Newton-Raphson Method | Numerical Computing in Python

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4 жыл бұрын

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Theory and coding of Newton-Raphson method to solve a nonlinear equation.
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Пікірлер: 29

@mechtutorcom 11 ай бұрын

If you are interested, here's a free NumPy mini-course link: rb.gy/pk99l ... I hope you'll find it useful.

@MathPhysicsFunwithGus 3 жыл бұрын

Amazing video thank you helped me so much in my computational class!

@souadmassi7683 4 жыл бұрын

It is a long time, welcome again on KZfaq. Please, make more videos on Matlab

@AJ-et3vf 2 жыл бұрын

Awesome video sir! Thank you!

@SS-zq5sc Жыл бұрын

thank you very much that was so helpful and well explained.

@ascorbic_b Жыл бұрын

You are the best man in the world. I spent one week for doing this shit. I love so much. Thank you

@amirahsalsabila7782 3 жыл бұрын

Terimakasih bapak

@googlegoogle1610 2 жыл бұрын

the best explanation ever :)

@amalabdallah568 3 жыл бұрын

That was really useful 🧡👍🏼

@dellpi3911 3 жыл бұрын

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@fairouzbadis7975 4 жыл бұрын

YOU SAVED ME

@juliomurillo1850 4 жыл бұрын

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 2 жыл бұрын

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.

@bidhayakgoswami2564 3 жыл бұрын

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

@ccuuttww 3 жыл бұрын

how do u add the bee sound every time u run the program

@alvarocarrillo4410 4 жыл бұрын

How can I do it but for the unreal roots?

@mechtutorcom 4 жыл бұрын

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.

@fahadtufailsahib Жыл бұрын

How to solve an equation with logx

@biswarajpalit 4 жыл бұрын

Zoom ur code...really appearing small

@mechtutorcom Жыл бұрын

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@Dusadof 2 жыл бұрын

What if I have 7 nonlinear equations

@kaasbaas5906 3 жыл бұрын

any laptopassignment 2 calculusers?

@TPLCompany 3 жыл бұрын

did you just like your own comment?

@kaasbaas5906 3 жыл бұрын

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@TPLCompany 3 жыл бұрын

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@ulisesalejandrogutierrezsa8334 3 жыл бұрын

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@thshm2938 Жыл бұрын

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 Жыл бұрын

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 Жыл бұрын

Do not forget to "import sympy as sym"