Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the implicit equation $ f^{-1} (x)=g(x)$. The function $g(x)$ is known and at least can be computed numerically. It may be piecewise continous or oscillating but it is always positive $ g(x) \ge 0 $. Here $ f(x) $ is not known.

Could it be that is there a function $ g(x) $ so it is NEVER invertible and hence we cannot get $ f(x) $?

share|improve this question
6  
What about $g(x)=0$? –  Alex Becker Apr 29 '12 at 20:32
    
More generally: the inverse of a function is always one-to-one on its domain. –  Robert Israel Apr 29 '12 at 20:36
    
$ g(x)=0 $ it can be viewed as the set of points $ (x,0) $ so its inverse would be the set of points $ (0,x) $ or more generally perhaps $ x=0 $ –  Jose Garcia Apr 29 '12 at 20:41
1  
@JoseGarcia You seem to be confusing "inverse" with "pre-image". The former doesn't necessarily always exist (as in the case g(x) = 0) but the latter does. An "inverse" by definition must be a function, and in particular cannot be "one-to-many" –  Jonathan Apr 29 '12 at 21:45
1  
@JoseGarcia Note that $x=0$ is not a function. How would you write it as $y=\text{something}$? –  Pedro Tamaroff Apr 29 '12 at 22:31

1 Answer 1

This might be useful for you:

DEFINITION: Let $f:A\to B$ and $g:B \to A$ be given. The function $f$ is called the inverse of $g$ and the function $g$ is called the inverse of $f$ if $g(f(a))=a$ for each $a \in A$ and $f(g(b))=b$ for each $b \in B$. In this event we sall also say that $f$ and $g$ are inverse functions and that each of the is invertible.

It is a consequence of this definition that if $f$ and $g$ are inverses, then both of them are one-one and onto:

  1. $f$ is one-one if you have $x,y \in A$ then $f(x)=f(y)\Leftrightarrow x=y$.
  2. $f$ is onto if $f(A)=B$.

As a general result, it is necessary and sufficient that $f$ is onto and one-one for $f$ to be invertible.

An example is the definition of $\arcsin x$. To define it, we must first change

$$f:\mathbb R \to \mathbb R\text{ ; } x\mapsto\sin x$$

to

$$f:\left[-\frac {\pi} 2, \frac {\pi} 2\right] \to [-1,1]\text{ ; } x\mapsto\sin x$$

Since in such definition, $\sin x$ is onto and one-one, it follows we can define

$$g: [-1,1]\to \left[-\frac {\pi} 2, \frac {\pi} 2\right] \text{ ; } x\mapsto\arcsin x$$

Directly answering your question. Let

  1. $p$: $f$ is invertible.
  2. $q$: $f$ is onto.
  3. $r$: $f$ is one-one.

Then

$$(q\wedge r) \equiv p $$ or

$$(-q\vee -r) \equiv -p $$ I reccomend you read Chapter 1 of Introduction to Topology by Bert Mendelson.

share|improve this answer
    
Reading a topology book seems a little overkill! –  The Chaz 2.0 Apr 29 '12 at 22:55
    
@TheChaz Not really! The first chapter is merely Theory of Sets and gives a great introduction to functions and relations in general. –  Pedro Tamaroff Apr 29 '12 at 22:56
    
Ah. Key words - first chapter :D –  The Chaz 2.0 Apr 29 '12 at 23:10
    
@TheChaz Indeed. Chapter 1. Are you still around mymathforum? I have stopped visiting it. –  Pedro Tamaroff Apr 29 '12 at 23:10
    
I mostly just moderate and add snide comments (which might be counterproductive...). I've learned a WHOLE lot more from MSE in the past year! –  The Chaz 2.0 Apr 29 '12 at 23:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.