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The section As a parity sequence above gives a way to speed up simulation of the sequence. To jump ahead k steps on each iteration (using the f function from that section), break up the current number into two parts, b (the k least significant bits, interpreted as an integer), and a (the rest of the bits as an integer
Given a function: from a set X (the domain) to a set Y (the codomain), the graph of the function is the set [4] = {(, ()):}, which is a subset of the Cartesian product.In the definition of a function in terms of set theory, it is common to identify a function with its graph, although, formally, a function is formed by the triple consisting of its domain, its codomain and its graph.
To solve this kind of equation, the technique is add, subtract, multiply, or divide both sides of the equation by the same number in order to isolate the variable on one side of the equation. Once the variable is isolated, the other side of the equation is the value of the variable. [37] This problem and its solution are as follows: Solving for x
In the simple case of a function of one variable, say, h(x), we can solve an equation of the form h(x) = c for some constant c by considering what is known as the inverse function of h. Given a function h : A → B , the inverse function, denoted h −1 and defined as h −1 : B → A , is a function such that
In mathematics, a polynomial is a mathematical expression consisting of indeterminates (also called variables) and coefficients, that involves only the operations of addition, subtraction, multiplication and exponentiation to nonnegative integer powers, and has a finite number of terms.
Divide the highest term of the remainder by the highest term of the divisor (3x ÷ x = 3). Place the result (+3) below the bar. 3x has been divided leaving no remainder, and can therefore be marked as used. The result 3 is then multiplied by the second term in the divisor −3 = −9. Determine the partial remainder by subtracting −4 − (− ...
One of the basic principles of algebra is that one can multiply both sides of an equation by the same expression without changing the equation's solutions. However, strictly speaking, this is not true, in that multiplication by certain expressions may introduce new solutions that were not present before. For example, consider the following ...
graph intersection: G 1 ∩ G 2 = (V 1 ∩ V 2, E 1 ∩ E 2); [1] graph join: . Graph with all the edges that connect the vertices of the first graph with the vertices of the second graph. It is a commutative operation (for unlabelled graphs); [2] graph products based on the cartesian product of the vertex sets: