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Given the difficulty of constructing explicit small families of solutions, much less presenting something like a "general" solution to the Einstein field equation, or even a "general" solution to the vacuum field equation, a very reasonable approach is to try to find qualitative properties which hold for all solutions, or at least for all ...
Einstein's equations can also be solved on a computer using sophisticated numerical methods. [1] [2] [3] Given sufficient computer power, such solutions can be more accurate than post-Newtonian solutions. However, such calculations are demanding because the equations must generally be solved in a four-dimensional space.
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]
Solving the geodesic equations is a procedure used in mathematics, particularly Riemannian geometry, and in physics, particularly in general relativity, that results in obtaining geodesics. Physically, these represent the paths of (usually ideal) particles with no proper acceleration , their motion satisfying the geodesic equations.
The extended Debye–Hückel equation provides accurate results for μ ≤ 0.1. For solutions of greater ionic strengths, the Pitzer equations should be used. In these solutions the activity coefficient may actually increase with ionic strength. The Debye–Hückel plot with different values for ion charge Z and ion diameter a
One particular solution is x = 0, y = 0, z = 0. Two other solutions are x = 3, y = 6, z = 1, and x = 8, y = 9, z = 2. There is a unique plane in three-dimensional space which passes through the three points with these coordinates, and this plane is the set of all points whose coordinates are solutions of the equation.
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 ...
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 ]