If you take m steps, and update the Jacobian every t steps, the time complexity will be O ( m N 2 + ( m / t) N 3). So the time taken per step is O ( N 2 + N 3 / t). You're reducing the amount of work you do by a factor of 1 / t, and it's O ( N 2) when t ≥ N.
18.02.2016 · My understanding is that the time complexity of Newton's method is $O(log(N)F(N))$ where $F(N)$ is the cost of calculating $\frac{f(x)}{f'(x)}$ with $N$-digit precision. But given the architecture of the bisection method, which halves the search interval at each iteration, I was under the impression that its time complexity was also logarithmic.
Newton's method has time complexity log(p) * (time to calculate derivative ratio) for a given precision p, assuming reasonable seeds. If multivariate Newton Raphson time complexity does not scale with the number of dimensions, then complexity at a given p will still depend on the gradient/hessian calculations and the linear solver.
From Newton's method - encyclopedia article - Citizendium: Using Newton's method as described above, the time complexity of calculating a root of a function f (x) with n-digit precision, provided that a good initial approximation is known, is O ( (\log n) F (n)) where F (n) is the cost of calculating f (x)/f' (x)\, with n-digit precision.
Answer (1 of 2): Hello Quora, I am not a math student, so I don’t fully understand Tom’s solution. But I wrote a computer program for Newton-Raphson’s method and tried to run on increasing values.
14.02.2011 · Using Newton's method as described above, the time complexity of calculating a root of a function f (x) with n-digit precision, provided that a good initial approximation is known, is O ( (\log n) F (n)) where F (n) is the cost of calculating f (x)/f' (x)\, with n-digit precision.
In numerical analysis, Newton's method, also known as the Newton–Raphson method, named after Isaac Newton and Joseph Raphson, is a root-finding algorithm ...
The Newton-Raphson algorithm uses, for computation of A1/p, the sequence u0=A, un+1=un−upn−Apup−1n, whose speed of convergence , always quadratic, is ...
What is the computational complexity of Newton Raphson method to find square root. ; The Newton-Raphson method for finding the root of an equation f (x) · 1.7k ...
Feb 15, 2011 · Using Newton's method as described above, the time complexity of calculating a root of a function f (x) with n-digit precision, provided that a good initial approximation is known, is O ( (\log n) F (n)) where F (n) is the cost of calculating f (x)/f' (x)\, with n-digit precision. If f (x) can be evaluated with variable precision, the algorithm ...
If you take m steps, and update the Jacobian every t steps, the time complexity will be O ( m N 2 + ( m / t) N 3). So the time taken per step is O ( N 2 + N 3 / t). You're reducing the amount of work you do by a factor of 1 / t, and it's O ( N 2) when t ≥ N. But t is determined adaptively by the behaviour of the loss function, so the point is ...
Using Newton's method as described above, the time complexity of calculating a root of a function f(x) with n-digit precision, provided that a good initial ...