Mixing
Which functions satisfy $f^n(x) = f(x)^n$ for some $n ge 2$?
Clash Royale CLAN TAG#URR8PPP
up vote
30
down vote
favorite
Let $n$ be an integer greater than $1$. The notation $f^n$ is notoriously ambiguous: it means either the $n$-th iterate of $f$ or its $n$-th power.
I was wondering when the two interpretations are in fact the same. In other words, if we write $f^n(x)$ for $f(f(dotsb f(x) dotsb))$ and $f(x)^n$ for $f(x) cdot f(x) dotsb f(x)$:
For which functions $f colon I to mathbb R$ is $f^n(x) = f(x)^n$ for all $x in I$?
So far I have been able to show that:
- The constant functions $f(x) = 0$ and $f(x) = 1$ satisfy the condition for all $n$ and $f(x) = -1$ satisfies the condition for all odd $n$. Also, if $f$ satisfies the condition for $n$, then the only fixed points of $f$ can be either $0$, $1$, or if $n$ is odd also $-1$.
- The squaring function $f(x) = x^2$ satisfies the condition for $n = 2$ and is essentially the only non-constant differentiable function to do so. Indeed,
$$f(f(x)) = f(x)^2 implies f'(f(x)) f'(x) = 2 f(x) f'(x)$$
so if we assume $f'(x) neq 0$ and let $y = f(x)$, we have that
$$f'(y) = 2 y implies f(y) = y^2 + c$$
and furthermore
$$(y^2 + c)^2 + c = (y^2 + c)^2 implies c = 0.$$
(Of course we could then consider also, e.g., $f(x) = x^2$ for $x ge 0$ and $f(x) = 0$ for $x < 0$.) - More generally, $f(x) = x^sqrt[n-1]n$ satisfies the condition for any $n > 2$. These functions are only defined on $mathbb R^+$: this is why I chose an interval $I$ instead of all $mathbb R$ as the domain, but I actually don't care if a solution is defined on any other non-trivial subset.
Are there any other solutions? If not, how can we prove so?
Remark: The question can be generalized to $n in mathbb Z$ if we assume $f$ to be invertible and denote by $f^-n$ either the $n$-th iterate of its compositional inverse or the $n$-th power of its multiplicative inverse (the cases $n = 0, 1$ are trivial). But this seems to be an even harder problem. For the case $n = -1$ see this question.
functions functional-equations function-and-relation-composition
asked Aug 3 at 19:34
Luca Bressan
3,7172835
add a comment |Â
up vote
30
down vote
favorite
Let $n$ be an integer greater than $1$. The notation $f^n$ is notoriously ambiguous: it means either the $n$-th iterate of $f$ or its $n$-th power.
I was wondering when the two interpretations are in fact the same. In other words, if we write $f^n(x)$ for $f(f(dotsb f(x) dotsb))$ and $f(x)^n$ for $f(x) cdot f(x) dotsb f(x)$:
For which functions $f colon I to mathbb R$ is $f^n(x) = f(x)^n$ for all $x in I$?
So far I have been able to show that:
- The constant functions $f(x) = 0$ and $f(x) = 1$ satisfy the condition for all $n$ and $f(x) = -1$ satisfies the condition for all odd $n$. Also, if $f$ satisfies the condition for $n$, then the only fixed points of $f$ can be either $0$, $1$, or if $n$ is odd also $-1$.
- The squaring function $f(x) = x^2$ satisfies the condition for $n = 2$ and is essentially the only non-constant differentiable function to do so. Indeed,
$$f(f(x)) = f(x)^2 implies f'(f(x)) f'(x) = 2 f(x) f'(x)$$
so if we assume $f'(x) neq 0$ and let $y = f(x)$, we have that
$$f'(y) = 2 y implies f(y) = y^2 + c$$
and furthermore
$$(y^2 + c)^2 + c = (y^2 + c)^2 implies c = 0.$$
(Of course we could then consider also, e.g., $f(x) = x^2$ for $x ge 0$ and $f(x) = 0$ for $x < 0$.) - More generally, $f(x) = x^sqrt[n-1]n$ satisfies the condition for any $n > 2$. These functions are only defined on $mathbb R^+$: this is why I chose an interval $I$ instead of all $mathbb R$ as the domain, but I actually don't care if a solution is defined on any other non-trivial subset.
Are there any other solutions? If not, how can we prove so?
Remark: The question can be generalized to $n in mathbb Z$ if we assume $f$ to be invertible and denote by $f^-n$ either the $n$-th iterate of its compositional inverse or the $n$-th power of its multiplicative inverse (the cases $n = 0, 1$ are trivial). But this seems to be an even harder problem. For the case $n = -1$ see this question.
functions functional-equations function-and-relation-composition
asked Aug 3 at 19:34
Luca Bressan
3,7172835
2
Isn't $f^n(x)$ sometimes used for the nth derivative ?
– herb steinberg
Aug 3 at 23:44
3
That would usually be with parentheses: $f^(n)(x)$.
– Zachary
Aug 4 at 0:59
1
I'll mention an identity. If we assume $f$ is invertible, then $f^n-1(y)=y^n$. Letting $y=z^p$ and exponentiating both sides by $q$ gives $$[f^n-1(z^p)]^q=z^npq=z^nqp=[f^n-1(z^q)]^p$$.
– user254433
Aug 4 at 1:52
add a comment |Â
up vote
30
down vote
favorite
up vote
30
down vote
favorite
Let $n$ be an integer greater than $1$. The notation $f^n$ is notoriously ambiguous: it means either the $n$-th iterate of $f$ or its $n$-th power.
I was wondering when the two interpretations are in fact the same. In other words, if we write $f^n(x)$ for $f(f(dotsb f(x) dotsb))$ and $f(x)^n$ for $f(x) cdot f(x) dotsb f(x)$:
For which functions $f colon I to mathbb R$ is $f^n(x) = f(x)^n$ for all $x in I$?
So far I have been able to show that:
- The constant functions $f(x) = 0$ and $f(x) = 1$ satisfy the condition for all $n$ and $f(x) = -1$ satisfies the condition for all odd $n$. Also, if $f$ satisfies the condition for $n$, then the only fixed points of $f$ can be either $0$, $1$, or if $n$ is odd also $-1$.
- The squaring function $f(x) = x^2$ satisfies the condition for $n = 2$ and is essentially the only non-constant differentiable function to do so. Indeed,
$$f(f(x)) = f(x)^2 implies f'(f(x)) f'(x) = 2 f(x) f'(x)$$
so if we assume $f'(x) neq 0$ and let $y = f(x)$, we have that
$$f'(y) = 2 y implies f(y) = y^2 + c$$
and furthermore
$$(y^2 + c)^2 + c = (y^2 + c)^2 implies c = 0.$$
(Of course we could then consider also, e.g., $f(x) = x^2$ for $x ge 0$ and $f(x) = 0$ for $x < 0$.) - More generally, $f(x) = x^sqrt[n-1]n$ satisfies the condition for any $n > 2$. These functions are only defined on $mathbb R^+$: this is why I chose an interval $I$ instead of all $mathbb R$ as the domain, but I actually don't care if a solution is defined on any other non-trivial subset.
Are there any other solutions? If not, how can we prove so?
Remark: The question can be generalized to $n in mathbb Z$ if we assume $f$ to be invertible and denote by $f^-n$ either the $n$-th iterate of its compositional inverse or the $n$-th power of its multiplicative inverse (the cases $n = 0, 1$ are trivial). But this seems to be an even harder problem. For the case $n = -1$ see this question.
functions functional-equations function-and-relation-composition
asked Aug 3 at 19:34
Luca Bressan
3,7172835
Let $n$ be an integer greater than $1$. The notation $f^n$ is notoriously ambiguous: it means either the $n$-th iterate of $f$ or its $n$-th power.
I was wondering when the two interpretations are in fact the same. In other words, if we write $f^n(x)$ for $f(f(dotsb f(x) dotsb))$ and $f(x)^n$ for $f(x) cdot f(x) dotsb f(x)$:
For which functions $f colon I to mathbb R$ is $f^n(x) = f(x)^n$ for all $x in I$?
So far I have been able to show that:
- The constant functions $f(x) = 0$ and $f(x) = 1$ satisfy the condition for all $n$ and $f(x) = -1$ satisfies the condition for all odd $n$. Also, if $f$ satisfies the condition for $n$, then the only fixed points of $f$ can be either $0$, $1$, or if $n$ is odd also $-1$.
- The squaring function $f(x) = x^2$ satisfies the condition for $n = 2$ and is essentially the only non-constant differentiable function to do so. Indeed,
$$f(f(x)) = f(x)^2 implies f'(f(x)) f'(x) = 2 f(x) f'(x)$$
so if we assume $f'(x) neq 0$ and let $y = f(x)$, we have that
$$f'(y) = 2 y implies f(y) = y^2 + c$$
and furthermore
$$(y^2 + c)^2 + c = (y^2 + c)^2 implies c = 0.$$
(Of course we could then consider also, e.g., $f(x) = x^2$ for $x ge 0$ and $f(x) = 0$ for $x < 0$.) - More generally, $f(x) = x^sqrt[n-1]n$ satisfies the condition for any $n > 2$. These functions are only defined on $mathbb R^+$: this is why I chose an interval $I$ instead of all $mathbb R$ as the domain, but I actually don't care if a solution is defined on any other non-trivial subset.
Are there any other solutions? If not, how can we prove so?
Remark: The question can be generalized to $n in mathbb Z$ if we assume $f$ to be invertible and denote by $f^-n$ either the $n$-th iterate of its compositional inverse or the $n$-th power of its multiplicative inverse (the cases $n = 0, 1$ are trivial). But this seems to be an even harder problem. For the case $n = -1$ see this question.
functions functional-equations function-and-relation-composition
asked Aug 3 at 19:34
Luca Bressan
3,7172835
edited yesterday
asked Aug 3 at 19:34
Luca Bressan
3,7172835
asked Aug 3 at 19:34
Luca Bressan
3,7172835
asked Aug 3 at 19:34
Luca Bressan
3,7172835
3,7172835
2
Isn't $f^n(x)$ sometimes used for the nth derivative ?
– herb steinberg
Aug 3 at 23:44
3
That would usually be with parentheses: $f^(n)(x)$.
– Zachary
Aug 4 at 0:59
1
I'll mention an identity. If we assume $f$ is invertible, then $f^n-1(y)=y^n$. Letting $y=z^p$ and exponentiating both sides by $q$ gives $$[f^n-1(z^p)]^q=z^npq=z^nqp=[f^n-1(z^q)]^p$$.
– user254433
Aug 4 at 1:52
add a comment |Â
2
Isn't $f^n(x)$ sometimes used for the nth derivative ?
– herb steinberg
Aug 3 at 23:44
3
That would usually be with parentheses: $f^(n)(x)$.
– Zachary
Aug 4 at 0:59
1
I'll mention an identity. If we assume $f$ is invertible, then $f^n-1(y)=y^n$. Letting $y=z^p$ and exponentiating both sides by $q$ gives $$[f^n-1(z^p)]^q=z^npq=z^nqp=[f^n-1(z^q)]^p$$.
– user254433
Aug 4 at 1:52
2
2
Isn't $f^n(x)$ sometimes used for the nth derivative ?
– herb steinberg
Aug 3 at 23:44
Isn't $f^n(x)$ sometimes used for the nth derivative ?
– herb steinberg
Aug 3 at 23:44
3
3
That would usually be with parentheses: $f^(n)(x)$.
– Zachary
Aug 4 at 0:59
That would usually be with parentheses: $f^(n)(x)$.
– Zachary
Aug 4 at 0:59
1
1
I'll mention an identity. If we assume $f$ is invertible, then $f^n-1(y)=y^n$. Letting $y=z^p$ and exponentiating both sides by $q$ gives $$[f^n-1(z^p)]^q=z^npq=z^nqp=[f^n-1(z^q)]^p$$.
– user254433
Aug 4 at 1:52
I'll mention an identity. If we assume $f$ is invertible, then $f^n-1(y)=y^n$. Letting $y=z^p$ and exponentiating both sides by $q$ gives $$[f^n-1(z^p)]^q=z^npq=z^nqp=[f^n-1(z^q)]^p$$.
– user254433
Aug 4 at 1:52
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
9
down vote
Something non-constant and discontinuous everywhere:
$$f(x) = left{beginarrayll
x^2 & textif x text is an integer \
lceil x rceil & textif x text is rational but not an integer\
lfloor x rfloor & textif x text is irrational
endarray right.$$
seems to satisfy $f(f(x))=f(x)cdot f(x)$
answered Aug 3 at 20:53
Henry
92.6k469146
should it be $f(x)$ not $f(m)$?
– user254433
Aug 4 at 0:54
@user254433 Yes it should - and now does - thank you
– Henry
2 days ago
4
I like this example because together with user254433's comment it suggests a way to make a new function $g$ out of a known invertible one $f$. First we need to find a subset $A$ such that $f(A) subseteq A$. Then we need to find a subset $B$ disjoint from $A$ and a function $g colon B to A$. Finally we extend $g$ to $A cup B$ by letting $g(x) = f(x)$ for $x in A$. It follows that $g^n(x) = g^n-1(g(x)) = f^n-1(g(x)) = g(x)^n$ for all $x in A cup B$.
– Luca Bressan
2 days ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
Something non-constant and discontinuous everywhere:
$$f(x) = left{beginarrayll
x^2 & textif x text is an integer \
lceil x rceil & textif x text is rational but not an integer\
lfloor x rfloor & textif x text is irrational
endarray right.$$
seems to satisfy $f(f(x))=f(x)cdot f(x)$
answered Aug 3 at 20:53
Henry
92.6k469146
should it be $f(x)$ not $f(m)$?
– user254433
Aug 4 at 0:54
@user254433 Yes it should - and now does - thank you
– Henry
2 days ago
4
I like this example because together with user254433's comment it suggests a way to make a new function $g$ out of a known invertible one $f$. First we need to find a subset $A$ such that $f(A) subseteq A$. Then we need to find a subset $B$ disjoint from $A$ and a function $g colon B to A$. Finally we extend $g$ to $A cup B$ by letting $g(x) = f(x)$ for $x in A$. It follows that $g^n(x) = g^n-1(g(x)) = f^n-1(g(x)) = g(x)^n$ for all $x in A cup B$.
– Luca Bressan
2 days ago
add a comment |Â
up vote
9
down vote
Something non-constant and discontinuous everywhere:
$$f(x) = left{beginarrayll
x^2 & textif x text is an integer \
lceil x rceil & textif x text is rational but not an integer\
lfloor x rfloor & textif x text is irrational
endarray right.$$
seems to satisfy $f(f(x))=f(x)cdot f(x)$
answered Aug 3 at 20:53
Henry
92.6k469146
should it be $f(x)$ not $f(m)$?
– user254433
Aug 4 at 0:54
@user254433 Yes it should - and now does - thank you
– Henry
2 days ago
4
I like this example because together with user254433's comment it suggests a way to make a new function $g$ out of a known invertible one $f$. First we need to find a subset $A$ such that $f(A) subseteq A$. Then we need to find a subset $B$ disjoint from $A$ and a function $g colon B to A$. Finally we extend $g$ to $A cup B$ by letting $g(x) = f(x)$ for $x in A$. It follows that $g^n(x) = g^n-1(g(x)) = f^n-1(g(x)) = g(x)^n$ for all $x in A cup B$.
– Luca Bressan
2 days ago
add a comment |Â
up vote
9
down vote
up vote
9
down vote
Something non-constant and discontinuous everywhere:
$$f(x) = left{beginarrayll
x^2 & textif x text is an integer \
lceil x rceil & textif x text is rational but not an integer\
lfloor x rfloor & textif x text is irrational
endarray right.$$
seems to satisfy $f(f(x))=f(x)cdot f(x)$
answered Aug 3 at 20:53
Henry
92.6k469146
Something non-constant and discontinuous everywhere:
$$f(x) = left{beginarrayll
x^2 & textif x text is an integer \
lceil x rceil & textif x text is rational but not an integer\
lfloor x rfloor & textif x text is irrational
endarray right.$$
seems to satisfy $f(f(x))=f(x)cdot f(x)$
answered Aug 3 at 20:53
Henry
92.6k469146
edited 2 days ago
answered Aug 3 at 20:53
Henry
92.6k469146
answered Aug 3 at 20:53
Henry
92.6k469146
answered Aug 3 at 20:53
Henry
92.6k469146
92.6k469146
should it be $f(x)$ not $f(m)$?
– user254433
Aug 4 at 0:54
@user254433 Yes it should - and now does - thank you
– Henry
2 days ago
4
I like this example because together with user254433's comment it suggests a way to make a new function $g$ out of a known invertible one $f$. First we need to find a subset $A$ such that $f(A) subseteq A$. Then we need to find a subset $B$ disjoint from $A$ and a function $g colon B to A$. Finally we extend $g$ to $A cup B$ by letting $g(x) = f(x)$ for $x in A$. It follows that $g^n(x) = g^n-1(g(x)) = f^n-1(g(x)) = g(x)^n$ for all $x in A cup B$.
– Luca Bressan
2 days ago
add a comment |Â
should it be $f(x)$ not $f(m)$?
– user254433
Aug 4 at 0:54
@user254433 Yes it should - and now does - thank you
– Henry
2 days ago
4
I like this example because together with user254433's comment it suggests a way to make a new function $g$ out of a known invertible one $f$. First we need to find a subset $A$ such that $f(A) subseteq A$. Then we need to find a subset $B$ disjoint from $A$ and a function $g colon B to A$. Finally we extend $g$ to $A cup B$ by letting $g(x) = f(x)$ for $x in A$. It follows that $g^n(x) = g^n-1(g(x)) = f^n-1(g(x)) = g(x)^n$ for all $x in A cup B$.
– Luca Bressan
2 days ago
should it be $f(x)$ not $f(m)$?
– user254433
Aug 4 at 0:54
should it be $f(x)$ not $f(m)$?
– user254433
Aug 4 at 0:54
@user254433 Yes it should - and now does - thank you
– Henry
2 days ago
@user254433 Yes it should - and now does - thank you
– Henry
2 days ago
4
4
I like this example because together with user254433's comment it suggests a way to make a new function $g$ out of a known invertible one $f$. First we need to find a subset $A$ such that $f(A) subseteq A$. Then we need to find a subset $B$ disjoint from $A$ and a function $g colon B to A$. Finally we extend $g$ to $A cup B$ by letting $g(x) = f(x)$ for $x in A$. It follows that $g^n(x) = g^n-1(g(x)) = f^n-1(g(x)) = g(x)^n$ for all $x in A cup B$.
– Luca Bressan
2 days ago
I like this example because together with user254433's comment it suggests a way to make a new function $g$ out of a known invertible one $f$. First we need to find a subset $A$ such that $f(A) subseteq A$. Then we need to find a subset $B$ disjoint from $A$ and a function $g colon B to A$. Finally we extend $g$ to $A cup B$ by letting $g(x) = f(x)$ for $x in A$. It follows that $g^n(x) = g^n-1(g(x)) = f^n-1(g(x)) = g(x)^n$ for all $x in A cup B$.
– Luca Bressan
2 days ago
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2871419%2fwhich-functions-satisfy-fnx-fxn-for-some-n-ge-2%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
Isn't $f^n(x)$ sometimes used for the nth derivative ?
– herb steinberg
Aug 3 at 23:44
3
That would usually be with parentheses: $f^(n)(x)$.
– Zachary
Aug 4 at 0:59
1
I'll mention an identity. If we assume $f$ is invertible, then $f^n-1(y)=y^n$. Letting $y=z^p$ and exponentiating both sides by $q$ gives $$[f^n-1(z^p)]^q=z^npq=z^nqp=[f^n-1(z^q)]^p$$.
– user254433
Aug 4 at 1:52