Mixing
A function satisfying $f left ( frac 1 f(x) right ) = x$ [duplicate]
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This question already has an answer here:
Is there a function whose inverse is exactly the reciprocal of the function, that is $f^-1 = frac1f$?
6 answers
Background. This question originates from the problem of finding a function $f$ such that its $n$-th iterate is equal to its $n$-th power, which I asked about here. Now I would like to focus on the related case $n = -1$, because it seems to be quite a different problem. In other words, I am looking for a function $f$ such that its compositional inverse is equal to its multiplicative inverse, that is,
$$f^-1(x) = frac 1 f(x).$$
It is immediate to see that this is equivalent to
$$f left ( frac 1 f(x) right ) = x$$
which is the equation in the title.
I believe I have found an example of a function $f colon mathbb Q^+ to mathbb Q^+$ satisfying the equation, but it is quite convoluted. The construction is as follows:
Let $x in mathbb Q^+$. Then we can write
$$x = left ( frac p_1^a_1 dotsb p_h^a_h q_1^b_1 dotsb q_k^b_k right )^n$$
where $p_1, dotsc, p_h, q_1, dotsc, q_k$ are distinct prime factors, $p_1^a_1 dotsb p_h^a_h le q_1^b_1 dotsb q_k^b_k$, the exponents $a_1, dotsc, a_h, b_1, dotsc, b_k$ are coprime positive integers, and $n in mathbb Z$. Then we define
$$rho colon mathbb Q^+ to mathbb Q^+ qquad rho (x) = frac p_1^a_1 dotsb p_h^a_h q_1^b_1 dotsb q_k^b_k.$$ Clearly the image $rho(mathbb Q^+)$ is countably infinite, so we can fix two sequences $(r_m)_m in mathbb N$ and $(s_m)_m in mathbb N$ such that:
$$r_0, s_0, r_1, s_1, dotsc, r_m, s_m, dotsc$$
is an injective enumeration of $rho(mathbb Q^+)$.
Notice then that each $x in mathbb Q^+$ is either $r_m^n$ or $s_m^n$ for some $m in mathbb N$ and $n in mathbb Z$, in a unique way. Therefore, we define
$$f(r_m^n) = s_m^n qquad f(s_m^n) = r_m^-n$$
for all $m in mathbb N$ and $n in mathbb Z$. With this definition, we can check that the above property is satisfied:
$$f left ( frac 1 f(r_m^n) right ) = f left ( frac 1 s_m^n right ) = f (s_m^-n) = r_m^n$$
$$f left ( frac 1 f(s_m^n) right ) = f left ( frac 1 r_m^-n right ) = f (r_m^n) = s_m^n.$$
My questions are:
- Are there any less artificial functions satisfying the equation?
- If so, is there such a function that is defined on an interval of $mathbb R$?
Remark. If $f$ is defined on an interval $I$ of $mathbb R$ and is always positive (or negative), then it is monotonic. Indeed, if $f$ were increasing we would have
$$x_1 < x_2 implies f(x_1) < f(x_2) implies frac 1 f(x_1) > frac 1 f(x_2) implies x_1 > x_2$$
and similarly if $f$ were decreasing.
functions functional-equations inverse-function
asked Aug 5 at 21:43
Luca Bressan
3,86621037
marked as duplicate by 6005, Brahadeesh, Community♦ Aug 6 at 6:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
11
down vote
favorite
This question already has an answer here:
Is there a function whose inverse is exactly the reciprocal of the function, that is $f^-1 = frac1f$?
6 answers
Background. This question originates from the problem of finding a function $f$ such that its $n$-th iterate is equal to its $n$-th power, which I asked about here. Now I would like to focus on the related case $n = -1$, because it seems to be quite a different problem. In other words, I am looking for a function $f$ such that its compositional inverse is equal to its multiplicative inverse, that is,
$$f^-1(x) = frac 1 f(x).$$
It is immediate to see that this is equivalent to
$$f left ( frac 1 f(x) right ) = x$$
which is the equation in the title.
I believe I have found an example of a function $f colon mathbb Q^+ to mathbb Q^+$ satisfying the equation, but it is quite convoluted. The construction is as follows:
Let $x in mathbb Q^+$. Then we can write
$$x = left ( frac p_1^a_1 dotsb p_h^a_h q_1^b_1 dotsb q_k^b_k right )^n$$
where $p_1, dotsc, p_h, q_1, dotsc, q_k$ are distinct prime factors, $p_1^a_1 dotsb p_h^a_h le q_1^b_1 dotsb q_k^b_k$, the exponents $a_1, dotsc, a_h, b_1, dotsc, b_k$ are coprime positive integers, and $n in mathbb Z$. Then we define
$$rho colon mathbb Q^+ to mathbb Q^+ qquad rho (x) = frac p_1^a_1 dotsb p_h^a_h q_1^b_1 dotsb q_k^b_k.$$ Clearly the image $rho(mathbb Q^+)$ is countably infinite, so we can fix two sequences $(r_m)_m in mathbb N$ and $(s_m)_m in mathbb N$ such that:
$$r_0, s_0, r_1, s_1, dotsc, r_m, s_m, dotsc$$
is an injective enumeration of $rho(mathbb Q^+)$.
Notice then that each $x in mathbb Q^+$ is either $r_m^n$ or $s_m^n$ for some $m in mathbb N$ and $n in mathbb Z$, in a unique way. Therefore, we define
$$f(r_m^n) = s_m^n qquad f(s_m^n) = r_m^-n$$
for all $m in mathbb N$ and $n in mathbb Z$. With this definition, we can check that the above property is satisfied:
$$f left ( frac 1 f(r_m^n) right ) = f left ( frac 1 s_m^n right ) = f (s_m^-n) = r_m^n$$
$$f left ( frac 1 f(s_m^n) right ) = f left ( frac 1 r_m^-n right ) = f (r_m^n) = s_m^n.$$
My questions are:
- Are there any less artificial functions satisfying the equation?
- If so, is there such a function that is defined on an interval of $mathbb R$?
Remark. If $f$ is defined on an interval $I$ of $mathbb R$ and is always positive (or negative), then it is monotonic. Indeed, if $f$ were increasing we would have
$$x_1 < x_2 implies f(x_1) < f(x_2) implies frac 1 f(x_1) > frac 1 f(x_2) implies x_1 > x_2$$
and similarly if $f$ were decreasing.
functions functional-equations inverse-function
asked Aug 5 at 21:43
Luca Bressan
3,86621037
marked as duplicate by 6005, Brahadeesh, Community♦ Aug 6 at 6:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
11
down vote
favorite
up vote
11
down vote
favorite
This question already has an answer here:
Is there a function whose inverse is exactly the reciprocal of the function, that is $f^-1 = frac1f$?
6 answers
Background. This question originates from the problem of finding a function $f$ such that its $n$-th iterate is equal to its $n$-th power, which I asked about here. Now I would like to focus on the related case $n = -1$, because it seems to be quite a different problem. In other words, I am looking for a function $f$ such that its compositional inverse is equal to its multiplicative inverse, that is,
$$f^-1(x) = frac 1 f(x).$$
It is immediate to see that this is equivalent to
$$f left ( frac 1 f(x) right ) = x$$
which is the equation in the title.
I believe I have found an example of a function $f colon mathbb Q^+ to mathbb Q^+$ satisfying the equation, but it is quite convoluted. The construction is as follows:
Let $x in mathbb Q^+$. Then we can write
$$x = left ( frac p_1^a_1 dotsb p_h^a_h q_1^b_1 dotsb q_k^b_k right )^n$$
where $p_1, dotsc, p_h, q_1, dotsc, q_k$ are distinct prime factors, $p_1^a_1 dotsb p_h^a_h le q_1^b_1 dotsb q_k^b_k$, the exponents $a_1, dotsc, a_h, b_1, dotsc, b_k$ are coprime positive integers, and $n in mathbb Z$. Then we define
$$rho colon mathbb Q^+ to mathbb Q^+ qquad rho (x) = frac p_1^a_1 dotsb p_h^a_h q_1^b_1 dotsb q_k^b_k.$$ Clearly the image $rho(mathbb Q^+)$ is countably infinite, so we can fix two sequences $(r_m)_m in mathbb N$ and $(s_m)_m in mathbb N$ such that:
$$r_0, s_0, r_1, s_1, dotsc, r_m, s_m, dotsc$$
is an injective enumeration of $rho(mathbb Q^+)$.
Notice then that each $x in mathbb Q^+$ is either $r_m^n$ or $s_m^n$ for some $m in mathbb N$ and $n in mathbb Z$, in a unique way. Therefore, we define
$$f(r_m^n) = s_m^n qquad f(s_m^n) = r_m^-n$$
for all $m in mathbb N$ and $n in mathbb Z$. With this definition, we can check that the above property is satisfied:
$$f left ( frac 1 f(r_m^n) right ) = f left ( frac 1 s_m^n right ) = f (s_m^-n) = r_m^n$$
$$f left ( frac 1 f(s_m^n) right ) = f left ( frac 1 r_m^-n right ) = f (r_m^n) = s_m^n.$$
My questions are:
- Are there any less artificial functions satisfying the equation?
- If so, is there such a function that is defined on an interval of $mathbb R$?
Remark. If $f$ is defined on an interval $I$ of $mathbb R$ and is always positive (or negative), then it is monotonic. Indeed, if $f$ were increasing we would have
$$x_1 < x_2 implies f(x_1) < f(x_2) implies frac 1 f(x_1) > frac 1 f(x_2) implies x_1 > x_2$$
and similarly if $f$ were decreasing.
functions functional-equations inverse-function
asked Aug 5 at 21:43
Luca Bressan
3,86621037
This question already has an answer here:
Is there a function whose inverse is exactly the reciprocal of the function, that is $f^-1 = frac1f$?
6 answers
Background. This question originates from the problem of finding a function $f$ such that its $n$-th iterate is equal to its $n$-th power, which I asked about here. Now I would like to focus on the related case $n = -1$, because it seems to be quite a different problem. In other words, I am looking for a function $f$ such that its compositional inverse is equal to its multiplicative inverse, that is,
$$f^-1(x) = frac 1 f(x).$$
It is immediate to see that this is equivalent to
$$f left ( frac 1 f(x) right ) = x$$
which is the equation in the title.
I believe I have found an example of a function $f colon mathbb Q^+ to mathbb Q^+$ satisfying the equation, but it is quite convoluted. The construction is as follows:
Let $x in mathbb Q^+$. Then we can write
$$x = left ( frac p_1^a_1 dotsb p_h^a_h q_1^b_1 dotsb q_k^b_k right )^n$$
where $p_1, dotsc, p_h, q_1, dotsc, q_k$ are distinct prime factors, $p_1^a_1 dotsb p_h^a_h le q_1^b_1 dotsb q_k^b_k$, the exponents $a_1, dotsc, a_h, b_1, dotsc, b_k$ are coprime positive integers, and $n in mathbb Z$. Then we define
$$rho colon mathbb Q^+ to mathbb Q^+ qquad rho (x) = frac p_1^a_1 dotsb p_h^a_h q_1^b_1 dotsb q_k^b_k.$$ Clearly the image $rho(mathbb Q^+)$ is countably infinite, so we can fix two sequences $(r_m)_m in mathbb N$ and $(s_m)_m in mathbb N$ such that:
$$r_0, s_0, r_1, s_1, dotsc, r_m, s_m, dotsc$$
is an injective enumeration of $rho(mathbb Q^+)$.
Notice then that each $x in mathbb Q^+$ is either $r_m^n$ or $s_m^n$ for some $m in mathbb N$ and $n in mathbb Z$, in a unique way. Therefore, we define
$$f(r_m^n) = s_m^n qquad f(s_m^n) = r_m^-n$$
for all $m in mathbb N$ and $n in mathbb Z$. With this definition, we can check that the above property is satisfied:
$$f left ( frac 1 f(r_m^n) right ) = f left ( frac 1 s_m^n right ) = f (s_m^-n) = r_m^n$$
$$f left ( frac 1 f(s_m^n) right ) = f left ( frac 1 r_m^-n right ) = f (r_m^n) = s_m^n.$$
My questions are:
- Are there any less artificial functions satisfying the equation?
- If so, is there such a function that is defined on an interval of $mathbb R$?
Remark. If $f$ is defined on an interval $I$ of $mathbb R$ and is always positive (or negative), then it is monotonic. Indeed, if $f$ were increasing we would have
$$x_1 < x_2 implies f(x_1) < f(x_2) implies frac 1 f(x_1) > frac 1 f(x_2) implies x_1 > x_2$$
and similarly if $f$ were decreasing.
This question already has an answer here:
Is there a function whose inverse is exactly the reciprocal of the function, that is $f^-1 = frac1f$?
6 answers
functions functional-equations inverse-function
asked Aug 5 at 21:43
Luca Bressan
3,86621037
edited Aug 6 at 6:43
asked Aug 5 at 21:43
Luca Bressan
3,86621037
asked Aug 5 at 21:43
Luca Bressan
3,86621037
asked Aug 5 at 21:43
Luca Bressan
3,86621037
3,86621037
marked as duplicate by 6005, Brahadeesh, Community♦ Aug 6 at 6:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by 6005, Brahadeesh, Community♦ Aug 6 at 6:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
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up vote
1
down vote
accepted
Piecewise examples are possible. For example, we can define $f: [frac13,3] to [frac13,3]$ as follows: $f(1) = 1$, and
for all $frac13 le x < frac12$,
beginalign*
f(x) &= 3x - frac12 &&in [tfrac12, 1)\
f(f(x)) &= frac1x &&in(2,3] \
f(f(f(x))) &= frac13x - frac12 &&in(1,2] \
f(f(f(f(x)) &= x &&in[tfrac13,tfrac12)
endalign*
Here is a plot:
The idea is that we send $[frac13,frac12)$ to $[frac12,1)$ in an increasing manner, then send $[frac12,1)$ to $(2,3]$, then send $(2,3]$ to $(1,2]$, and finally send $(1,2]$ back to $[frac13,frac12)$. It's a 4-cycle.
I'm not sure this example is any less artificial, but it's piecewise continuous (I think being continuous outright is impossible).
The example is derived from the following more general necessary and sufficient condition:
Proposition. Assume that $f$'s domain is some set $A subseteq mathbbR$ which is closed under multiplicative inverse and does not contain $0$. A necessary and sufficient condition for $f$ to satisfy this equation is then that $f(1) = 1$, and that the remaining elements of $A$ are partitioned into quadruples
$$
left a, frac1a, b, frac1b right,
$$
where $a, b < 1$, $a ne b$, and $f$ behaves on this set as follows: $f(a) = b$, $f(b) = frac1a$, $f(frac1a) = frac1b$, and $f(frac1b) = a$.
As a consequence of this, $f(f(f(f(x)))) = x$ for all $x$.
Proof: consider where $f$ maps a pair of elements $x, frac1x$, for any $x ne 1$. Letting $y = f(x)$, we consider the pair $y, frac1y$, and notice that $f$ maps $frac1y$ back to $x$ by the equation $f(1/f(x)) = x$. But then plugging in $frac1y$ to the equation instead gives that $f$ maps $frac1x$ to $frac1y$. Finally, plugging in $frac1x$ to the equation instead we get that $f$ maps $y$ to $frac1x$. So $f$ sends $x mapsto y mapsto frac1x mapsto frac1y mapsto x$.
If $x = 1$ or $y = 1$ then this all collapses to $f(1) = 1$. Otherwise, $x, y$ must be distinct, and we get the claim by setting $a$ to be whichever of $x, y, frac1x, frac1y$ is smaller than $1$ and maps to something smaller than $1$, and $b$ to whichever of $x, y, frac1x, frac1y$ is smaller than $1$ and maps to something larger than $1$.
P.S.: The example in this answer is essentially the same as this answer in the duplicate.
answered Aug 6 at 1:01
6005
34.7k750123
add a comment |Â
up vote
8
down vote
(I would put this in a comment but don't have enough reputation.)
The complex function $fracz+iiz+1$ seems to satisfy your criterion. There might be a way of making this real?
answered Aug 5 at 22:32
AdamW
813
1
good. No version with just real numbers.
– Will Jagy
Aug 5 at 22:51
1
There are no real homographies satisfying the equation. But nice catch on this one.
– Arnaud Mortier
Aug 5 at 22:52
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Piecewise examples are possible. For example, we can define $f: [frac13,3] to [frac13,3]$ as follows: $f(1) = 1$, and
for all $frac13 le x < frac12$,
beginalign*
f(x) &= 3x - frac12 &&in [tfrac12, 1)\
f(f(x)) &= frac1x &&in(2,3] \
f(f(f(x))) &= frac13x - frac12 &&in(1,2] \
f(f(f(f(x)) &= x &&in[tfrac13,tfrac12)
endalign*
Here is a plot:
The idea is that we send $[frac13,frac12)$ to $[frac12,1)$ in an increasing manner, then send $[frac12,1)$ to $(2,3]$, then send $(2,3]$ to $(1,2]$, and finally send $(1,2]$ back to $[frac13,frac12)$. It's a 4-cycle.
I'm not sure this example is any less artificial, but it's piecewise continuous (I think being continuous outright is impossible).
The example is derived from the following more general necessary and sufficient condition:
Proposition. Assume that $f$'s domain is some set $A subseteq mathbbR$ which is closed under multiplicative inverse and does not contain $0$. A necessary and sufficient condition for $f$ to satisfy this equation is then that $f(1) = 1$, and that the remaining elements of $A$ are partitioned into quadruples
$$
left a, frac1a, b, frac1b right,
$$
where $a, b < 1$, $a ne b$, and $f$ behaves on this set as follows: $f(a) = b$, $f(b) = frac1a$, $f(frac1a) = frac1b$, and $f(frac1b) = a$.
As a consequence of this, $f(f(f(f(x)))) = x$ for all $x$.
Proof: consider where $f$ maps a pair of elements $x, frac1x$, for any $x ne 1$. Letting $y = f(x)$, we consider the pair $y, frac1y$, and notice that $f$ maps $frac1y$ back to $x$ by the equation $f(1/f(x)) = x$. But then plugging in $frac1y$ to the equation instead gives that $f$ maps $frac1x$ to $frac1y$. Finally, plugging in $frac1x$ to the equation instead we get that $f$ maps $y$ to $frac1x$. So $f$ sends $x mapsto y mapsto frac1x mapsto frac1y mapsto x$.
If $x = 1$ or $y = 1$ then this all collapses to $f(1) = 1$. Otherwise, $x, y$ must be distinct, and we get the claim by setting $a$ to be whichever of $x, y, frac1x, frac1y$ is smaller than $1$ and maps to something smaller than $1$, and $b$ to whichever of $x, y, frac1x, frac1y$ is smaller than $1$ and maps to something larger than $1$.
P.S.: The example in this answer is essentially the same as this answer in the duplicate.
answered Aug 6 at 1:01
6005
34.7k750123
add a comment |Â
up vote
1
down vote
accepted
Piecewise examples are possible. For example, we can define $f: [frac13,3] to [frac13,3]$ as follows: $f(1) = 1$, and
for all $frac13 le x < frac12$,
beginalign*
f(x) &= 3x - frac12 &&in [tfrac12, 1)\
f(f(x)) &= frac1x &&in(2,3] \
f(f(f(x))) &= frac13x - frac12 &&in(1,2] \
f(f(f(f(x)) &= x &&in[tfrac13,tfrac12)
endalign*
Here is a plot:
The idea is that we send $[frac13,frac12)$ to $[frac12,1)$ in an increasing manner, then send $[frac12,1)$ to $(2,3]$, then send $(2,3]$ to $(1,2]$, and finally send $(1,2]$ back to $[frac13,frac12)$. It's a 4-cycle.
I'm not sure this example is any less artificial, but it's piecewise continuous (I think being continuous outright is impossible).
The example is derived from the following more general necessary and sufficient condition:
Proposition. Assume that $f$'s domain is some set $A subseteq mathbbR$ which is closed under multiplicative inverse and does not contain $0$. A necessary and sufficient condition for $f$ to satisfy this equation is then that $f(1) = 1$, and that the remaining elements of $A$ are partitioned into quadruples
$$
left a, frac1a, b, frac1b right,
$$
where $a, b < 1$, $a ne b$, and $f$ behaves on this set as follows: $f(a) = b$, $f(b) = frac1a$, $f(frac1a) = frac1b$, and $f(frac1b) = a$.
As a consequence of this, $f(f(f(f(x)))) = x$ for all $x$.
Proof: consider where $f$ maps a pair of elements $x, frac1x$, for any $x ne 1$. Letting $y = f(x)$, we consider the pair $y, frac1y$, and notice that $f$ maps $frac1y$ back to $x$ by the equation $f(1/f(x)) = x$. But then plugging in $frac1y$ to the equation instead gives that $f$ maps $frac1x$ to $frac1y$. Finally, plugging in $frac1x$ to the equation instead we get that $f$ maps $y$ to $frac1x$. So $f$ sends $x mapsto y mapsto frac1x mapsto frac1y mapsto x$.
If $x = 1$ or $y = 1$ then this all collapses to $f(1) = 1$. Otherwise, $x, y$ must be distinct, and we get the claim by setting $a$ to be whichever of $x, y, frac1x, frac1y$ is smaller than $1$ and maps to something smaller than $1$, and $b$ to whichever of $x, y, frac1x, frac1y$ is smaller than $1$ and maps to something larger than $1$.
P.S.: The example in this answer is essentially the same as this answer in the duplicate.
answered Aug 6 at 1:01
6005
34.7k750123
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Piecewise examples are possible. For example, we can define $f: [frac13,3] to [frac13,3]$ as follows: $f(1) = 1$, and
for all $frac13 le x < frac12$,
beginalign*
f(x) &= 3x - frac12 &&in [tfrac12, 1)\
f(f(x)) &= frac1x &&in(2,3] \
f(f(f(x))) &= frac13x - frac12 &&in(1,2] \
f(f(f(f(x)) &= x &&in[tfrac13,tfrac12)
endalign*
Here is a plot:
The idea is that we send $[frac13,frac12)$ to $[frac12,1)$ in an increasing manner, then send $[frac12,1)$ to $(2,3]$, then send $(2,3]$ to $(1,2]$, and finally send $(1,2]$ back to $[frac13,frac12)$. It's a 4-cycle.
I'm not sure this example is any less artificial, but it's piecewise continuous (I think being continuous outright is impossible).
The example is derived from the following more general necessary and sufficient condition:
Proposition. Assume that $f$'s domain is some set $A subseteq mathbbR$ which is closed under multiplicative inverse and does not contain $0$. A necessary and sufficient condition for $f$ to satisfy this equation is then that $f(1) = 1$, and that the remaining elements of $A$ are partitioned into quadruples
$$
left a, frac1a, b, frac1b right,
$$
where $a, b < 1$, $a ne b$, and $f$ behaves on this set as follows: $f(a) = b$, $f(b) = frac1a$, $f(frac1a) = frac1b$, and $f(frac1b) = a$.
As a consequence of this, $f(f(f(f(x)))) = x$ for all $x$.
Proof: consider where $f$ maps a pair of elements $x, frac1x$, for any $x ne 1$. Letting $y = f(x)$, we consider the pair $y, frac1y$, and notice that $f$ maps $frac1y$ back to $x$ by the equation $f(1/f(x)) = x$. But then plugging in $frac1y$ to the equation instead gives that $f$ maps $frac1x$ to $frac1y$. Finally, plugging in $frac1x$ to the equation instead we get that $f$ maps $y$ to $frac1x$. So $f$ sends $x mapsto y mapsto frac1x mapsto frac1y mapsto x$.
If $x = 1$ or $y = 1$ then this all collapses to $f(1) = 1$. Otherwise, $x, y$ must be distinct, and we get the claim by setting $a$ to be whichever of $x, y, frac1x, frac1y$ is smaller than $1$ and maps to something smaller than $1$, and $b$ to whichever of $x, y, frac1x, frac1y$ is smaller than $1$ and maps to something larger than $1$.
P.S.: The example in this answer is essentially the same as this answer in the duplicate.
answered Aug 6 at 1:01
6005
34.7k750123
Piecewise examples are possible. For example, we can define $f: [frac13,3] to [frac13,3]$ as follows: $f(1) = 1$, and
for all $frac13 le x < frac12$,
beginalign*
f(x) &= 3x - frac12 &&in [tfrac12, 1)\
f(f(x)) &= frac1x &&in(2,3] \
f(f(f(x))) &= frac13x - frac12 &&in(1,2] \
f(f(f(f(x)) &= x &&in[tfrac13,tfrac12)
endalign*
Here is a plot:
The idea is that we send $[frac13,frac12)$ to $[frac12,1)$ in an increasing manner, then send $[frac12,1)$ to $(2,3]$, then send $(2,3]$ to $(1,2]$, and finally send $(1,2]$ back to $[frac13,frac12)$. It's a 4-cycle.
I'm not sure this example is any less artificial, but it's piecewise continuous (I think being continuous outright is impossible).
The example is derived from the following more general necessary and sufficient condition:
Proposition. Assume that $f$'s domain is some set $A subseteq mathbbR$ which is closed under multiplicative inverse and does not contain $0$. A necessary and sufficient condition for $f$ to satisfy this equation is then that $f(1) = 1$, and that the remaining elements of $A$ are partitioned into quadruples
$$
left a, frac1a, b, frac1b right,
$$
where $a, b < 1$, $a ne b$, and $f$ behaves on this set as follows: $f(a) = b$, $f(b) = frac1a$, $f(frac1a) = frac1b$, and $f(frac1b) = a$.
As a consequence of this, $f(f(f(f(x)))) = x$ for all $x$.
Proof: consider where $f$ maps a pair of elements $x, frac1x$, for any $x ne 1$. Letting $y = f(x)$, we consider the pair $y, frac1y$, and notice that $f$ maps $frac1y$ back to $x$ by the equation $f(1/f(x)) = x$. But then plugging in $frac1y$ to the equation instead gives that $f$ maps $frac1x$ to $frac1y$. Finally, plugging in $frac1x$ to the equation instead we get that $f$ maps $y$ to $frac1x$. So $f$ sends $x mapsto y mapsto frac1x mapsto frac1y mapsto x$.
If $x = 1$ or $y = 1$ then this all collapses to $f(1) = 1$. Otherwise, $x, y$ must be distinct, and we get the claim by setting $a$ to be whichever of $x, y, frac1x, frac1y$ is smaller than $1$ and maps to something smaller than $1$, and $b$ to whichever of $x, y, frac1x, frac1y$ is smaller than $1$ and maps to something larger than $1$.
P.S.: The example in this answer is essentially the same as this answer in the duplicate.
answered Aug 6 at 1:01
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edited Aug 6 at 1:07
answered Aug 6 at 1:01
6005
34.7k750123
answered Aug 6 at 1:01
6005
34.7k750123
answered Aug 6 at 1:01
6005
34.7k750123
34.7k750123
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add a comment |Â
up vote
8
down vote
(I would put this in a comment but don't have enough reputation.)
The complex function $fracz+iiz+1$ seems to satisfy your criterion. There might be a way of making this real?
answered Aug 5 at 22:32
AdamW
813
1
good. No version with just real numbers.
– Will Jagy
Aug 5 at 22:51
1
There are no real homographies satisfying the equation. But nice catch on this one.
– Arnaud Mortier
Aug 5 at 22:52
add a comment |Â
up vote
8
down vote
(I would put this in a comment but don't have enough reputation.)
The complex function $fracz+iiz+1$ seems to satisfy your criterion. There might be a way of making this real?
answered Aug 5 at 22:32
AdamW
813
1
good. No version with just real numbers.
– Will Jagy
Aug 5 at 22:51
1
There are no real homographies satisfying the equation. But nice catch on this one.
– Arnaud Mortier
Aug 5 at 22:52
add a comment |Â
up vote
8
down vote
up vote
8
down vote
(I would put this in a comment but don't have enough reputation.)
The complex function $fracz+iiz+1$ seems to satisfy your criterion. There might be a way of making this real?
answered Aug 5 at 22:32
AdamW
813
(I would put this in a comment but don't have enough reputation.)
The complex function $fracz+iiz+1$ seems to satisfy your criterion. There might be a way of making this real?
answered Aug 5 at 22:32
AdamW
813
answered Aug 5 at 22:32
AdamW
813
answered Aug 5 at 22:32
AdamW
813
answered Aug 5 at 22:32
AdamW
813
813
1
good. No version with just real numbers.
– Will Jagy
Aug 5 at 22:51
1
There are no real homographies satisfying the equation. But nice catch on this one.
– Arnaud Mortier
Aug 5 at 22:52
add a comment |Â
1
good. No version with just real numbers.
– Will Jagy
Aug 5 at 22:51
1
There are no real homographies satisfying the equation. But nice catch on this one.
– Arnaud Mortier
Aug 5 at 22:52
1
1
good. No version with just real numbers.
– Will Jagy
Aug 5 at 22:51
good. No version with just real numbers.
– Will Jagy
Aug 5 at 22:51
1
1
There are no real homographies satisfying the equation. But nice catch on this one.
– Arnaud Mortier
Aug 5 at 22:52
There are no real homographies satisfying the equation. But nice catch on this one.
– Arnaud Mortier
Aug 5 at 22:52
add a comment |Â