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The derivative of a constant term is 0, so when a term containing a constant term is differentiated, the constant term vanishes, regardless of its value. Therefore the antiderivative is only determined up to an unknown constant term, which is called "the constant of integration" and added in symbolic form (usually denoted as ).
The names for the degrees may be applied to the polynomial or to its terms. For example, the term 2x in x 2 + 2x + 1 is a linear term in a quadratic polynomial. The polynomial 0, which may be considered to have no terms at all, is called the zero polynomial. Unlike other constant polynomials, its degree is not zero.
An ordered pair of vertices, such as an edge in a directed graph. An arrow (x, y) has a tail x, a head y, and a direction from x to y; y is said to be the direct successor to x and x the direct predecessor to y. The arrow (y, x) is the inverted arrow of the arrow (x, y). articulation point A vertex in a connected graph whose removal would ...
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.
The y-intercept is the initial value = = at =. The slope a measures the rate of change of the output y per unit change in the input x. In the graph, moving one unit to the right (increasing x by 1) moves the y-value up by a: that is, (+) = +.
the exponents of y in the terms are 0, 1, 2, ..., n − 1, n (the first term implicitly contains y 0 = 1); the coefficients form the n th row of Pascal's triangle; before combining like terms, there are 2 n terms x i y j in the expansion (not shown); after combining like terms, there are n + 1 terms, and their coefficients sum to 2 n.
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Taylor's theorem [4] [5] [6] — Let k ≥ 1 be an integer and let the function f : R → R be k times differentiable at the point a ∈ R. Then there exists a function h k : R → R such that f ( x ) = ∑ i = 0 k f ( i ) ( a ) i !