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If p ≤ 0, then the nth-term test identifies the series as divergent. If 0 < p ≤ 1, then the nth-term test is inconclusive, but the series is divergent by the integral test for convergence. If 1 < p, then the nth-term test is inconclusive, but the series is convergent by the integral test for convergence.
For example, for Newton's method as applied to a function f to oscillate between 0 and 1, it is only necessary that the tangent line to f at 0 intersects the x-axis at 1 and that the tangent line to f at 1 intersects the x-axis at 0. [19] This is the case, for example, if f(x) = x 3 − 2x + 2.
The series can be compared to an integral to establish convergence or divergence. Let : [,) + be a non-negative and monotonically decreasing function such that () =.If = <, then the series converges.
The geometric series is an infinite series derived from a special type of sequence called a geometric progression.This means that it is the sum of infinitely many terms of geometric progression: starting from the initial term , and the next one being the initial term multiplied by a constant number known as the common ratio .
Two other well-known examples are when integration by parts is applied to a function expressed as a product of 1 and itself. This works if the derivative of the function is known, and the integral of this derivative times is also known. The first example is (). We write this as:
It is assumed that the value of a function f defined on [,] is known at + equally spaced points: < < <.There are two classes of Newton–Cotes quadrature: they are called "closed" when = and =, i.e. they use the function values at the interval endpoints, and "open" when > and <, i.e. they do not use the function values at the endpoints.
A famous example is the recurrence for the Fibonacci numbers, = + where the order is two and the linear function merely adds the two previous terms. This example is a linear recurrence with constant coefficients , because the coefficients of the linear function (1 and 1) are constants that do not depend on n . {\displaystyle n.}
Then, applying the induction hypothesis, (+) = (+) + = [(+) +]. Note that the term within square bracket has n-times successive integration, and upper limit of outermost integral inside the square bracket is .