反函數

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在數學裡，反函數為對一給定函數做逆運算的函數。更正式些地說，若f為一定義域為X的函數，則f ⁻¹為其反函數若且唯若對每一 $x \in X$ ，都會有：

$f^{-1}(f(x))=f(f^{-1}(x))=x.\,$

例如，若給定一函數x → 3x + 2，則其反函數為x → (x−2) / 3。這通常寫成：

$f\colon x\to 3x+2$

$f^{-1}\colon x\to(x-2)/3$

上標"−1"指的並不是冪。類似地，只要不在三角學或微積分裡，f ²(x)會是指「作用f兩次」，即為f(f(x))，而不是指f(x)的平方。例如，若f : x → 3x + 2，則f ² : x = 3 ((3x + 2)) + 2 = 9x + 8。但在三角學裡，因為歷史上的原因，sin²(x)通常確實是指sin(x)的平方。而字首arc有時則被用來標記反三角函數，如arcsin x為sin(x)的逆運算。在微積分裡，f ⁽ⁿ⁾(x)是用來指f的n次微分的。

若一函數有反函數，此函數便稱為可逆的。

[编辑] 簡單規則

一般而言，當f(x)為一任意函數，且g為其反函數，則g(f(x)) = x且f(g(x)) = x。換句話說，一反函數會取消原函數的作用。在上述例子，可以證明f⁻¹確為反函數，以將(x − 2) / 3代入f的方式，如此

3(x − 2) / 3 + 2 = x。

相似地，也可以將f代入f⁻¹來證明。

確實，f的反函數g的一等價定義，為需要g o f為於f定義域上的恆等函數，且f o g為f陪域上的恆等函數，其中的"o"表示函數複合。

[编辑] 存在性

一函數f若要是一明確的反函數，它必須是一雙射函數，即：

(滿射)陪域上的每一元素都必須被f映射到：不然將沒有辦法對某些元素定義f的反函數。
(單射)陪域上的每一元素都必須只被f映射到一次：不然其反函數將必須將元素映射到超到一個的值上去。

若f為一實變函數，則若f有一明確反函數，它必通過水平線測試，即一放在f圖上的水平線 $y = k$ 必對所有實數k，通過且只通過一次。

It is possible to work around this condition, by redefining f's codomain to be precisely its range, and by admitting a multi-valued function as an inverse.

If one represents the function f graphically in an x-y coordinate system, then the graph of f ⁻¹ is the reflection of the graph of f across the line y = x.

Algebraically, one computes the inverse function of f by solving the equation

y = f (x)

for x, and then exchanging y and x to get

y = f - 1 (x)

This is not always easy; if the function f(x) is analytic, the Lagrange inversion theorem may be used.

The symbol f ⁻¹ is also used for the (set valued) function associating to an element or a subset of the codomain, the inverse image of this subset (or element, seen as a singleton).

[编辑] 性質

When an inverse function exists, it is unique.
$(f\circ g)^{-1}=g^{-1}\circ f^{-1}$ provided all indicated compositions and inverses exist.
The inverse function and the inverse image of a set coincide in the following sense. Suppose $f - 1 (A)$ is the inverse image of a set $A\subset Y$ under a function $f:X\to Y.$ If $f$ is a bijection, then $f - 1 (y) = f - 1 ({y}).$
A linear mapping between vector spaces is invertible if and only if the determinant of the mapping is nonzero.
For functions between Euclidean spaces, the inverse function theorem gives a sufficient condition for the inverse to exist.

[编辑] Left inverses, right inverses, and partial functions

A function f has at least one "left inverse" if and only if it is an injection. A left inverse is a function g such that
$g (f (x)) = x$ .
If f is not a surjection, we obtain g by setting g(f(x)) = x for each element in the range of x, and g(y) = z, where z is any element whatever, for any y in the codomain of f but not in its range.

The same way, f has at least one "right inverse" ( $f (g (x)) = x$ ) if and only if it is a surjection. Here, for each x, g assigns one of the elements in the domain of f which "produce" x. For example, we know that $f (x) = x 2$ is a surjection from $\R$ to $\R^+$ . Then, $g(x) = -\sqrt{x}$ is a famous right inverse to $x 2$ , because $(-\sqrt{x})^2 = x$ for all $x \in R^+$ . But it is not a left inverse: $-\sqrt{x^2} = -x$ for $x \in R^+$ .

If f is a bijection, then the (unique) right inverse equals the left inverse, and we have come again to the ordinary inverse described above.

Using this definition, we can view any partial function as a left inverse of an injection. Because the range of a left inverse is not restricted, we can adjoin to the domain of this injection an element "undefined", which we then assign to every element of the codomain which is not in the range.