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The problem with equispaced points is that the interpolation error polynomial, i.e. $$f(x) - P_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!} \prod_{i=0}^n (x - x_i),\quad \xi\in[x_0,x_n]$$ behaves differently for different sets of nodes $x_i$. In the case of equispaced points, this polynomial blows up at the edges. If you use Gauss-Legendre points, the error ...

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Error estimates usually have the form $$\|u - u_h\| \leq C(h),$$ where $u$ is the exact solution you are interested in, $u_h$ is a computed approximate solution, $h$ is an approximation parameter you can control, and $C(h)$ is some function of $h$ (among other things). In finite element methods, $u$ is the solution of a partial differential equation and $... 22 This is a really interesting question, and there are a lot of possible explanations. If we are attempting to use a polynomial interpolation, then note that polynomial satisfy the following annoying inequality Given a polynomial$P$of degree not exceeding$N$we have $$|P^{\prime}(x)| \leq \frac{N}{\sqrt{1-x^2}}\max _x |P(x) |$$ for every$x \in (-1,1)$... 20 I think this is not quite what you had in mind, but for the sake of completeness, let's start with some basics. Most quadrature formulas such as Newton-Cotes and Gauss are based on the idea that in order to evaluate the integral of a function approximately, you can approximate the function by, e.g., a polynomial that you can then integrate exactly: $$\int_a^... 17 Both norms are similar in that they are induced by the scalar product of the respective Hilbert space, but they differ because the different spaces are endowed with different inner products: For \mathbb{R}^N, the Euclidean norm of v = (v_1,\dots,v_N)^T\in \mathbb{R}^N is defined by$$\|v\|_{2}^2 = (v,v)_{2} = \sum_{i=1}^N v_i^2.$$For \ell^2 (the ... 14 Chapter 8 of Brenner and Scott's Mathematical Theory of Finite Element Methods is devoted to this subject. In particular, theorem 8.1.11 and the corollary give you that \|u - u_h\|_{W^1_\infty} \le C h^{k - 1}\|u\|_{W^k_\infty} for linear elliptic problems with sufficiently smooth coefficients, provided that the finite element space satisfies some other ... 12 Adams-Moulton method is significantly more stable. The analogy used when I was taught the difference is the same as extrapolation and interpolation. Interpolation is relatively safe numerically. Extrapolation can blow up if you happen to have an asymptote or some other odd feature. For instance, solving the ode y'(t) = -y(t) with y(0) = 1 using ... 11 And, it is my understanding that the 4 and the 5 are for the order of the global and local error, respectively. Your understanding is wrong. The local error of a Runge–Kutta method of order n is proportional to h^n. What ode45 does is to estimate the solution (of one step) with two Runge–Kutta methods with local orders of 4 and 5, respectively (hence ... 10 It's not implementation-dependent in the sense that this is a mathematical operation performed on your matrix. However, it is very much matrix-dependent. If your matrix is diagonalizable and A=XDX^{-1}, then zeroing out some element adds a small perturbation matrix E, so the new eigenvalues will be (assuming the matrix X does not change much)$$ X^{-1}... 10 The term is commonly used in the following way in the finite element context: Let's assume that$u\in V$is the exact solution of the PDE, and$u_h\in V_h$the finite element approximation. Then we ask for the convergence order of the norm of the error$\|u-u_h\|_X$with regard to some norm$\|\cdot\|_X$. We then say that the approximate solution "converges ... 9 I guess, you start by collecting error rates of components, such as DRAM, like this Google research on DRAM Errors in the Wild: A Large-Scale Field Study They found ~1% chance to get one uncorrectable error per year. I'm not sure if that's what you're interested. I'd be more interested in undetectable errors. Errors such that typical error checking methods ... 8 To extend VorKir's basically correct answer, to derive a priori error estimate for finite element approximations, you need to stack the following three Lego blocks: The well-posedness of the weak formulation of the PDE. The (global) conformity of the Galerkin approximation. The (local) approximation properties of your finite element space. (Only the last ... 7 Using product quadrature rules for multi-dimensional integrals suffers from the so-called curse of dimensionality. An$O(N^{-2})$accurate rule using N evaluations in one-dimension is generally$O(N^{-2})$accurate when applied as a product rule in multiple dimensions, but there will be$M = N^d$evaluations required. So the accuracy is$O(M^{-2/d}).$The ... 7 For a polynomial interpolant we have the formula for the pointwise error $$E_n(x) = f(x) - p_n(x) = \frac{f^{(n+1)}}{(n+1)!}\prod_{i=0}^n(x-x_i),$$ where the$x_i$are the interpolation knots. In general, we want to work with the right hand side of the error formula (a derivation appears in the book by Quarteroni et. al.) as the function we interpolate is ... 6 There are a number of "corrected" integration rules which invoke derivatives of the endpoints. One simple example is the corrected trapezoidal rule. Suppose we wish to approximate the integral $$\int_a^bf(x)\,dx.$$ Let$n$be an integer, and$h=(b-a)/n$. Then the trapezoidal rule $$T = \frac{h}{2}\bigl(f(a)+2f(a+h)+2f(a+2h)+\cdots+2f(a+(n-1)h)+f(b)\... 6 There is a field of study known as eigenvalue sensitivity analysis or eigenvalue perturbation analysis that allows you to estimate the effect of small matrix perturbations on the eigenvalues and eigenvectors. The basic technique used for this is differentiating the eigenvalue matrix equation,$$AX = X\Lambda.$$For situations where the eigenvalues of the ... 6 Use equation (2), equation (1) is wrong. Technically the L^2 norm (upper case "L") is an integral norm of a function defined as$$ \left|\left|f(x)\right|\right|_2 = \sqrt{ \int_\Omega |f(x)|^2 dx}. $$I'm sure you meant l^2. What you typically want to compute in the context of convergence of numerical methods is the finite dimensional analog of the L^... 6 Assuming that you mean the following inequality in your prompt$$ |F_\mathrm{true}(x) - F(x)| \le |dF(x)| \qquad \forall x,$$a simple bound for dM is the following$$ dM \equiv | M_\mathrm{true} - M | = \left|\int_{x_1}^{x_2} F_\mathrm{true}(x) - F(x) dx\right| \le \int_{x_1}^{x_2} |dF(x)| dx. $$This also assumes that dF is absolute integrable. 5 It may look like the error is getting larger, but I think you are confusing multiple issues here. The first is that you are looking at finite approximating Taylor polynomials p_3(x), p_5(x), p_7(x) evaluated at x=\pi/2. Taylor's theorem says that for analytic functions (which the sine is), p_N(\pi/2)\rightarrow 1 as N\to \infty. Your results appear ... 5 Have you looked at the various exascale reports that have come out? Hardward failures are not a significant concern today -- sure, they happen, but their frequency is not sufficiently high to cause grave worry. But they are estimated to be sufficiently frequent on exascale systems with O(10^8) or more cores that codes need to be prepared to react ... 5 Apart from the methods based on Newton-Cotes mentioned in the other answers, there is what is now called Gauss-Turán quadrature (see e.g this and this, as well as the venerable reference by Walter Gautschi). This is a generalization of the usual Gaussian quadrature, where one can now exploit the knowledge of a function's derivatives in finding an optimal set ... 5 The 2-norm for sequences is denoted by \ell^2. For functions on the real line L^2 is the standard notation of the 2-norm. 5 Continuous It looks like you only need 2d curl, so let's start with a simpler continuous definition:$$ \omega = \nabla \times \mathbf{u} = \frac{\delta v}{\delta x} - \frac{\delta u}{\delta y}$$where 2d vector field$\mathbf{u}=(u,v)$(same as your$\mathbf F = (F_1,F_2,F_3)$, dropping$F_3$). Note that curl is a vector and in the 2d version, and is ... 5 First of all I don't see how this estimates the local error defined as the error we make in a single step using correct previous values. What does a lower order method have anything to do with previous correct values? I can accept that this is the error we make using a lower order method instead of a higher order method, but what is the relevance of this? ... 5 Like any other integral, you evaluate the error integral through quadrature. Your suggestion at the end of the question is exactly what you would do: The norm of the gradient error square is a sum of two integrals (or one integral whose integrand is a sum of two terms) which you would evaluate through quadrature. That is, you will have to evaluate the two ... 5 In general, the initial error will already grow exponentially even if you solved the ODE exactly. The Lyapunov exponent (related to the Lipschitz condition) will tell you have fast. Separately, though, the Gronwall lemma is exactly what you are looking for. It allows you to estimate the error at the end time as a function of the errors introduced in each ... 4 Although this thread is quite old, I thought it might be useful to have a reference to a peer-reviewed paper for generalizations of some common quadrature rules. Nenad Ujevic, "A generalization of the modified Simpson’s rule and error bounds", ANZIAM Journal, Vol. 47, 2005. http://journal.austms.org.au/ojs/index.php/ANZIAMJ/article/view/2/1268 I thought ... 4 The two basic approaches I can think of are: interval arithmetic something like polynomial chaos expansions I'm sure there are other approaches out there. In brief, interval arithmetic redefines common arithmetic operations (addition, subtraction, multiplication, etc.) over intervals, so that given an interval$I$, a function$f\$, and the interval ...

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I'm using double precision values in the calculation. Is it feasible of me to demand of my integrator that the difference between the 4th and 5th order estimates be < 1 x 10 ^-16 if machine precision is that of a double? Assuming you're talking about absolute error tolerances, in general, yes, it makes sense.. A unit in the last place (ulp) would be the ...

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After a many days of discussions, your problems are 1) you choose too simple problem for which you reach the machine zero and therefore cannot observe normal properties 2) perhaps you do not use the boundary values correctly. With the scheme you work the values should be added to the RHS, but since they are zero there is nothing to add. However, one thing ...

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