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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
As with the (non-q) Chu–Vandermonde identity, there are several possible proofs of the q-Vandermonde identity.The following proof uses the q-binomial theorem.. One standard proof of the Chu–Vandermonde identity is to expand the product (+) (+) in two different ways.
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 ...
Pascal's triangle, rows 0 through 7. The hockey stick identity confirms, for example: for n=6, r=2: 1+3+6+10+15=35.. In combinatorics, the hockey-stick identity, [1] Christmas stocking identity, [2] boomerang identity, Fermat's identity or Chu's Theorem, [3] states that if are integers, then
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
In mathematics, Pascal's rule (or Pascal's formula) is a combinatorial identity about binomial coefficients.It states that for positive natural numbers n and k, + = (), where () is a binomial coefficient; one interpretation of the coefficient of the x k term in the expansion of (1 + x) n.
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} .
Since a binomial coefficient is always an integer, the nth binomial coefficient is divisible by p and hence equal to 0 in the ring. We are left with the zeroth and pth coefficients, which both equal 1, yielding the desired equation. Thus in characteristic p the freshman's dream is a valid identity.