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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]
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.
In general relativity, an exact solution is a (typically closed form) solution of the Einstein field equations whose derivation does not invoke simplifying approximations of the equations, though the starting point for that derivation may be an idealized case like a perfectly spherical shape of matter.
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.
The solutions that are not exact are called non-exact solutions. Such solutions mainly arise due to the difficulty of solving the EFE in closed form and often take the form of approximations to ideal systems. Many non-exact solutions may be devoid of physical content, but serve as useful counterexamples to theoretical conjectures.
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.
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.