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If a real function has a domain that is self-symmetric with respect to the origin, it may be uniquely decomposed as the sum of an even and an odd function, which are called respectively the even part (or the even component) and the odd part (or the odd component) of the function, and are defined by = + (), and = ().
An odd function, such as an odd power of a variable, gives for any argument the negation of its result when given the negation of that argument. It is possible for a function to be neither odd nor even, and for the case f ( x ) = 0, to be both odd and even. [ 20 ]
In Boolean algebra, a parity function is a Boolean function whose value is one if and only if the input vector has an odd number of ones. The parity function of two inputs is also known as the XOR function. The parity function is notable for its role in theoretical investigation of circuit complexity of Boolean functions.
The permutation is odd if and only if this factorization contains an odd number of even-length cycles. Another method for determining whether a given permutation is even or odd is to construct the corresponding permutation matrix and compute its determinant. The value of the determinant is the same as the parity of the permutation.
The cube operation can also be defined for any other mathematical expression, for example (x + 1) 3. The cube is also the number multiplied by its square: n 3 = n × n 2 = n × n × n. The cube function is the function x ↦ x 3 (often denoted y = x 3) that maps a number to its cube. It is an odd function, as (−n) 3 = −(n 3).
This is useful, for example, ... To use these approximations for negative x, use the fact that erf x is an odd function, so erf x = −erf ...
The Borsuk–Ulam theorem is equivalent to the following statement: A continuous odd function from an n-sphere into Euclidean n-space has a zero. PROOF: If the theorem is correct, then it is specifically correct for odd functions, and for an odd function, () = iff () =. Hence every odd continuous function has a zero.
The choice between odd and even is typically motivated by boundary conditions associated with a differential equation satisfied by (). Example Calculate the half range Fourier sine series for the function f ( x ) = cos ( x ) {\displaystyle f(x)=\cos(x)} where 0 < x < π {\displaystyle 0<x<\pi } .