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The relation between local and global truncation errors is slightly different from in the simpler setting of one-step methods. For linear multistep methods, an additional concept called zero-stability is needed to explain the relation between local and global truncation errors.
1 Examples. Toggle Examples subsection. 1.1 Infinite series. 1.2 Differentiation. ... Example A: Find the truncation in calculating the first derivative of () = ...
Hilbert matrix — example of a matrix which is extremely ill-conditioned (and thus difficult to handle) Wilkinson matrix — example of a symmetric tridiagonal matrix with pairs of nearly, but not exactly, equal eigenvalues; Convergent matrix — square matrix whose successive powers approach the zero matrix; Algorithms for matrix multiplication:
Computing the square root of 2 (which is roughly 1.41421) is a well-posed problem. Many algorithms solve this problem by starting with an initial approximation x 0 to , for instance x 0 = 1.4, and then computing improved guesses x 1, x 2, etc. One such method is the famous Babylonian method, which is given by x k+1 = (x k + 2/x k)/2.
called the local Artin symbol, the local reciprocity map or the norm residue symbol. [4] [5] Let L⁄K be a Galois extension of global fields and C L stand for the idèle class group of L. The maps θ v for different places v of K can be assembled into a single global symbol map by multiplying the local
For example, consider the ordinary differential equation ′ = + The Euler method for solving this equation uses the finite difference quotient (+) ′ to approximate the differential equation by first substituting it for u'(x) then applying a little algebra (multiplying both sides by h, and then adding u(x) to both sides) to get (+) + (() +).
In statistics, truncation results in values that are limited above or below, resulting in a truncated sample. [1] A random variable y {\displaystyle y} is said to be truncated from below if, for some threshold value c {\displaystyle c} , the exact value of y {\displaystyle y} is known for all cases y > c {\displaystyle y>c} , but unknown for ...
The exact result is 10005.85987, which rounds to 10005.9. With a plain summation, each incoming value would be aligned with sum, and many low-order digits would be lost (by truncation or rounding). The first result, after rounding, would be 10003.1. The second result would be 10005.81828 before rounding and 10005.8 after rounding. This is not ...