 In-Class Exercise
Name = Krishna Teja Vojjila the update rule is x(n+1) = x(n) - f(xn)/f'(xn) f'(x) = 2*(x-2)*(x-3)*(2x-5) thus x0 = 5 and we compute and resubstitute in the equation hence x1 = 4.4 x2 = 3.95 x3 = 3.63 x4 = 3.40
(Edited: 2017-11-01)
Name = Krishna Teja Vojjila the update rule is x(n+1) = x(n) - f(xn)/f'(xn) f'(x) = 2*(x-2)*(x-3)*(2x-5) thus x0 = 5 and we compute and resubstitute in the equation hence x1 = 4.4 x2 = 3.95 x3 = 3.63 x4 = 3.40
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