Understanding Inverse Functions and Notation Confusions

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Discussion Overview

The discussion revolves around the properties and implications of inverse functions, particularly focusing on the notation f^{-1}(x) and its potential equivalence to 1/f(x). Participants explore whether there exist functions for which f^{-1}(x) equals 1/f(x), delving into the implications of such relationships and the conditions under which they hold.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses initial confusion about the notation f^{-1}(x), mistakenly equating it with 1/f(x) and questions if any function satisfies f^{-1}(x) = 1/f(x).
  • Another participant suggests that exploring the equation could lead to finding such functions, hinting at the relationship between f(a), f(b), and their inverses.
  • Discussion includes the assertion that if f(a) = b, then f^{-1}(b) = a, and participants explore the implications of this relationship.
  • One participant proposes that if f^{-1}(b) = 1/f(b), then it leads to a series of equations involving f(a) and f(b), but expresses uncertainty about how to derive values directly.
  • Another participant emphasizes the need to assume the domain of f does not include zero, as it leads to contradictions in the equations.
  • Participants compute various values of f and its inverse based on earlier discussions, but some express uncertainty about the rigor of their conclusions.
  • There is a suggestion to explore orderings of values to determine the continuity and invertibility of f, leading to further questions about specific values of f(1) and f(-1).
  • One participant concludes that organizing values into groups could help construct a function satisfying the discussed properties.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether a function exists that satisfies f^{-1}(x) = 1/f(x). The discussion remains unresolved, with various hypotheses and explorations presented without definitive conclusions.

Contextual Notes

Participants note the importance of domain restrictions, particularly the exclusion of zero, which affects the validity of the equations discussed. There are also indications of missing rigorous proofs for some claims made during the discussion.

murshid_islam
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there have been a thread going on about worst/best notations. as i said there, i was confused with [tex]f^{-1}(x)[/tex] when i first came across it in high school. i thought [tex]f^{-1}(x)[/tex] is the same as [tex]{1}\over{f(x)}[/tex]

but now i am wondering, is there any function for which
[tex]f^{-1}(x)[/tex] = [tex]{1}\over{f(x)}[/tex]?
 
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If you just play with that equation, I think it's easy enough to see how to find such functions.

i.e. if f(a) = b, then what do you know about f^-1(b)? An about f(b)?

also, if you pick an ordering on things, to see that f cannot be a continuous function.
 
Hurkyl said:
if f(a) = b, then what do you know about f^-1(b)? An about f(b)?
if f(a) = b, then f^-1(b) = a.
and f(b) = f(f(a))
but where is it going?

Hurkyl said:
also, if you pick an ordering on things, to see that f cannot be a continuous function.
i am afraid i didn't get it.
 
murshid_islam said:
and f(b) = f(f(a))
You can do better than that. (And once you see it, keep going for a while)
 
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f(b)=f(f(f^-1(b)))
perhaps this what hurkyl meant.
 
Nope. He has the lovely equation

[tex]f^{-1}(x) = \frac{1}{f(x)}[/tex]

which I expect him to use sometime while doing this problem.
 
ok, if f^-1(b) = 1/f(b), then
a = 1/f(f(a))
f(f(a)) = 1/a
f(a) = f^-1(1/a)

but i am still in the dark.
 
With the work you've done there, you should be able to tell me the values of

f(b)

and

f^(-1) (1/a)

directly, rather than leave them in terms of other values of f.
 
Hurkyl said:
With the work you've done there, you should be able to tell me the values of f(b) and f^(-1) (1/a) directly, rather than leave them in terms of other values of f.
in my last post, i got
f(a) = f^-1(1/a)
then f(b) = f(f(a)) = f(f^-1(1/a)) = 1/a = 1/f^-1(b)

now we have f(b) = 1/f^-1(b)

but, how can i tell the values of f(b) and f^-1(1/a) directly, rather than leave them in terms of other values of f?
 
  • #10
Argh, I have to go; I was hoping to keep giving you hints so you could figure out all the important steps yourself. I'll write out what I had been planning to say with answers... but please try to answer the questions yourself before reading my answer, and moving onto the next question! (I'll put the answers in white -- highlight for easier reading)


(From here on, I'm going to set g = f^-1, to make it easier to write)

There is a loose end I missed before -- we have to assume the domain of our function does not contain zero. Do you see why?

--------------

f(0) = a
g(a) = 0
g(a) = 1/f(a)
1/f(a) = 0 <==== there is no possible value of f(a) that works here
[/color]
--------------

So, for the rest of this post, let's assume that we do not require 0 to be in the domain of f.



We know f(a) = b and g(b) = a. I've already said you can tell me the values of f(b) and g(1/a). What are they?

----------

f(b) = 1/a
g(1/a) = b
[/color]
----------


Are there any other values of f or g that you can compute?

----------

yes: we can compute both f and g at the points a, b, 1/a, 1/b.
[/color]----------



What are those values?

----------

f(a) = b
f(b) = 1/a
f(1/a) = 1/b
f(1/b) = a
g(a) = 1/b
g(b) = a
g(1/a) = b
g(1/b) = 1/a
[/color]----------


Do you think your equation g(x) = 1/f(x) can be used to relate the values we just computed to anything else?

----------

No. I don't have a rigorous proof of this, but it seems clear to be just from playing with the equation.
----------


So, we know that if we pick the value b for f(a), then this forces upon us a particular choice of f(b), f(1/a), and f(1/b), and nothing else.

Play with some possible orderings of a and b, such as:

1/b < 1/a < a < b

and plot the values of f at those points. If f is continuous, can it possibly be invertible?



Can you tell me the possible values of f(1) and f(-1)? (Hint: we've already done the work: set a = 1, and see what that tells us)

----------

f(a) = b, so f(1) = b
f(1/a) = 1/b, so f(1) = 1/b
So, b = 1/b, and thus b = 1, or b = -1.
The same is true if we set a = -1.
So, we can either choose f(1) = 1 and f(-1) = -1, or we can choose f(1) = -1 and f(-1) = 1
[/color]
----------


Now, what about the rest of the values of f? All we have to do is to organize the rest of the numbers into groups of four, of the form: {a, b, 1/a, 1/b}, then define f(a) = b, et cetera. It seems clear to me that there is no problem in doing this organization, and thus we've constructed an f satisfying your equation.
 
  • #11
thanks. i also have to go now. i will sit with your hints tomorrow and try to solve it myself. wish me luck :)
 
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