Newton-Raphson Method | Numerical Computing in Python

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Theory and coding of Newton-Raphson method to solve a nonlinear equation.
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[@mechtutorcom]

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[@AJ-et3vf]

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[@bidhayakgoswami2564]

Once I wrote this function to find the zero of any continuous and differentiable function def newton(fun,y): #A function_name and an initial guess is provided as arguments of the function 'newton' n=len(y);epsilon=10**(-12); e=np.eye(n)*epsilon; fval=fun(y);i=1 while (np.linalg.norm(fval))>1e-10 and i

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[@juliomurillo1850]

Dear mechtutor, I would like to apply the same code, but for a list of elements, It means my initial values are X0 = {1,2,3,4,5,6} and my functions give me a result for each of these initial values it means f = f(x0), same my derivative f'=f'(x0). So as you see at the end I will get a list of values. Do you think is possible?

[@AJ-et3vf]

Yes, it is possible. You will have to modify your code such that if a list of values were given, the code would loop through the values given. You would also need to allocate for another list/array where you would store the roots that were found. Your use case is interesting, but most root-solver codes don't have a "vectorized" implementation as the one you're asking. They only take one input guess at a time. So if you want to find the root for a given initial values, just put the Newton solver function inside a loop that loops through your given initial values.

[@fahadtufailsahib]

How to solve an equation with logx

[@alvarocarrillo4410]

How can I do it but for the unreal roots?

[@mechtutorcom]

Because this method works only in the real domain, you cannot use it for complex roots. The alternative way is to use the function Scipy.optimize.fsolve() Thanks for your comment.

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What if I have 7 nonlinear equations

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[@thshm2938]

import numpy as np from scipy.misc import derivative def f(x): return 2*x**3-9.5*x+7.5 x1 = 2 x = np.linspace(0.2, 2, 500) m = derivative(f, x1, dx=0.1) for n in range(20): xn = x1 -( f(x1) / m) print(xn) x1 = xn

[@shardulpingale2431]

def newton_approximation(f,appr): x = sym.Symbol('x') fprime = sym.diff(f,x) f = sym.lambdify(x,f) fprime = sym.lambdify(x,fprime) steps = 0 x = appr for i in range(0,100): new_x = x - (f(x) / fprime(x)) steps+=1 if round(x,8) == round(new_x, 8): break else: x = new_x return new_x, steps x = sym.Symbol('x') f = x**2 - 104 print("Total steps taken in order to approximate the root {0} are {1}".format(*newton_approximation(f,3)))

[@shardulpingale2431]

Do not forget to "import sympy as sym"