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

xilinix Group TitleBest 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

xilinix Group TitleBest 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

bwCA Group TitleBest 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

lopus Group TitleBest ResponseYou've already chosen the best response.1
yes epsilon is the same
 one year ago

lopus Group TitleBest 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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