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Find a function such that $f^-1=f'$
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Let $f:BbbR^+rightarrowBbbR^+$ be a differentiable bijection and let $f$ satisfy: $f'=f^-1$ (where $f^-1$ denotes the inverse of $f$). Find $f$.
This comes from a facebook page "Mathematical theorems you had no idea existed, cause they're false". The negation of this statemed is given here, that is: There is no such function that satisfies $f'=f^-1$, but in the comments, there is a counterexample given:
$$g(x)=varphi^1-varphix^varphi$$
where $varphi=frac1+sqrt52$ is the golden ratio.
It's straightforward to check that $g'=g^-1$ holds. The OP claims that this solution is unique. Can someone come up with a way to derive the function $g$ or more functions satisfying this property? Also IS this really the unique solution?
differential-equations functions functional-equations
asked Jul 27 at 7:16
Michal Dvořák
48912
 |Â
show 1 more comment
up vote
11
down vote
favorite
Let $f:BbbR^+rightarrowBbbR^+$ be a differentiable bijection and let $f$ satisfy: $f'=f^-1$ (where $f^-1$ denotes the inverse of $f$). Find $f$.
This comes from a facebook page "Mathematical theorems you had no idea existed, cause they're false". The negation of this statemed is given here, that is: There is no such function that satisfies $f'=f^-1$, but in the comments, there is a counterexample given:
$$g(x)=varphi^1-varphix^varphi$$
where $varphi=frac1+sqrt52$ is the golden ratio.
It's straightforward to check that $g'=g^-1$ holds. The OP claims that this solution is unique. Can someone come up with a way to derive the function $g$ or more functions satisfying this property? Also IS this really the unique solution?
differential-equations functions functional-equations
asked Jul 27 at 7:16
Michal Dvořák
48912
2
See math.stackexchange.com/questions/2651188/… for an (almost) equivalent question.
– Kavi Rama Murthy
Jul 27 at 7:23
If you replace $f$ by its inverse and write it in the integrated form you will get Jacky Chong's question in above comment. However, a complete answer for that question has not been given. If someone can come up with a uniqueness proof (which looks pretty hard) it would be nice.
– Kavi Rama Murthy
Jul 27 at 7:27
1
I think the idea behind this is as follows: 1) Let's try to find a solution among monomials. 2) The inverse of $f_a(x) = x^a$ is $f_a^-1(x)= x^1/a$. At the same time, $f_a'(x) = a x^a - 1$. If $f' = f^-1$, then we need something like $a - 1 = 1/a$. It is known that $varphi$ is a solution of this quadratic equation. Now, we only need an appropriate constant $c$, such that $f(x) = c x^varphi$ is indeed the solution. 3) Compute $c$ from $f' = f^-1$
– Antoine
Jul 27 at 7:51
5
mathoverflow.net/questions/34052/function-satisfying-f-1-f
– Brevan Ellefsen
Jul 27 at 8:11
1
To "derive" $g$ just make the ansatz $g(x) = c x^alpha$ and find what $c$ and $alpha$ must be.
– md2perpe
Jul 27 at 8:19
 |Â
show 1 more comment
up vote
11
down vote
favorite
up vote
11
down vote
favorite
Let $f:BbbR^+rightarrowBbbR^+$ be a differentiable bijection and let $f$ satisfy: $f'=f^-1$ (where $f^-1$ denotes the inverse of $f$). Find $f$.
This comes from a facebook page "Mathematical theorems you had no idea existed, cause they're false". The negation of this statemed is given here, that is: There is no such function that satisfies $f'=f^-1$, but in the comments, there is a counterexample given:
$$g(x)=varphi^1-varphix^varphi$$
where $varphi=frac1+sqrt52$ is the golden ratio.
It's straightforward to check that $g'=g^-1$ holds. The OP claims that this solution is unique. Can someone come up with a way to derive the function $g$ or more functions satisfying this property? Also IS this really the unique solution?
differential-equations functions functional-equations
asked Jul 27 at 7:16
Michal Dvořák
48912
Let $f:BbbR^+rightarrowBbbR^+$ be a differentiable bijection and let $f$ satisfy: $f'=f^-1$ (where $f^-1$ denotes the inverse of $f$). Find $f$.
This comes from a facebook page "Mathematical theorems you had no idea existed, cause they're false". The negation of this statemed is given here, that is: There is no such function that satisfies $f'=f^-1$, but in the comments, there is a counterexample given:
$$g(x)=varphi^1-varphix^varphi$$
where $varphi=frac1+sqrt52$ is the golden ratio.
It's straightforward to check that $g'=g^-1$ holds. The OP claims that this solution is unique. Can someone come up with a way to derive the function $g$ or more functions satisfying this property? Also IS this really the unique solution?
differential-equations functions functional-equations
asked Jul 27 at 7:16
Michal Dvořák
48912
edited Jul 27 at 7:30
asked Jul 27 at 7:16
Michal Dvořák
48912
asked Jul 27 at 7:16
Michal Dvořák
48912
asked Jul 27 at 7:16
Michal Dvořák
48912
48912
2
See math.stackexchange.com/questions/2651188/… for an (almost) equivalent question.
– Kavi Rama Murthy
Jul 27 at 7:23
If you replace $f$ by its inverse and write it in the integrated form you will get Jacky Chong's question in above comment. However, a complete answer for that question has not been given. If someone can come up with a uniqueness proof (which looks pretty hard) it would be nice.
– Kavi Rama Murthy
Jul 27 at 7:27
1
I think the idea behind this is as follows: 1) Let's try to find a solution among monomials. 2) The inverse of $f_a(x) = x^a$ is $f_a^-1(x)= x^1/a$. At the same time, $f_a'(x) = a x^a - 1$. If $f' = f^-1$, then we need something like $a - 1 = 1/a$. It is known that $varphi$ is a solution of this quadratic equation. Now, we only need an appropriate constant $c$, such that $f(x) = c x^varphi$ is indeed the solution. 3) Compute $c$ from $f' = f^-1$
– Antoine
Jul 27 at 7:51
5
mathoverflow.net/questions/34052/function-satisfying-f-1-f
– Brevan Ellefsen
Jul 27 at 8:11
1
To "derive" $g$ just make the ansatz $g(x) = c x^alpha$ and find what $c$ and $alpha$ must be.
– md2perpe
Jul 27 at 8:19
 |Â
show 1 more comment
2
See math.stackexchange.com/questions/2651188/… for an (almost) equivalent question.
– Kavi Rama Murthy
Jul 27 at 7:23
If you replace $f$ by its inverse and write it in the integrated form you will get Jacky Chong's question in above comment. However, a complete answer for that question has not been given. If someone can come up with a uniqueness proof (which looks pretty hard) it would be nice.
– Kavi Rama Murthy
Jul 27 at 7:27
1
I think the idea behind this is as follows: 1) Let's try to find a solution among monomials. 2) The inverse of $f_a(x) = x^a$ is $f_a^-1(x)= x^1/a$. At the same time, $f_a'(x) = a x^a - 1$. If $f' = f^-1$, then we need something like $a - 1 = 1/a$. It is known that $varphi$ is a solution of this quadratic equation. Now, we only need an appropriate constant $c$, such that $f(x) = c x^varphi$ is indeed the solution. 3) Compute $c$ from $f' = f^-1$
– Antoine
Jul 27 at 7:51
5
mathoverflow.net/questions/34052/function-satisfying-f-1-f
– Brevan Ellefsen
Jul 27 at 8:11
1
To "derive" $g$ just make the ansatz $g(x) = c x^alpha$ and find what $c$ and $alpha$ must be.
– md2perpe
Jul 27 at 8:19
2
2
See math.stackexchange.com/questions/2651188/… for an (almost) equivalent question.
– Kavi Rama Murthy
Jul 27 at 7:23
See math.stackexchange.com/questions/2651188/… for an (almost) equivalent question.
– Kavi Rama Murthy
Jul 27 at 7:23
If you replace $f$ by its inverse and write it in the integrated form you will get Jacky Chong's question in above comment. However, a complete answer for that question has not been given. If someone can come up with a uniqueness proof (which looks pretty hard) it would be nice.
– Kavi Rama Murthy
Jul 27 at 7:27
If you replace $f$ by its inverse and write it in the integrated form you will get Jacky Chong's question in above comment. However, a complete answer for that question has not been given. If someone can come up with a uniqueness proof (which looks pretty hard) it would be nice.
– Kavi Rama Murthy
Jul 27 at 7:27
1
1
I think the idea behind this is as follows: 1) Let's try to find a solution among monomials. 2) The inverse of $f_a(x) = x^a$ is $f_a^-1(x)= x^1/a$. At the same time, $f_a'(x) = a x^a - 1$. If $f' = f^-1$, then we need something like $a - 1 = 1/a$. It is known that $varphi$ is a solution of this quadratic equation. Now, we only need an appropriate constant $c$, such that $f(x) = c x^varphi$ is indeed the solution. 3) Compute $c$ from $f' = f^-1$
– Antoine
Jul 27 at 7:51
I think the idea behind this is as follows: 1) Let's try to find a solution among monomials. 2) The inverse of $f_a(x) = x^a$ is $f_a^-1(x)= x^1/a$. At the same time, $f_a'(x) = a x^a - 1$. If $f' = f^-1$, then we need something like $a - 1 = 1/a$. It is known that $varphi$ is a solution of this quadratic equation. Now, we only need an appropriate constant $c$, such that $f(x) = c x^varphi$ is indeed the solution. 3) Compute $c$ from $f' = f^-1$
– Antoine
Jul 27 at 7:51
5
5
mathoverflow.net/questions/34052/function-satisfying-f-1-f
– Brevan Ellefsen
Jul 27 at 8:11
mathoverflow.net/questions/34052/function-satisfying-f-1-f
– Brevan Ellefsen
Jul 27 at 8:11
1
1
To "derive" $g$ just make the ansatz $g(x) = c x^alpha$ and find what $c$ and $alpha$ must be.
– md2perpe
Jul 27 at 8:19
To "derive" $g$ just make the ansatz $g(x) = c x^alpha$ and find what $c$ and $alpha$ must be.
– md2perpe
Jul 27 at 8:19
 |Â
show 1 more comment
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2
See math.stackexchange.com/questions/2651188/… for an (almost) equivalent question.
– Kavi Rama Murthy
Jul 27 at 7:23
If you replace $f$ by its inverse and write it in the integrated form you will get Jacky Chong's question in above comment. However, a complete answer for that question has not been given. If someone can come up with a uniqueness proof (which looks pretty hard) it would be nice.
– Kavi Rama Murthy
Jul 27 at 7:27
1
I think the idea behind this is as follows: 1) Let's try to find a solution among monomials. 2) The inverse of $f_a(x) = x^a$ is $f_a^-1(x)= x^1/a$. At the same time, $f_a'(x) = a x^a - 1$. If $f' = f^-1$, then we need something like $a - 1 = 1/a$. It is known that $varphi$ is a solution of this quadratic equation. Now, we only need an appropriate constant $c$, such that $f(x) = c x^varphi$ is indeed the solution. 3) Compute $c$ from $f' = f^-1$
– Antoine
Jul 27 at 7:51
5
mathoverflow.net/questions/34052/function-satisfying-f-1-f
– Brevan Ellefsen
Jul 27 at 8:11
1
To "derive" $g$ just make the ansatz $g(x) = c x^alpha$ and find what $c$ and $alpha$ must be.
– md2perpe
Jul 27 at 8:19