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The Bogacki–Shampine method is a Runge–Kutta method of order three with four stages with the First Same As Last (FSAL) property, so that it uses approximately three function evaluations per step. It has an embedded second-order method which can be used to implement adaptive step size. The Bogacki–Shampine method is implemented in the ode3 ...
History. A simple arithmetic calculator was first included with Windows 1.0.. In Windows 3.0, a scientific mode was added, which included exponents and roots, logarithms, factorial-based functions, trigonometry (supports radian, degree and gradians angles), base conversions (2, 8, 10, 16), logic operations, statistical functions such as single variable statistics and linear regression.
Adaptive step size. In mathematics and numerical analysis, an adaptive step size is used in some methods for the numerical solution of ordinary differential equations (including the special case of numerical integration) in order to control the errors of the method and to ensure stability properties such as A-stability.
Robot in a wooden maze. A maze-solving algorithm is an automated method for solving a maze.The random mouse, wall follower, Pledge, and Trémaux's algorithms are designed to be used inside the maze by a traveler with no prior knowledge of the maze, whereas the dead-end filling and shortest path algorithms are designed to be used by a person or computer program that can see the whole maze at once.
In mathematics, a system of linear equations (or linear system) is a collection of one or more linear equations involving the same variables. [1] For example, is a system of three equations in the three variables x, y, z. A solution to a linear system is an assignment of values to the variables such that all the equations are simultaneously ...
There are many programs that, for some inputs, return a correct answer to the halting problem, while for other inputs they do not return an answer at all. However the problem "given program p, is it a partial halting solver" (in the sense described) is at least as hard as the halting problem. To see this, assume that there is an algorithm PHSR ...
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