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Simplification is the process of replacing a mathematical expression by an equivalent one that is simpler (usually shorter), according to a well-founded ordering. Examples include:
As there is zero X n+1 or X −1 in (1 + X) n, one might extend the definition beyond the above boundaries to include () = when either k > n or k < 0. This recursive formula then allows the construction of Pascal's triangle , surrounded by white spaces where the zeros, or the trivial coefficients, would be.
The inverse Langevin function (L −1 (x)) is without an explicit analytical form, but there exist several approximations. [21] The inverse Langevin function L −1 (x) is defined on the open interval (−1, 1). For small values of x, it can be approximated by a truncation of its Taylor series [22]
The Barzilai-Borwein method [1] is an iterative gradient descent method for unconstrained optimization using either of two step sizes derived from the linear trend of the most recent two iterates.
Trigonometric identities may help simplify the answer. [ 1 ] [ 2 ] Like other methods of integration by substitution, when evaluating a definite integral, it may be simpler to completely deduce the antiderivative before applying the boundaries of integration.
In complex analysis of one and several complex variables, Wirtinger derivatives (sometimes also called Wirtinger operators [1]), named after Wilhelm Wirtinger who introduced them in 1927 in the course of his studies on the theory of functions of several complex variables, are partial differential operators of the first order which behave in a very similar manner to the ordinary derivatives ...
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 n ≥ r ≥ 0 {\displaystyle n\geq r\geq 0} are integers, then
The idea becomes clearer by considering the general series 1 − 2x + 3x 2 − 4x 3 + 5x 4 − 6x 5 + &c. that arises while expanding the expression 1 ⁄ (1+x) 2, which this series is indeed equal to after we set x = 1.