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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]
In decision problem versions of the art gallery problem, one is given as input both a polygon and a number k, and must determine whether the polygon can be guarded with k or fewer guards. This problem is ∃ R {\displaystyle \exists \mathbb {R} } -complete , as is the version where the guards are restricted to the edges of the polygon. [ 10 ]
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.
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 ...
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 ″.
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.
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.
The Banach fixed point theorem is then invoked to show that there exists a unique fixed point, which is the solution of the initial value problem. An older proof of the Picard–Lindelöf theorem constructs a sequence of functions which converge to the solution of the integral equation, and thus, the solution of the initial value problem.