# Tag Info

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Your question is asking about model Verification. You can find numerous resources on methods and standards by searching for Verification and Validation (Roache 1997, 2002, 2004, Oberkampf & Trucano 2002, Salari & Knupp 2000, Babuska & Oden 2004), as well as the broader topic of Uncertainty Quantification. Rather than elaborate on methods, I would ...

26

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 ...

23

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: $$\int_a^... 19 No such standards exist, as reliable error estimates often cost much more than the approximate calculations. Basically there are four kinds of error estimates: (i) Theoretical analyses proving that a numerical method is numerically stable. This doesn't really give an error bar as the analysis only guarantees that the error made is not worse than a ... 16 Both norms are similar in that they are induced by the scalar product of the respective Hilbert space, but they differ because the different spaces are endowed with different inner products: For \mathbb{R}^N, the Euclidean norm of v = (v_1,\dots,v_N)^T\in \mathbb{R}^N is defined by$$\|v\|_{2}^2 = (v,v)_{2} = \sum_{i=1}^N v_i^2.$$For \ell^2 (the ... 14 Chapter 8 of Brenner and Scott's Mathematical Theory of Finite Element Methods is devoted to this subject. In particular, theorem 8.1.11 and the corollary give you that \|u - u_h\|_{W^1_\infty} \le C h^{k - 1}\|u\|_{W^k_\infty} for linear elliptic problems with sufficiently smooth coefficients, provided that the finite element space satisfies some other ... 13 Sort of. There are theoretical error bounds that have been derived by numerical analysts that are usually overestimates, and may not be as useful in practice, because they may involve information that is difficult to obtain for problems in practice. A good example would be the bounds on numerical errors in the solution of ordinary equations, which you can ... 12 Adams-Moulton method is significantly more stable. The analogy used when I was taught the difference is the same as extrapolation and interpolation. Interpolation is relatively safe numerically. Extrapolation can blow up if you happen to have an asymptote or some other odd feature. For instance, solving the ode y'(t) = -y(t) with y(0) = 1 using ... 11 For measuring the error in the solution of PDE, it is quite natural to choose the norm of the space in which the solution lies. For example, for elliptic PDEs, the solution lies in H^1 and so it is natural to choose the H^1 norm to measure the error. This makes sense because, for example, the solution does not lie in the space W^{1,\infty} and so it ... 10 It's not implementation-dependent in the sense that this is a mathematical operation performed on your matrix. However, it is very much matrix-dependent. If your matrix is diagonalizable and A=XDX^{-1}, then zeroing out some element adds a small perturbation matrix E, so the new eigenvalues will be (assuming the matrix X does not change much)$$ X^{-1}... 10 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 ... 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 ... 9 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 ... 8 The reason why people prefer to use the first estimate, in my opinion, is that the first one arises naturally from the Galerkin orthogonality of the FEM, interpolation approximation property, and most importantly the coercivity of the bilinear form(for Poisson equation's boundary value problem, it is equivalent with the Poincaré/Friedrichs inequality for$H^...

8

The Euclidean norm is often used based on the assumption that the Euclidean distance of two points is a reasonable measure of distance. But unless this is the case, this choice is not preferable to a problem-adapted choice. For example if the typical size of the components of a vector are very different (since they mean very different things), the Euclidean ...

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

In addition to the other answers, there are a few additional points to consider. Numerical discretization errors, or at least the order of the schemes, can be determined analytically. The discussion of these errors may be omitted from papers if they use a commonly known scheme. Grid refinement studies where the same problem, usually something simple, is ...

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

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

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

A few remarks: In general, which norm you choose depends on what you want to measure. It is that simple. For numerical pde, the $L^2$-norm has the convenient property of providing a Hilbert space structure. A natural reason to use this norm comes from the treatment of measurement errors, as described in https://scicomp.stackexchange.com/a/2763/238. I do ...

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 ...

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