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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 ...
Relationship to the binomial theorem [ edit ] The Leibniz rule bears a strong resemblance to the binomial theorem , and in fact the binomial theorem can be proven directly from the Leibniz rule by taking f ( x ) = e a x {\displaystyle f(x)=e^{ax}} and g ( x ) = e b x , {\displaystyle g(x)=e^{bx},} which gives
The generalized binomial theorem gives (+) = = = + + +.A proof for this identity can be obtained by showing that it satisfies the differential equation ...
The Gaussian binomial coefficient, written as () or [], is a polynomial in q with integer coefficients, whose value when q is set to a prime power counts the number of subspaces of dimension k in a vector space of dimension n over , a finite field with q elements; i.e. it is the number of points in the finite Grassmannian (,).
2.2.2 Proof by binomial theorem ... Download as PDF; Printable version; ... This can be generalized to rational exponents of the form / ...
The binomial series is therefore sometimes referred to as Newton's binomial theorem. Newton gives no proof and is not explicit about the nature of the series. Later, on 1826 Niels Henrik Abel discussed the subject in a paper published on Crelle's Journal, treating notably questions of convergence. [4]
in which form it is clearly recognizable as an umbral variant of the binomial theorem (for more on umbral variants of the binomial theorem, see binomial type). The Chu–Vandermonde identity can also be seen to be a special case of Gauss's hypergeometric theorem, which states that
These "generalized binomial coefficients" appear in Newton's generalized binomial theorem. For each k, the polynomial () can be characterized as the unique degree k polynomial p(t) satisfying p(0) = p(1) = ⋯ = p(k − 1) = 0 and p(k) = 1. Its coefficients are expressible in terms of Stirling numbers of the first kind: