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i have a question in the code newton method say that answer is:
>>> poly =(13.39, 0.0, 17.5, 3.0, 1.0) #x4 + 3.0x3 + 17.5x2  13.39
>>> x_0 = 0.1
>>> epsilon = .0001
>>> print compute_root(poly, x_0, epsilon)
(0.80679075379635201, 8)
my answer is:
(0.806790753796352, 7) why 8?
 one year ago
 one year ago
i have a question in the code newton method say that answer is: >>> poly =(13.39, 0.0, 17.5, 3.0, 1.0) #x4 + 3.0x3 + 17.5x2  13.39 >>> x_0 = 0.1 >>> epsilon = .0001 >>> print compute_root(poly, x_0, epsilon) (0.80679075379635201, 8) my answer is: (0.806790753796352, 7) why 8?
 one year ago
 one year ago

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lopusBest ResponseYou've already chosen the best response.1
this is my code: http://dpaste.com/790245/
 one year ago

xilinixBest ResponseYou've already chosen the best response.0
yo creo que tu codigo esta bueno, porque haciendo el seguimiento si se cuenta desde 0 llega hasta 6 y si se itera desde 1 es hasta 7
 one year ago

xilinixBest ResponseYou've already chosen the best response.0
(3.7760990539788537, 1) (2.513165507392411, 2) (1.6266862300590645, 3) (1.0874006562988667, 4) (0.8541676766719186, 5) (0.8084748052947073, 6) (0.806790753796352, 7)
 one year ago

bwCABest ResponseYou've already chosen the best response.0
it took 8 iterations instead of 7. was epsilon the same? is anything different in the code
 one year ago

lopusBest ResponseYou've already chosen the best response.1
i think found the solution: def compute_root(poly,x_0,epsilon): i=1 while epsilon<=abs(Evaluate_poly(poly,x_0)): derive=compute_derive(poly) x1= x_0(Evaluate_poly(poly,x_0)/Evaluate_poly(derive,x_0)) i+=1 # change that why? x_0=x1 result=(x1,i) print result
 one year ago
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