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[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} .
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 ...
When is a positive integer, () gives the number of n-permutations (sequences of distinct elements) from an x-element set, or equivalently the number of injective functions from a set of size to a set of size .
For example, antiderivatives of x 2 + 1 have the form 1 / 3 x 3 + x + c. For polynomials whose coefficients come from more abstract settings (for example, if the coefficients are integers modulo some prime number p , or elements of an arbitrary ring), the formula for the derivative can still be interpreted formally, with the coefficient ...
Yet another variant of the Stirling polynomials is considered in [3] (see also the subsection on Stirling convolution polynomials below). In particular, the article by I. Gessel and R. P. Stanley defines the modified Stirling polynomial sequences, ():= (+,) and ():= (,) where (,):= (,) are the unsigned Stirling numbers of the first kind, in terms of the two Stirling number triangles for non ...
In algebra, the partial fraction decomposition or partial fraction expansion of a rational fraction (that is, a fraction such that the numerator and the denominator are both polynomials) is an operation that consists of expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.
P(x) will be divided by Q(x) using Ruffini's rule. The main problem is that Q(x) is not a binomial of the form x − r, but rather x + r. Q(x) must be rewritten as = + = (). Now the algorithm is applied: Write down the coefficients and r. Note that, as P(x) didn't contain a coefficient for x, 0 is written:
The polynomial x 2 + 2x + 2, on the other hand, is primitive. Denote one of its roots by α. Then, because the natural numbers less than and relatively prime to 3 2 − 1 = 8 are 1, 3, 5, and 7, the four primitive roots in GF(3 2) are α, α 3 = 2α + 1, α 5 = 2α, and α 7 = α + 2. The primitive roots α and α 3 are algebraically conjugate.