Search results
Results From The WOW.Com Content Network
There is a symmetry between a function and its inverse. Specifically, if f is an invertible function with domain X and codomain Y, then its inverse f −1 has domain Y and image X, and the inverse of f −1 is the original function f. In symbols, for functions f:X → Y and f −1:Y → X, [13]
If an invertible function is C k with >, then so too is its inverse. This follows by induction using the fact that the map F ( A ) = A − 1 {\displaystyle F(A)=A^{-1}} on operators is C k for any k {\displaystyle k} (in the finite-dimensional case this is an elementary fact because the inverse of a matrix is given as the adjugate matrix ...
In a semigroup, a left-invertible element is left-cancellative, and analogously for right and two-sided. If a −1 is the left inverse of a, then a ∗ b = a ∗ c implies a −1 ∗ (a ∗ b) = a −1 ∗ (a ∗ c), which implies b = c by associativity. For example, every quasigroup, and thus every group, is cancellative.
In calculus, the inverse function rule is a formula that expresses the derivative of the inverse of a bijective and differentiable function f in terms of the derivative of f. More precisely, if the inverse of f {\displaystyle f} is denoted as f − 1 {\displaystyle f^{-1}} , where f − 1 ( y ) = x {\displaystyle f^{-1}(y)=x} if and only if f ...
By convention, f 0 is defined as the identity map on f 's domain, id X. If Y = X and f: X → X admits an inverse function f −1, negative functional powers f −n are defined for n > 0 as the negated power of the inverse function: f −n = (f −1) n. [12] [10] [11]
Since and the inverse function : are continuous, they have antiderivatives by the fundamental theorem of calculus. Laisant proved that if F {\displaystyle F} is an antiderivative of f {\displaystyle f} , then the antiderivatives of f − 1 {\displaystyle f^{-1}} are:
If an open map f has an inverse function, that inverse is continuous, and if a continuous map g has an inverse, that inverse is open. Given a bijective function f between two topological spaces, the inverse function f − 1 {\displaystyle f^{-1}} need not be continuous.
A T ∈ L(H) is a Fredholm operator if and only if T is invertible modulo compact perturbation, i.e. TS = I + C 1 and ST = I + C 2 for some bounded operator S and compact operators C 1 and C 2. In other words, an operator T ∈ L(H) is Fredholm, in the classical sense, if and only if its projection in the Calkin algebra is invertible.