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Iterative methods in a nutshell: Stationary methods are in essence fixed point iterations: To solve $Ax=b$, you pick an invertible matrix $C$ and find a fixed point of $$x = x + Cb- CAx$$ This converges by Banach's fixed point theorem if $\|I-CA\|<1$. The various methods then correspond to a specific choice of $C$ (e.g., for Jacobi iteration, $C=D^{-1}... 12 Yes, this is the standard Aubin-Nitsche (or duality) trick. The idea is to use the fact that$L^2$is its own dual space to write the$L^2$-norm as an operator norm $$\|u\|_{L^2} = \sup_{\phi\in L^2\setminus\{0\}} \frac{(u,\phi)}{\|\phi\|_{L^2}}.$$ We thus have to estimate$(u-u_h,\phi)$for arbitrary$\phi\in L^2$. To do that, we "lift"$u-u_h$to$H^1_0$... 11 In addition to Christian's answer, it's also worth noting that for linear convergence you have$e_{k+1} \le \lambda_1 e_k$where you have$\lambda_1<1$if the method converges. On the other hand, for quadratic convergence you have$e_{k+1} \le \lambda_2 e_k^2$and the fact that a method converges does not necessarily imply that$\lambda_2$must be smaller ... 10 Let me first answer all the questions: What is the theoretical convergence rate for an FFT Poison solver? The theoretical convergence is exponential as long as the solution is sufficiently smooth. How fast should this energy converge? The Hartree energy$E_H$should converge exponentially for a sufficiently smooth solution. If the solution is less ... 10 It isn't so much that we want to compare the$p$-refinement and$h$-refinement errors directly, instead we want to compare the convergence properties (e.g. speed) of each refinement strategy. This requires more knowledge of the constant in the apriori error estimate. We'll illustrate by looking at the apriori error estimate of a discontinuous Galerkin ... 10 It is because we typically neglect higher order terms in error estimates. For example, we can show that $$\|e\| \le C(u) h^2 + {\cal O}(h^3).$$ The point is that when$h$is small, the cubic term is small and can be neglected. In fact, when$h$is small, you can observe quadratic convergence. But whenever$h$is not small (where "small" is relative to ... 10 Jean-Michel Muller, et. al., "Handbook of Floating-Point Arithmetic 2nd ed.", Birkhäuser 2018, gives the following example due to Muller, specifically constructed to deliver incorrect results with floating-point evaluation: $${u_{0} = 2,\\ u_{1} = -4,\\ u_{n} = 111 - \frac{1130}{u_{n-1}} + \frac{3000}{u_{n-1}u_{n-2}},\>\>\>\>n \ge 2.}$$ ... 9 In practice, yes. While$e_k$is still large, the rate coefficient$\lambda$will dominate the error rather than the q-rate. (Note that these are asymptotic rates, so the statements you linked to only hold for the limit as$k\to\infty$.) For example, for first order methods in optimization you often observe an initially fast decrease in error, which then ... 9$\|x^{(k)}-x^*\|$is the error in the$k$th term, call it$E_k$. For a "good" numerical method, we want the approximation to get closer and closer to the desired result so$E_k$has to decrease to zero. If the error is guaranteed to reduce to at least a certain fraction$L$of the previous step, you have linear convergence: $$E_{k+1} \le L E_k.$$ This ... 8 For a single rational equation in the complex domain, the basin of attraction is fractal, the compelement of a so-called Julia set. http://en.wikipedia.org/wiki/Julia_set . For theory with some nice online figures, see, e.g., http://mathlab.mathlab.sunysb.edu/~scott/Papers/Newton/Published.pdf http://hera.ugr.es/doi/15019160.pdf Even the ''globalized'' ... 8 Let's examine the one-dimensional three-point stencil case in detail, because I think it's important to be clear just how this behaviour arises, and what it means to set a point to a certain value in a finite-difference grid when the underlying function is discontinuous. The equation will be $$u''(x) = \rho(x).$$ Instead of using the interval$[-1,1]$with ... 8 Yes: See Higham's book "Accuracy And Stability of Numerical Algorithms", second edition, chapter 25: Nonlinear Systems and Newton's Method. In particular, see the section on the "limiting residual" in terms of a condition number for the Jacobian. It may well be that since your system is ill-conditioned, that you quickly hit the limiting residual and your ... 8 It's fairly easy to evaluate, to do this expand the logs in Taylor series in$x=(k+1)^{-2}$: $$\log_2(1-x) = \frac{-1}{\log 2}\sum_{m\geq1}\frac{x^{m}}{m}$$ $$\log_2(-\log_2(1-x)) = \frac{\log x}{\log 2} - \frac{\log\log 2}{\log 2} + \sum_{n\geq 1}a_n x^n,$$ where$a_n$are Taylor series coefficients of the l.h.s. after the log-singularity is subtracted. ... 7 Convergence is a statement about the asymptotic limit$h\to0$. Therefore, it's not possible to "prove" convergence by any finite number of computations. The best you can do with computations is motivate the belief that an algorithm does converge. To prove convergence, you need to...write a proof. 7 One cannot conclude from a residual how accurate the solution is. Between the best and the worst case in norm, there is a factor of exactly the condition number. More precisely, if the residual norm is r and the error norm is e then$\|A\|^{-1}\le e/r \le \|A^{-1}\|$, and both bounds are attainable. Taking the quotient of the bounds proves the claim. The ... 7 If analytic techniques are disallowed but the periodic structure is known, here is one approach. Let $$g(x) = \frac{\cos x}{2-\cos x}$$ be periodic with period$2 \pi, so that $$g(x) = \sum_j w_j e^{ijx}$$ where $$w_j = \frac{1}{2\pi}\int_0^{2\pi} g(x) e^{-ijx} dx$$ Thus, \begin{aligned} f(x) &= \sum_{k \ge 1} \frac{g(kx)}{k^p} \\ &= \sum_{k ... 7 The Taylor-Hood approximation of the Stokes flow is a mixed finite element method, for which error estimates generally have the form \|u-u_h\|_V + \|p-p_h\|_M \leq C (\inf_{w_h\in V_h}\|u-w_h\|_V + \inf_{q_h\in M_h}\|p-q_h\|_M), \tag{1} $$where (u,p)\in V\times M is the exact solution and (u_h,p_h)\in V_h\times M_h is the approximation. For the ... 7 If your matrix is symmetric, positive definite, the CG method may converge slowly, but it converges for n\to\infty. The only reason it does not converge on a computer are round-off errors, in particular if the condition number of the matrix, the quotient of largest and smallest eigenvalue is large. Experience is, that with double precision arithmetic, ... 7 The two normalization formulas result in two different algorithms, that they both "normalize" a vector is not so relevant. As an example, consider the following transition matrix:$$M = \begin{pmatrix}0&1\\1&0\end{pmatrix}.$$Starting with r=(1,0), consecutive normalized vectors r will be (0,1) and (1,0), so the algorithm clearly will not ... 7 The issues you're running into now are not a failing of Newton-Raphson, but a question of coupling. You're doing iterated sequential coupling -- solving each equation sequentially and then iterating until (hopeful) convergence. No solver choice in place of NR is going to fix this lack of convergence, as long as you are doing iterated sequential coupling. ... 7 High order methods are not necessarily always more accurate than low order methods; they simply have a more rapid convergence rate. For example, if you take the Taylor expansion of a function at a point x and evaluate it at x+h, you have$$f(x+h) = f(x) + f'(x)h + \frac{f''(x)}{2}h^2 + \ldots$$The order of a finite difference method is determined by ... 7 Standard terminology in nonlinear (i.e., derivative-based) optimization is that "global convergence" means "convergence to a stationary point no matter where you start from" (where the limit may in fact depend on the starting point). This is in contrast to local convergence, which requires that you start sufficiently close to a stationary point; if your ... 7 Unfortunately, convergence of GMRES does not have a clear dependence on the distribution of eigenvalues. It was proved by Greenbaum, Ptak and Strakos in 1996 that you can construct examples with an arbitrary spectrum and an arbitrary convergence history: that is, give me any n nonzero complex numbers, and any decreasing sequence \|r_k\|, and I can ... 6 The interpretation is qualitatively correct. Note that linear and quadratic convergence are with respect to the worst case, the situation in a particular algorithm can be better than what you get from the worst case analysis given by Wolfgang Bangerth, though the qualitative situation usually corresponds to this analysis. In concrete algorithms (e.g., in ... 6 I don't think there's enough information to give a heuristic like, "in general, it takes k iterations for differential evolution to reach a global optimum". First, we don't know how many iterations deterministic global optimizers will take to converge to global \varepsilon-optimality in the general case. In limited special cases, we know that certain ... 6 First of all, rates of convergence are usually given in the form$$ \|u-u_h\| \leq C N^\alpha,$$rather than equality. Furthermore, rates are asymptotic, i.e., only have to hold for N\to \infty. This means that you're unlikely to find a single C and \alpha such that your equation holds. Another reason why your approach doesn't work is that what you're ... 6 One thing that makes your non-dimensionalized ODE confusing is that you use the same symbols for dimensionalized and nondimensionalized variables, even though they are different variables. Consider writing it in a different way: let there be a new set of variables, \tilde t, \tilde x,\ldots, all unitless, defined by$$ \tilde t = n_{\mathrm{time}}t, \... 6 In my opinion, an example could be the calculation of the minimal solution of a three term recurrence relation (TTRR) $$y_{n+1} +a_{n}y_{n}+b_{n}y_{n-1} = 0, \quad n=1,2,3,\ldots$$. For example, the BesselJ_{n}(x)$function for a fixed$x$satisfies the TTRR $$y_{n+1}-\frac{2n}{x}y_{n}+y_{n-1} = 0$$ Suppose you know$y_{0} = J_{0}(1)$and$y_{1} = J_{1}(...

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There is nothing that can guarantee converge of such a fixed point iteration, but there are criteria that if you can show that they hold, the iteration will converge. In particular, the fixed point iteration will converge if it is a contraction. I am quite sure, however, that it is not very difficult to find example of functions $f(x)$ where even with ...

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The purpose of automatic error estimation and step size control is to free you from the problem of determining manually what a sufficiently small step size is. So your question is a bit like asking "somebody gave me this automatic transmission car; how can I tell what gear I'm in?" The point is that you shouldn't need to know. Of course, if the ...

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