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Word problem from the Līlāvatī (12th century), with its English translation and solution. In science education, a word problem is a mathematical exercise (such as in a textbook, worksheet, or exam) where significant background information on the problem is presented in ordinary language rather than in mathematical notation.
The word problem for an algebra is then to determine, given two expressions (words) involving the generators and operations, whether they represent the same element of the algebra modulo the identities. The word problems for groups and semigroups can be phrased as word problems for algebras. [1]
Complementary antonyms are word pairs whose meanings are opposite but whose meanings do not lie on a continuous spectrum (push, pull). Relational antonyms are word pairs where opposite makes sense only in the context of the relationship between the two meanings (teacher, pupil). These more restricted meanings may not apply in all scholarly ...
The function r is called the stability function. [31] It follows from the formula that r is the quotient of two polynomials of degree s if the method has s stages. Explicit methods have a strictly lower triangular matrix A , which implies that det( I − zA ) = 1 and that the stability function is a polynomial.
In the mathematical field of real analysis, a simple function is a real (or complex)-valued function over a subset of the real line, similar to a step function. Simple functions are sufficiently "nice" that using them makes mathematical reasoning, theory, and proof easier. For example, simple functions attain only a finite number of values.
However, for any degree there are some polynomial equations that have algebraic solutions; for example, the equation = can be solved as =. The eight other solutions are nonreal complex numbers , which are also algebraic and have the form x = ± r 2 10 , {\displaystyle x=\pm r{\sqrt[{10}]{2}},} where r is a fifth root of unity , which can be ...
This represents the value (or values) of the argument x in the interval (−∞,−1] that minimizes (or minimize) the objective function x 2 + 1 (the actual minimum value of that function is not what the problem asks for). In this case, the answer is x = −1, since x = 0 is infeasible, that is, it does not belong to the feasible set. Similarly,
Technically, a point z 0 is a pole of a function f if it is a zero of the function 1/f and 1/f is holomorphic (i.e. complex differentiable) in some neighbourhood of z 0. A function f is meromorphic in an open set U if for every point z of U there is a neighborhood of z in which at least one of f and 1/f is holomorphic.