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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 ...
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
If is an odd prime and <, then (,) =. Proof: There are exactly two factors of in the numerator of the expression () = ()! / (!), coming from the two terms and in ()!, and also two factors of in the denominator from one copy of the term in each of the two factors of !.
Download as PDF; Printable version; ... Fermat's identity or Chu's Theorem, [3] ... Further, by the binomial theorem, we also find that
Thus many identities on binomial coefficients carry over to the falling and rising factorials. The rising and falling factorials are well defined in any unital ring , and therefore x {\displaystyle x} can be taken to be, for example, a complex number , including negative integers, or a polynomial with complex coefficients, or any complex-valued ...
In addition to algebra, coordinate geometry, Pythagorean theorem, trigonometry and calculus, which were on the previous specification, this course also includes: 'Enumeration' content, which expands the topic of the binomial distribution to include permutations and combinations
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
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