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[17] [18] For example, the fraction 1/(x 2 + 1) is not a polynomial, and it cannot be written as a finite sum of powers of the variable x. For polynomials in one variable, there is a notion of Euclidean division of polynomials, generalizing the Euclidean division of integers.
The unique pair of values a, b satisfying the first two equations is (a, b) = (1, 1); since these values also satisfy the third equation, there do in fact exist a, b such that a times the original first equation plus b times the original second equation equals the original third equation; we conclude that the third equation is linearly ...
Vertical line of equation x = a Horizontal line of equation y = b. Each solution (x, y) of a linear equation + + = may be viewed as the Cartesian coordinates of a point in the Euclidean plane. With this interpretation, all solutions of the equation form a line, provided that a and b are not both zero. Conversely, every line is the set of all ...
A linear multistep method is zero-stable if all roots of the characteristic equation that arises on applying the method to ′ = have magnitude less than or equal to unity, and that all roots with unit magnitude are simple. [2]
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]
Indeed, since y(x) is real, c 1 − c 2 must be imaginary or zero and c 1 + c 2 must be real, in order for both terms after the last equals sign to be real.) For example, if c 1 = c 2 = 1 / 2 , then the particular solution y 1 (x) = e ax cos bx is formed.
Then, f(x)g(x) = 4x 2 + 4x + 1 = 1. Thus deg(f⋅g) = 0 which is not greater than the degrees of f and g (which each had degree 1). Since the norm function is not defined for the zero element of the ring, we consider the degree of the polynomial f(x) = 0 to also be undefined so that it follows the rules of a norm in a Euclidean domain.
It is also possible to take the variable y to be the unknown, and then the equation is solved by y = x – 1. Or x and y can both be treated as unknowns, and then there are many solutions to the equation; a symbolic solution is (x, y) = (a + 1, a), where the variable a may take any value. Instantiating a symbolic solution with specific numbers ...