A special function

I need a function so that f^-1(x) (the inverse function) is 1/f(x). I've been thinking about it but can't find any.

The definition of an inverse function is if f^-1(x) is the inverse of f(x), then f(f^-1(x)) = f^-1(f(x)) = x

You're looking for a function for which:
f^-1(x) = 1/f(x)
f^-1(x)*f(x) = 1

If we use the one-to-one mapping x → f(x):

f^-1(f(x))*f(f(x)) = 1
x*f(f(x))= 1
f(f(x)) = 1/x
f(x) = f^-1(1/x)

If we use the one-to-one mapping x → 1/x, then:

f(1/x) = f^-1(x)

also holds true. Also, earlier we said that:

f^-1(x)*f(x) = 1

Therefore:

f(1/x)*f(x) = 1

I'm not sure how to isolate this, but for the case f(1/x) = f(x), then:

f(x)^2 = 1
f(x) = ±1

This fits your relation because the inverse of f(x) = 1 is x = 1, and the inverse of f(x) = -1 is x = -1. You wanted a function such that f^-1(x) = 1/f(x).

For the first case:
1 = 1/1

For the second case:
-1 = 1/(-1)

Note that this isn't a true inverse, since x = -1 is a vertical line. I'm not sure how to find other cases. :/

Edit: I think I've found another solution. Earlier, I arrived at f(f(x)) = 1/x = x^(-1). The only way this is possible is if we let f(x) = x^a, since the nested function must be of the same form. Therefore:

(x^(a))^a = x^(-1)
x^(a*a) = x^(-1)
a^2 = -1
a = ±i

Thus, f(x) = x^±i are solutions, though they are complex. Notice that they also fit this relation:

f(1/x)*f(x) = 1

For the first case:
(1/x)^i * x*i = x^(-i) * x^i = x^0 = 1

For the second case:
(1/x)^(-i) * x^(-i) = x^i * x^(-i) = x^0 = 1

Edit: Obviously, I'm too blind to read in the morning, I should get my cup of tea first! Thank you, shouri.

2³¹ - 1, you're very close!

is not correct. For x, you get . You're supposed to get x.

is not entirely correct, though there are minimal work to get it over to a correct solution. f or f⁻¹ is not defined for x = 0, so you have to specify that the domain for the functions are (-∞, 0) ∪ (0, ∞.

I'll have a look at it during breakfast.

Noah

I thought he was after a function and an inverse function which equal themselves. This will only limit your search for these functions (only to find no (correct) answers).

 Spoiler

You're doing some other fancy stuff too.

is not correct. For x, you get . You're supposed to get x.

is not correct either. For 2, you get (for both functions) . We're supposed to get 2.

Here's what he asked for:



Though, we need to have the inverse identities as you said.



The first obvious ones are f(x) = x and f(x) = -x, though I don't know if there are more simple ones. There are certainly infinitely many ones, though, like e.g.



Noah

I thought he wanted

f^-1 = 1/f

not
f^-1 = f

is not correct. For x, you get . You're supposed to get x.

Actually, it was f(x)=1 and f⁻¹(x) is the line x=1, so f⁻¹(x)≠1.

I did note that x=±1 are not true inverses of f(x)=±1, because x=±1 are vertical lines. However, here I used the fact that an inverse is really just a reflection of f(x) over the line f(x)=x, because it would be "silly" to exclude everything that is not one to one but can be represented in some way (such as parametrically).

x=1 is a reflection of f(x)=1 over the line f(x)=x, and even though f(f⁻¹(x))=1, keep in mind that the inverse of f(x) is the line x=1, which means that f(f⁻¹(x))=1=x. Better yet, we can use the limit process to see that it actually equals x. Let f(x) = Cx+h, where C is some constant. The inverse of f(x) is therefore f⁻¹(x) = (x-h)/C. Now observe that:

f(f⁻¹(x)) = f⁻¹(f(x)) = x

If we take the limit as C approaches 0, f(x) becomes the line f(x)=h and f⁻¹(x) becomes the line x=h. Therefore, technically this works. It's just not immediately apparent.

While this doesn't seem to satisfy the equation, we can observe that even if f⁻¹(x) is x=1 cannot truly be represented, it has a "form" of f⁻¹(x)=∞-y, where y is some impossible y-intercept. To see this, we can consider that even "better" technical solutions are:

1. f(x)=0 and f⁻¹(x)=∞, or x=0
2. f(x)=∞, or x=0, and f⁻¹(x)=0

Since it isn't too difficult to see that 1/∞=0, and vice versa.

Actually, it was f(x)=1 and f⁻¹(x) is the line x=1, so f⁻¹(x)≠1.

I did note that x=±1 are not true inverses of f(x)=±1, because x=±1 are vertical lines. However, here I used the fact that an inverse is really just a reflection of f(x) over the line f(x)=x, because it would be "silly" to exclude everything that is not one to one but can be represented in some way (such as parametrically).

x=1 is a reflection of f(x)=1 over the line f(x)=x, and even though f(f⁻¹(x))=1, keep in mind that the inverse of f(x) is the line x=1, which means that f(f⁻¹(x))=1=x. Better yet, we can use the limit process to see that it actually equals x. Let f(x) = Cx+h, where C is some constant. The inverse of f(x) is therefore f⁻¹(x) = (x-h)/C. Now observe that:

f(f⁻¹(x)) = f⁻¹(f(x)) = x

If we take the limit as C approaches 0, f(x) becomes the line f(x)=h and f⁻¹(x) becomes the line x=h. Therefore, technically this works. It's just not immediately apparent.

While this doesn't seem to satisfy the equation, we can observe that even if f⁻¹(x) is x=1 cannot truly be represented, it has a "form" of f⁻¹(x)=∞-y, where y is some impossible y-intercept. To see this, we can consider that even "better" technical solutions are:

1. f(x)=0 and f⁻¹(x)=∞, or x=0
2. f(x)=∞, or x=0, and f⁻¹(x)=0

Since it isn't too difficult to see that 1/∞=0, and vice versa.

I'm afraid you cannot use calculus in set theory, at least not that kind of calculus. I like your thoughts though!

A function f: X → Y has an inverse iff it has a one-to-one mapping from the set X to Y. If you change its set (as you would like to with parametrization), then it is not the same function (because, obviously, we're going from one set to another set not contained within the other set or visa versa.). They may be able to represent the same model, but they do not, at any point, equal each other, because they work in different sets and do different mappings.

An inverse is not a reflection, though it's nice to think of it as such for ℝ¹ and its subsets. It is, of course, okay to think of that function as a mapping from ℝ² → ℝ² (and subset friends), but then you have to specify that through the definition of f(x) in the ℝ²-subset, and not with f(x) in the ℝ¹-subset only. You also have to be very specific with the set, so that you map to the correct subsets. With the correct subsets, you see that they will not give you any more information than its ℝ¹-brother.

TL;DR: In order to have an invertible function:

• f must be an injective function - If f is continuous, then f must either be strictly increasing or decreasing in order to be injective.
  
• f(f⁻¹(x)) = f⁻¹(f(x)) = x, not their limits
  

Noah

Hmmm so there's no simple function that works like that? (Read: defined in Reals)

A function f: X → Y has an inverse iff it has a one-to-one mapping from the set X to Y.

If we're transforming a line, f(x)=1, we could reparametrize this line in vector form as [x,f(x)] = [t, 0]. Then if we subject this to a one-to-one matrix transformation A[x,f(x)]+B, where A and B are matrices with constant entries, we can arrive at [x,f(x)]=[0,t]. Even though x → f(x) is no longer one-to-one, the resulting matrix is one-to-one. :|

TL;DR: In order to have an invertible function:

• f must be an injective function - If f is continuous, then f must either be strictly increasing or decreasing in order to be injective.
  
• f(f⁻¹(x)) = f⁻¹(f(x)) = x, not their limits
  

What about periodic functions, like sin(x) and cos(x)? Are asin(x) and acos(x) somehow not true inverses in comparison to arcsin(x) and arccos(x)? (By asin(x) and acos(x), I mean the programming command that spans the entire range, unlike arcsin(x) and arccos(x), which restricts the domain.)

Hmmm so there's no simple function that works like that? (Read: defined in Reals)

 Spoiler

1. f(x)=0 and f⁻¹(x)=∞, or x=0
2. f(x)=∞, or x=0, and f⁻¹(x)=0

Alternatively, you could force f(x) = x^±i to return real numbers by having your domain be a set of complex numbers.

^You'd have to restrict the domain of sin(x) and cos(x) as well for arcsin(x) and arccos(x) to be one-one!

Hadriel

Hmm... I just realized something. If we restrict the domain of f(x)=x^2 to two sections, then we can find "two" inverses:

For f(x)=x^2, x>0
f⁻¹(x)=sqrt(x)

For f(x)=x^2, x<0
f⁻¹(x)= -sqrt(x)

Thus, when we put the two together, we once again see the reflection over the line f(x)=x. Since both of them are true inverses, the put together "pseudo"-inverse still as a whole satisfies being the inverse of f(x). If we had a periodic function such as f(x)=sin(x), then we can just restrict it infinitely many times so that the domains are one-to-one everywhere. When we put together all the pieces, we once again end up with the reflection across the line f(x)=x.

If we're transforming a line, f(x)=1, we could reparametrize this line in vector form as [x,f(x)] = [t, 0]. Then if we subject this to a one-to-one matrix transformation A[x,f(x)]+B, where A and B are matrices with constant entries, we can arrive at [x,f(x)]=[0,t]. Even though x → f(x) is no longer one-to-one, the resulting matrix is one-to-one. :|

The idea is that the injective function f(x) has an inverse so that f(x) = 1/f⁻¹(x). I don't see how this solves the problem, sadly.

 Spoiler

Alternatively, you could force f(x) = x^±i to return real numbers by having your domain be a set of complex numbers.

Well, infinity is not in ℝ The other idea of your works perfectly, though!

A simple function in a subset of ℝ is the function 2³¹ - 1 has mentioned for quite some time: f(x) = 1, f⁻¹(x) = 1 on the domain [1, 1]. It is unquestionably useless though.

Hmm... I just realized something. If we restrict the domain of f(x)=x^2 to two sections, then we can find "two" inverses:

For f(x)=x^2, x>0
f⁻¹(x)=sqrt(x)

For f(x)=x^2, x<0
f⁻¹(x)= -sqrt(x)

Thus, when we put the two together, we once again see the reflection over the line f(x)=x. Since both of them are true inverses, the put together "pseudo"-inverse still as a whole satisfies being the inverse of f(x). If we had a periodic function such as f(x)=sin(x), then we can just restrict it infinitely many times so that the domains are one-to-one everywhere. When we put together all the pieces, we once again end up with the reflection across the line f(x)=x.

Correct! Split the function at the local maximas and local minimas, as you'd then get a set of injective functions. Find the inverses, and voilà!

Noah