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The solution set for the equations x − y = −1 and 3x + y = 9 is the single point (2, 3). A solution of a linear system is an assignment of values to the variables ,, …, such that each of the equations is satisfied. The set of all possible solutions is called the solution set. [5]
To do so, the different variables in the equation are understood as coordinates and the values that solve the equation are interpreted as points of a graph. For example, if x {\displaystyle x} is set to zero in the equation y = 0.5 x − 1 {\displaystyle y=0.5x-1} , then y {\displaystyle y} must be −1 for the equation to be true.
Given two different points (x 1, y 1) and (x 2, y 2), there is exactly one line that passes through them. There are several ways to write a linear equation of this line. If x 1 ≠ x 2, the slope of the line is . Thus, a point-slope form is [3]
The roots of the quadratic function y = 1 / 2 x 2 − 3x + 5 / 2 are the places where the graph intersects the x-axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.
At any step in a Gauss-Seidel iteration, solve the first equation for in terms of , …,; then solve the second equation for in terms of just found and the remaining , …,; and continue to . Then, repeat iterations until convergence is achieved, or break if the divergence in the solutions start to diverge beyond a predefined level.
For example, to solve a system of n equations for n unknowns by performing row operations on the matrix until it is in echelon form, and then solving for each unknown in reverse order, requires n(n + 1)/2 divisions, (2n 3 + 3n 2 − 5n)/6 multiplications, and (2n 3 + 3n 2 − 5n)/6 subtractions, [10] for a total of approximately 2n 3 /3 operations.
Input: initial guess x (0) to the solution, (diagonal dominant) matrix A, right-hand side vector b, convergence criterion Output: solution when convergence is reached Comments: pseudocode based on the element-based formula above k = 0 while convergence not reached do for i := 1 step until n do σ = 0 for j := 1 step until n do if j ≠ i then ...
The artificial landscapes presented herein for single-objective optimization problems are taken from Bäck, [1] Haupt et al. [2] and from Rody Oldenhuis software. [3] Given the number of problems (55 in total), just a few are presented here. The test functions used to evaluate the algorithms for MOP were taken from Deb, [4] Binh et al. [5] and ...