Search results
Results from the WOW.Com Content Network
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]
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.
So, if the three non-monic coefficients of the depressed quartic equation, + + + =, in terms of the five coefficients of the general quartic equation are given as follows: =, = + and = +, then the criteria to identify a priori each case of quartic equations with multiple roots and their respective solutions are exposed below.
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.
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.
Pell's equation, also called the Pell–Fermat equation, is any Diophantine equation of the form =, where n is a given positive nonsquare integer, and integer solutions are sought for x and y. In Cartesian coordinates , the equation is represented by a hyperbola ; solutions occur wherever the curve passes through a point whose x and y ...
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.