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The Taylor polynomials for ln(1 + x) only provide accurate approximations in the range −1 < x ≤ 1. For x > 1, Taylor polynomials of higher degree provide worse approximations. The Taylor approximations for ln(1 + x) (black). For x > 1, the approximations diverge. Pictured is an accurate approximation of sin x around the point x = 0. The ...
In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial.According to the theorem, the power (+) expands into a polynomial with terms of the form , where the exponents and are nonnegative integers satisfying + = and the coefficient of each term is a specific positive integer ...
In probability theory, it is possible to approximate the moments of a function f of a random variable X using Taylor expansions, provided that f is sufficiently differentiable and that the moments of X are finite. A simulation-based alternative to this approximation is the application of Monte Carlo simulations.
[1] The approximation can be proven several ways, and is closely related to the binomial theorem . By Bernoulli's inequality , the left-hand side of the approximation is greater than or equal to the right-hand side whenever x > − 1 {\displaystyle x>-1} and α ≥ 1 {\displaystyle \alpha \geq 1} .
However, it holds also in the sense of Riemann integral provided the (k + 1)th derivative of f is continuous on the closed interval [a,x]. Integral form of the remainder [ 10 ] — Let f ( k ) {\textstyle f^{(k)}} be absolutely continuous on the closed interval between a {\textstyle a} and x {\textstyle x} .
2.3 Trigonometric, inverse trigonometric, hyperbolic, and inverse hyperbolic functions relationship. 2.4 Modified-factorial denominators. 2.5 Binomial coefficients.
In mathematics, an expansion of a product of sums expresses it as a sum of products by using the fact that multiplication distributes over addition. Expansion of a polynomial expression can be obtained by repeatedly replacing subexpressions that multiply two other subexpressions, at least one of which is an addition, by the equivalent sum of products, continuing until the expression becomes a ...
and can be found by examination of the coefficient of in the expansion of (1 + x) m (1 + x) n−m = (1 + x) n using equation . When m = 1, equation reduces to equation . In the special case n = 2m, k = m, using , the expansion becomes (as seen in Pascal's triangle at right)