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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).
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.
An example of using Newton–Raphson method to solve numerically the equation f(x) = 0. In mathematics, to solve an equation is to find its solutions, which are the values (numbers, functions, sets, etc.) that fulfill the condition stated by the equation, consisting generally of two expressions related by an equals sign.
Some solutions of a differential equation having a regular singular point with indicial roots = and .. In mathematics, the method of Frobenius, named after Ferdinand Georg Frobenius, is a way to find an infinite series solution for a linear second-order ordinary differential equation of the form ″ + ′ + = with ′ and ″.
Because is a linear differential operator, the solution () to a general system of this type can be written as an integral over a distribution of source given by (): = (, ′) (′) ′ where the Green's function for Laplacian in three variables (, ′) describes the response of the system at the point to a point source located at ...
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.
How to Solve It suggests the following steps when solving a mathematical problem: . First, you have to understand the problem. [2]After understanding, make a plan. [3]Carry out the plan.