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Find $f(x)$ satisfying $f(f(x))=x^x$
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By inspection my attempts are always wrong. I really have no idea and given up.
How to find $f(x)$ satisfying $f(f(x))=x^x$?
My attempts:
- $f(x)=x^x$
- $f(x)=x^1/x$
- $f(x)=frac1x^x$
My profession is not a mathematician so I am not well trained in mathematics beyond high school mathematics. If you know the solution, please give me a hint.
functional-equations fractional-iteration
edited Aug 6 at 15:05
Simply Beautiful Art
49.4k572172
asked Aug 6 at 13:29
Field Medalist
2078
 |Â
show 1 more comment
up vote
4
down vote
favorite
By inspection my attempts are always wrong. I really have no idea and given up.
How to find $f(x)$ satisfying $f(f(x))=x^x$?
My attempts:
- $f(x)=x^x$
- $f(x)=x^1/x$
- $f(x)=frac1x^x$
My profession is not a mathematician so I am not well trained in mathematics beyond high school mathematics. If you know the solution, please give me a hint.
functional-equations fractional-iteration
edited Aug 6 at 15:05
Simply Beautiful Art
49.4k572172
asked Aug 6 at 13:29
Field Medalist
2078
"By inspection my attempts are always wrong" Such as?
– Did
Aug 6 at 13:35
Is this called avoiding a question by asking another question or what? For the record, I do not call the lines you just added, "an attempt".
– Did
Aug 6 at 13:41
Where did you find this problem?
– TheSimpliFire
Aug 6 at 13:47
@TheSimpliFire: I just created it for fun.
– Field Medalist
Aug 6 at 13:48
I might be wrong but if you take the derivative of both sides then you get $f'(f(x)) cdot f'(x) = x^x(ln(x) + 1)$. Maybe you can solve this DE or prove that i has no solutions?
– Shrey Joshi
Aug 6 at 13:59
 |Â
show 1 more comment
up vote
4
down vote
favorite
up vote
4
down vote
favorite
By inspection my attempts are always wrong. I really have no idea and given up.
How to find $f(x)$ satisfying $f(f(x))=x^x$?
My attempts:
- $f(x)=x^x$
- $f(x)=x^1/x$
- $f(x)=frac1x^x$
My profession is not a mathematician so I am not well trained in mathematics beyond high school mathematics. If you know the solution, please give me a hint.
functional-equations fractional-iteration
edited Aug 6 at 15:05
Simply Beautiful Art
49.4k572172
asked Aug 6 at 13:29
Field Medalist
2078
By inspection my attempts are always wrong. I really have no idea and given up.
How to find $f(x)$ satisfying $f(f(x))=x^x$?
My attempts:
- $f(x)=x^x$
- $f(x)=x^1/x$
- $f(x)=frac1x^x$
My profession is not a mathematician so I am not well trained in mathematics beyond high school mathematics. If you know the solution, please give me a hint.
functional-equations fractional-iteration
edited Aug 6 at 15:05
Simply Beautiful Art
49.4k572172
asked Aug 6 at 13:29
Field Medalist
2078
edited Aug 6 at 15:05
Simply Beautiful Art
49.4k572172
edited Aug 6 at 15:05
Simply Beautiful Art
49.4k572172
edited Aug 6 at 15:05
Simply Beautiful Art
49.4k572172
49.4k572172
asked Aug 6 at 13:29
Field Medalist
2078
asked Aug 6 at 13:29
Field Medalist
2078
asked Aug 6 at 13:29
Field Medalist
2078
2078
"By inspection my attempts are always wrong" Such as?
– Did
Aug 6 at 13:35
Is this called avoiding a question by asking another question or what? For the record, I do not call the lines you just added, "an attempt".
– Did
Aug 6 at 13:41
Where did you find this problem?
– TheSimpliFire
Aug 6 at 13:47
@TheSimpliFire: I just created it for fun.
– Field Medalist
Aug 6 at 13:48
I might be wrong but if you take the derivative of both sides then you get $f'(f(x)) cdot f'(x) = x^x(ln(x) + 1)$. Maybe you can solve this DE or prove that i has no solutions?
– Shrey Joshi
Aug 6 at 13:59
 |Â
show 1 more comment
"By inspection my attempts are always wrong" Such as?
– Did
Aug 6 at 13:35
Is this called avoiding a question by asking another question or what? For the record, I do not call the lines you just added, "an attempt".
– Did
Aug 6 at 13:41
Where did you find this problem?
– TheSimpliFire
Aug 6 at 13:47
@TheSimpliFire: I just created it for fun.
– Field Medalist
Aug 6 at 13:48
I might be wrong but if you take the derivative of both sides then you get $f'(f(x)) cdot f'(x) = x^x(ln(x) + 1)$. Maybe you can solve this DE or prove that i has no solutions?
– Shrey Joshi
Aug 6 at 13:59
"By inspection my attempts are always wrong" Such as?
– Did
Aug 6 at 13:35
"By inspection my attempts are always wrong" Such as?
– Did
Aug 6 at 13:35
Is this called avoiding a question by asking another question or what? For the record, I do not call the lines you just added, "an attempt".
– Did
Aug 6 at 13:41
Is this called avoiding a question by asking another question or what? For the record, I do not call the lines you just added, "an attempt".
– Did
Aug 6 at 13:41
Where did you find this problem?
– TheSimpliFire
Aug 6 at 13:47
Where did you find this problem?
– TheSimpliFire
Aug 6 at 13:47
@TheSimpliFire: I just created it for fun.
– Field Medalist
Aug 6 at 13:48
@TheSimpliFire: I just created it for fun.
– Field Medalist
Aug 6 at 13:48
I might be wrong but if you take the derivative of both sides then you get $f'(f(x)) cdot f'(x) = x^x(ln(x) + 1)$. Maybe you can solve this DE or prove that i has no solutions?
– Shrey Joshi
Aug 6 at 13:59
I might be wrong but if you take the derivative of both sides then you get $f'(f(x)) cdot f'(x) = x^x(ln(x) + 1)$. Maybe you can solve this DE or prove that i has no solutions?
– Shrey Joshi
Aug 6 at 13:59
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
8
down vote
The key is to identify a fixed point of the function $x^x$. An obvious fixed point is $x=1.$ To simplify the work, we assume that $f$ has the same fixed point. We define $ g(x) := f(1+x)-1 $ where $ g(0) = 0 $ because $ f(1) = 1. $ Now using $ f(f(x)) = x^x $ we have
$$ g(g(x)) = f(1!+!g(x)) !-! 1 = f(f(1!+!x)) !-! 1 = (1!+!x)^1+x - 1 = x + x^2 + fracx^32 + fracx^43 + O(x^5). $$
Assuming a power series expansion for $ g(x), $ we can solve for its coefficients and get
$$ g(x) = x + fracx^22 + 0x^3 + frac548x^4 - frac1196x^5 + frac2571920x^6 - frac8515760x^7 + frac15751107520x^8 + O(x^9). $$
Now $ f(x) = 1 + g(x-1). $ I am not sure about the radius of convergence of the series. It may be zero. All the coefficients up to $x^18$ are less than $1$ in absolute value, but then they grow very rapidly. Still, for $ .7 < x < 1.2 $ the $ f(f(x)) $ is a close approximation to $ x^x $ but adding more terms in the series makes it worse. It reminds me of the asymptotic expansion of $ log Gamma (x). $
answered Aug 6 at 14:23
Somos
11.7k11033
1
You might consider adding the reasoning behind the assertion that $f(1)=1$. Obviously, $f(f(1))=1$, but how does that imply that $f(1)=1$?
– Mark Viola
Aug 6 at 14:54
@MarkViola Note that $f(1)^f(1)=f(f(f(1)))=f(1^1)$, so we don't have to assume this is the case (as per the last edit)
– Simply Beautiful Art
Aug 6 at 15:04
The procedure of Ecalle gives $f in C^infty$ for $x > frac1e$ and, except for the point $x=1$ itself, actually $C^omega$ See math.stackexchange.com/questions/208996/half-iterate-of-x2c/… and mathoverflow.net/questions/45608/…
– Will Jagy
Aug 6 at 16:36
1
Your series for $g$ has radius of convergence $0 ; .$
– Will Jagy
Aug 6 at 16:39
@SimplyBeautifulArt All that shows is that $f(1)=f(1)^f(1)$.
– Mark Viola
Aug 6 at 17:19
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
The key is to identify a fixed point of the function $x^x$. An obvious fixed point is $x=1.$ To simplify the work, we assume that $f$ has the same fixed point. We define $ g(x) := f(1+x)-1 $ where $ g(0) = 0 $ because $ f(1) = 1. $ Now using $ f(f(x)) = x^x $ we have
$$ g(g(x)) = f(1!+!g(x)) !-! 1 = f(f(1!+!x)) !-! 1 = (1!+!x)^1+x - 1 = x + x^2 + fracx^32 + fracx^43 + O(x^5). $$
Assuming a power series expansion for $ g(x), $ we can solve for its coefficients and get
$$ g(x) = x + fracx^22 + 0x^3 + frac548x^4 - frac1196x^5 + frac2571920x^6 - frac8515760x^7 + frac15751107520x^8 + O(x^9). $$
Now $ f(x) = 1 + g(x-1). $ I am not sure about the radius of convergence of the series. It may be zero. All the coefficients up to $x^18$ are less than $1$ in absolute value, but then they grow very rapidly. Still, for $ .7 < x < 1.2 $ the $ f(f(x)) $ is a close approximation to $ x^x $ but adding more terms in the series makes it worse. It reminds me of the asymptotic expansion of $ log Gamma (x). $
answered Aug 6 at 14:23
Somos
11.7k11033
1
You might consider adding the reasoning behind the assertion that $f(1)=1$. Obviously, $f(f(1))=1$, but how does that imply that $f(1)=1$?
– Mark Viola
Aug 6 at 14:54
@MarkViola Note that $f(1)^f(1)=f(f(f(1)))=f(1^1)$, so we don't have to assume this is the case (as per the last edit)
– Simply Beautiful Art
Aug 6 at 15:04
The procedure of Ecalle gives $f in C^infty$ for $x > frac1e$ and, except for the point $x=1$ itself, actually $C^omega$ See math.stackexchange.com/questions/208996/half-iterate-of-x2c/… and mathoverflow.net/questions/45608/…
– Will Jagy
Aug 6 at 16:36
1
Your series for $g$ has radius of convergence $0 ; .$
– Will Jagy
Aug 6 at 16:39
@SimplyBeautifulArt All that shows is that $f(1)=f(1)^f(1)$.
– Mark Viola
Aug 6 at 17:19
 |Â
show 5 more comments
up vote
8
down vote
The key is to identify a fixed point of the function $x^x$. An obvious fixed point is $x=1.$ To simplify the work, we assume that $f$ has the same fixed point. We define $ g(x) := f(1+x)-1 $ where $ g(0) = 0 $ because $ f(1) = 1. $ Now using $ f(f(x)) = x^x $ we have
$$ g(g(x)) = f(1!+!g(x)) !-! 1 = f(f(1!+!x)) !-! 1 = (1!+!x)^1+x - 1 = x + x^2 + fracx^32 + fracx^43 + O(x^5). $$
Assuming a power series expansion for $ g(x), $ we can solve for its coefficients and get
$$ g(x) = x + fracx^22 + 0x^3 + frac548x^4 - frac1196x^5 + frac2571920x^6 - frac8515760x^7 + frac15751107520x^8 + O(x^9). $$
Now $ f(x) = 1 + g(x-1). $ I am not sure about the radius of convergence of the series. It may be zero. All the coefficients up to $x^18$ are less than $1$ in absolute value, but then they grow very rapidly. Still, for $ .7 < x < 1.2 $ the $ f(f(x)) $ is a close approximation to $ x^x $ but adding more terms in the series makes it worse. It reminds me of the asymptotic expansion of $ log Gamma (x). $
answered Aug 6 at 14:23
Somos
11.7k11033
1
You might consider adding the reasoning behind the assertion that $f(1)=1$. Obviously, $f(f(1))=1$, but how does that imply that $f(1)=1$?
– Mark Viola
Aug 6 at 14:54
@MarkViola Note that $f(1)^f(1)=f(f(f(1)))=f(1^1)$, so we don't have to assume this is the case (as per the last edit)
– Simply Beautiful Art
Aug 6 at 15:04
The procedure of Ecalle gives $f in C^infty$ for $x > frac1e$ and, except for the point $x=1$ itself, actually $C^omega$ See math.stackexchange.com/questions/208996/half-iterate-of-x2c/… and mathoverflow.net/questions/45608/…
– Will Jagy
Aug 6 at 16:36
1
Your series for $g$ has radius of convergence $0 ; .$
– Will Jagy
Aug 6 at 16:39
@SimplyBeautifulArt All that shows is that $f(1)=f(1)^f(1)$.
– Mark Viola
Aug 6 at 17:19
 |Â
show 5 more comments
up vote
8
down vote
up vote
8
down vote
The key is to identify a fixed point of the function $x^x$. An obvious fixed point is $x=1.$ To simplify the work, we assume that $f$ has the same fixed point. We define $ g(x) := f(1+x)-1 $ where $ g(0) = 0 $ because $ f(1) = 1. $ Now using $ f(f(x)) = x^x $ we have
$$ g(g(x)) = f(1!+!g(x)) !-! 1 = f(f(1!+!x)) !-! 1 = (1!+!x)^1+x - 1 = x + x^2 + fracx^32 + fracx^43 + O(x^5). $$
Assuming a power series expansion for $ g(x), $ we can solve for its coefficients and get
$$ g(x) = x + fracx^22 + 0x^3 + frac548x^4 - frac1196x^5 + frac2571920x^6 - frac8515760x^7 + frac15751107520x^8 + O(x^9). $$
Now $ f(x) = 1 + g(x-1). $ I am not sure about the radius of convergence of the series. It may be zero. All the coefficients up to $x^18$ are less than $1$ in absolute value, but then they grow very rapidly. Still, for $ .7 < x < 1.2 $ the $ f(f(x)) $ is a close approximation to $ x^x $ but adding more terms in the series makes it worse. It reminds me of the asymptotic expansion of $ log Gamma (x). $
answered Aug 6 at 14:23
Somos
11.7k11033
The key is to identify a fixed point of the function $x^x$. An obvious fixed point is $x=1.$ To simplify the work, we assume that $f$ has the same fixed point. We define $ g(x) := f(1+x)-1 $ where $ g(0) = 0 $ because $ f(1) = 1. $ Now using $ f(f(x)) = x^x $ we have
$$ g(g(x)) = f(1!+!g(x)) !-! 1 = f(f(1!+!x)) !-! 1 = (1!+!x)^1+x - 1 = x + x^2 + fracx^32 + fracx^43 + O(x^5). $$
Assuming a power series expansion for $ g(x), $ we can solve for its coefficients and get
$$ g(x) = x + fracx^22 + 0x^3 + frac548x^4 - frac1196x^5 + frac2571920x^6 - frac8515760x^7 + frac15751107520x^8 + O(x^9). $$
Now $ f(x) = 1 + g(x-1). $ I am not sure about the radius of convergence of the series. It may be zero. All the coefficients up to $x^18$ are less than $1$ in absolute value, but then they grow very rapidly. Still, for $ .7 < x < 1.2 $ the $ f(f(x)) $ is a close approximation to $ x^x $ but adding more terms in the series makes it worse. It reminds me of the asymptotic expansion of $ log Gamma (x). $
answered Aug 6 at 14:23
Somos
11.7k11033
edited Aug 6 at 18:16
answered Aug 6 at 14:23
Somos
11.7k11033
answered Aug 6 at 14:23
Somos
11.7k11033
answered Aug 6 at 14:23
Somos
11.7k11033
11.7k11033
1
You might consider adding the reasoning behind the assertion that $f(1)=1$. Obviously, $f(f(1))=1$, but how does that imply that $f(1)=1$?
– Mark Viola
Aug 6 at 14:54
@MarkViola Note that $f(1)^f(1)=f(f(f(1)))=f(1^1)$, so we don't have to assume this is the case (as per the last edit)
– Simply Beautiful Art
Aug 6 at 15:04
The procedure of Ecalle gives $f in C^infty$ for $x > frac1e$ and, except for the point $x=1$ itself, actually $C^omega$ See math.stackexchange.com/questions/208996/half-iterate-of-x2c/… and mathoverflow.net/questions/45608/…
– Will Jagy
Aug 6 at 16:36
1
Your series for $g$ has radius of convergence $0 ; .$
– Will Jagy
Aug 6 at 16:39
@SimplyBeautifulArt All that shows is that $f(1)=f(1)^f(1)$.
– Mark Viola
Aug 6 at 17:19
 |Â
show 5 more comments
1
You might consider adding the reasoning behind the assertion that $f(1)=1$. Obviously, $f(f(1))=1$, but how does that imply that $f(1)=1$?
– Mark Viola
Aug 6 at 14:54
@MarkViola Note that $f(1)^f(1)=f(f(f(1)))=f(1^1)$, so we don't have to assume this is the case (as per the last edit)
– Simply Beautiful Art
Aug 6 at 15:04
The procedure of Ecalle gives $f in C^infty$ for $x > frac1e$ and, except for the point $x=1$ itself, actually $C^omega$ See math.stackexchange.com/questions/208996/half-iterate-of-x2c/… and mathoverflow.net/questions/45608/…
– Will Jagy
Aug 6 at 16:36
1
Your series for $g$ has radius of convergence $0 ; .$
– Will Jagy
Aug 6 at 16:39
@SimplyBeautifulArt All that shows is that $f(1)=f(1)^f(1)$.
– Mark Viola
Aug 6 at 17:19
1
1
You might consider adding the reasoning behind the assertion that $f(1)=1$. Obviously, $f(f(1))=1$, but how does that imply that $f(1)=1$?
– Mark Viola
Aug 6 at 14:54
You might consider adding the reasoning behind the assertion that $f(1)=1$. Obviously, $f(f(1))=1$, but how does that imply that $f(1)=1$?
– Mark Viola
Aug 6 at 14:54
@MarkViola Note that $f(1)^f(1)=f(f(f(1)))=f(1^1)$, so we don't have to assume this is the case (as per the last edit)
– Simply Beautiful Art
Aug 6 at 15:04
@MarkViola Note that $f(1)^f(1)=f(f(f(1)))=f(1^1)$, so we don't have to assume this is the case (as per the last edit)
– Simply Beautiful Art
Aug 6 at 15:04
The procedure of Ecalle gives $f in C^infty$ for $x > frac1e$ and, except for the point $x=1$ itself, actually $C^omega$ See math.stackexchange.com/questions/208996/half-iterate-of-x2c/… and mathoverflow.net/questions/45608/…
– Will Jagy
Aug 6 at 16:36
The procedure of Ecalle gives $f in C^infty$ for $x > frac1e$ and, except for the point $x=1$ itself, actually $C^omega$ See math.stackexchange.com/questions/208996/half-iterate-of-x2c/… and mathoverflow.net/questions/45608/…
– Will Jagy
Aug 6 at 16:36
1
1
Your series for $g$ has radius of convergence $0 ; .$
– Will Jagy
Aug 6 at 16:39
Your series for $g$ has radius of convergence $0 ; .$
– Will Jagy
Aug 6 at 16:39
@SimplyBeautifulArt All that shows is that $f(1)=f(1)^f(1)$.
– Mark Viola
Aug 6 at 17:19
@SimplyBeautifulArt All that shows is that $f(1)=f(1)^f(1)$.
– Mark Viola
Aug 6 at 17:19
 |Â
show 5 more comments
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"By inspection my attempts are always wrong" Such as?
– Did
Aug 6 at 13:35
Is this called avoiding a question by asking another question or what? For the record, I do not call the lines you just added, "an attempt".
– Did
Aug 6 at 13:41
Where did you find this problem?
– TheSimpliFire
Aug 6 at 13:47
@TheSimpliFire: I just created it for fun.
– Field Medalist
Aug 6 at 13:48
I might be wrong but if you take the derivative of both sides then you get $f'(f(x)) cdot f'(x) = x^x(ln(x) + 1)$. Maybe you can solve this DE or prove that i has no solutions?
– Shrey Joshi
Aug 6 at 13:59