Reciprocal application

Example

Let us consider the function

f: X \ mapsto 3x+2

One poses:

y = 3x + 2

One reverses the couple (X, there) :

x = 3y + 2

And one isolates there :

x - 2 = 3y \ Leftrightarrow \ frac {x-2} {3} = y

The reciprocal application is thus:

f^ {- 1}: X \ mapsto \ frac {x-2} {3}

The exhibitor “-1” is not a power and $f^ {- 1}$ does not correspond contrary to a function for the multiplication, but contrary for the Composition to functions. One finds also the notations ${} ^r \! f$ and $f^r$ which raises this ambiguity.

In fact, so that a function F admits a reciprocal application, it must be bijective:

each element of the whole of arrival must be reached by F : if not it would not be possible to define the image by $f^ {- 1}$ of certain elements.
each element of the whole of arrival must be reached only once by F : if not the reciprocal application would send this element on more than only one value.

Formal definition

Formally, the reciprocal application of a bijective application F of a unit X on a unit Y , is the noted application F ^-1 which with an element there of the Ensemble of arrival Y , associates single the previous X of there by F .

for all X of X , $f^ {- 1} (F (X)) = X \,$ , because $f (X)$ has for single antecedent X
for all there in Y , $f (f^ {- 1} (there)) = there \,$ , because F sends the single antecedent of there on there .

What we can write:

f^ {- 1} \ circ f= {\ rm Id} _ {X}

and

f \ circ f^ {- 1} = {\ rm Id} _ {Y}

It is possible to define the reciprocal application of a function not inevitably bijective, by considering the application G in the same way together of definition as F whose whole of arrival is restricted with the image of F and who sends an element on the image of this element by F ; the reciprocal application is then the multiform application which with an element of the image of F associates its antecedents by F .

Are I and J two parts of $\ R$ and $f: I \ rightarrow J \,$ a bijective function. If we represent graphically the function F in a Cartesian reference mark, then the graph of $f^ {- 1}$ is the symmetrical orthogonal one of that of F compared to the line of equation there = X .

Algebraically, we determine the reciprocal application of F by solving the equation

y = F (X)

of unknown factor X , and by exchanging there and X to obtain

y = f^ {- 1} (X)

That is not always easy or possible.

If the function F is analytical, then the Théorème of inversion of Lagrange can be used.

Derivation

That is to say F a bijection of I in J . If F admits a nonnull derivative

f'

on I, then its reciprocal function

f^ {- 1}

for derived:

(f^ {- 1}) ': \ begin {array} {lcl} J & \ rightarrow & I \ \

X & \ mapsto & \ frac {1} {(f' \ circ f^ {- 1}) (X)} \end{array}

Demonstration: For any X of J,

(F \ circ f^ {- 1}) (X) = x

While deriving from the two with dimensions ones:

(f' \ circ f^ {- 1}) (X) (f^ {- 1}) '(X) = 1

That is to say

(f^ {- 1}) “= \ frac {1} {(F” \ circ f^ {- 1}) (X)}

One thus easily notices the need for having

f'

nonnull, if not one would find oneself with a division by 0.

See too

Simple: Opposite function

Random links: