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The elements of an arithmetico-geometric sequence () are the products of the elements of an arithmetic progression (in blue) with initial value and common difference , = + (), with the corresponding elements of a geometric progression (in green) with initial value and common ratio , =, so that [4]
The result, x 2, is a "better" approximation to the system's solution than x 1 and x 0. If exact arithmetic were to be used in this example instead of limited-precision, then the exact solution would theoretically have been reached after n = 2 iterations ( n being the order of the system).
It is generally used in solving non-linear equations like Euler's equations in computational fluid dynamics. Matrix-free conjugate gradient method has been applied in the non-linear elasto-plastic finite element solver. [7] Solving these equations requires the calculation of the Jacobian which is costly in terms of CPU time and storage. To ...
If an equation can be put into the form f(x) = x, and a solution x is an attractive fixed point of the function f, then one may begin with a point x 1 in the basin of attraction of x, and let x n+1 = f(x n) for n ≥ 1, and the sequence {x n} n ≥ 1 will converge to the solution x.
In mathematics, a collocation method is a method for the numerical solution of ordinary differential equations, partial differential equations and integral equations.The idea is to choose a finite-dimensional space of candidate solutions (usually polynomials up to a certain degree) and a number of points in the domain (called collocation points), and to select that solution which satisfies the ...
The MacCormack method is well suited for nonlinear equations (Inviscid Burgers equation, Euler equations, etc.) The order of differencing can be reversed for the time step (i.e., forward/backward followed by backward/forward). For nonlinear equations, this procedure provides the best results.
Indeed, multiplying each equation of the second auxiliary system by , adding with the corresponding equation of the first auxiliary system and using the representation = +, we immediately see that equations number 2 through n of the original system are satisfied; it only remains to satisfy equation number 1.
Since z = 1 − x, the solution of the hypergeometric equation at x = 1 is the same as the solution for this equation at z = 0. But the solution at z = 0 is identical to the solution we obtained for the point x = 0, if we replace each γ by α + β − γ + 1. Hence, to get the solutions, we just make this substitution in the previous results.