Mixing
Fractional Composite of Functions
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I would like to know how I can calculate a fractional composition of a function. Let be $f(x)$, where $x in R$ and $f(x) in R$. I now how to do $f(f(x))=f^2(x)$. Now suppose I would like to do $f^{frac{1}{2}}(x)$. Any tip? In other words, $f^n(x)$ is a $n$ composition of a function. Usually, $n in N$. Supose now I would like to calculate a result for $n=1.2$.
function-and-relation-composition tetration
edited Jul 3 '17 at 18:49
Gottfried Helms
23.7k245101
asked Feb 14 '14 at 10:56
Erivelton Geraldo NepomucenoErivelton Geraldo Nepomuceno
836
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add a comment |
$begingroup$
I would like to know how I can calculate a fractional composition of a function. Let be $f(x)$, where $x in R$ and $f(x) in R$. I now how to do $f(f(x))=f^2(x)$. Now suppose I would like to do $f^{frac{1}{2}}(x)$. Any tip? In other words, $f^n(x)$ is a $n$ composition of a function. Usually, $n in N$. Supose now I would like to calculate a result for $n=1.2$.
function-and-relation-composition tetration
edited Jul 3 '17 at 18:49
Gottfried Helms
23.7k245101
asked Feb 14 '14 at 10:56
Erivelton Geraldo NepomucenoErivelton Geraldo Nepomuceno
836
$endgroup$
$begingroup$
One can look at a collection of Q&A in mathoverflow: mathoverflow.net/questions/tagged/fractional-iteration or here: math.stackexchange.com/questions/tagged/tetration
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– Gottfried Helms
Jul 3 '17 at 18:52
add a comment |
$begingroup$
I would like to know how I can calculate a fractional composition of a function. Let be $f(x)$, where $x in R$ and $f(x) in R$. I now how to do $f(f(x))=f^2(x)$. Now suppose I would like to do $f^{frac{1}{2}}(x)$. Any tip? In other words, $f^n(x)$ is a $n$ composition of a function. Usually, $n in N$. Supose now I would like to calculate a result for $n=1.2$.
function-and-relation-composition tetration
edited Jul 3 '17 at 18:49
Gottfried Helms
23.7k245101
asked Feb 14 '14 at 10:56
Erivelton Geraldo NepomucenoErivelton Geraldo Nepomuceno
836
$endgroup$
I would like to know how I can calculate a fractional composition of a function. Let be $f(x)$, where $x in R$ and $f(x) in R$. I now how to do $f(f(x))=f^2(x)$. Now suppose I would like to do $f^{frac{1}{2}}(x)$. Any tip? In other words, $f^n(x)$ is a $n$ composition of a function. Usually, $n in N$. Supose now I would like to calculate a result for $n=1.2$.
function-and-relation-composition tetration
function-and-relation-composition tetration
edited Jul 3 '17 at 18:49
Gottfried Helms
23.7k245101
asked Feb 14 '14 at 10:56
Erivelton Geraldo NepomucenoErivelton Geraldo Nepomuceno
836
edited Jul 3 '17 at 18:49
Gottfried Helms
23.7k245101
asked Feb 14 '14 at 10:56
Erivelton Geraldo NepomucenoErivelton Geraldo Nepomuceno
836
edited Jul 3 '17 at 18:49
Gottfried Helms
23.7k245101
edited Jul 3 '17 at 18:49
Gottfried Helms
23.7k245101
edited Jul 3 '17 at 18:49
Gottfried Helms
23.7k245101
23.7k245101
asked Feb 14 '14 at 10:56
Erivelton Geraldo NepomucenoErivelton Geraldo Nepomuceno
836
asked Feb 14 '14 at 10:56
Erivelton Geraldo NepomucenoErivelton Geraldo Nepomuceno
836
asked Feb 14 '14 at 10:56
Erivelton Geraldo NepomucenoErivelton Geraldo Nepomuceno
836
836
$begingroup$
One can look at a collection of Q&A in mathoverflow: mathoverflow.net/questions/tagged/fractional-iteration or here: math.stackexchange.com/questions/tagged/tetration
$endgroup$
– Gottfried Helms
Jul 3 '17 at 18:52
add a comment |
$begingroup$
One can look at a collection of Q&A in mathoverflow: mathoverflow.net/questions/tagged/fractional-iteration or here: math.stackexchange.com/questions/tagged/tetration
$endgroup$
– Gottfried Helms
Jul 3 '17 at 18:52
$begingroup$
One can look at a collection of Q&A in mathoverflow: mathoverflow.net/questions/tagged/fractional-iteration or here: math.stackexchange.com/questions/tagged/tetration
$endgroup$
– Gottfried Helms
Jul 3 '17 at 18:52
$begingroup$
One can look at a collection of Q&A in mathoverflow: mathoverflow.net/questions/tagged/fractional-iteration or here: math.stackexchange.com/questions/tagged/tetration
$endgroup$
– Gottfried Helms
Jul 3 '17 at 18:52
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
I put a number of relevant articles at http://zakuski.utsa.edu/~jagy/other.html I tell people to start with the obituary of Baker.
The short version is this:
- if your function has no fixpoints, you can mostly manage this by Kneser's method.
- If there is just one fixpoint,
- and the modulus of the derivative there is not $1,$ fairly easy by Schroder's equation.
- However, if the derivative there is $1,$ it is a big song and dance invented by Ecalle in the 1970's. I have only second-hand accounts, but enough to apply, see https://mathoverflow.net/questions/45608/formal-power-series-convergence including my own answer.
edited Jun 15 '17 at 8:15
Adam
1,1951919
answered Feb 14 '14 at 11:32
Will JagyWill Jagy
104k5103202
$endgroup$
$begingroup$
It applies for $n in R$?
$endgroup$
– Erivelton Geraldo Nepomuceno
Feb 14 '14 at 11:42
1
$begingroup$
Only under the most fortunate circumstances does it apply for all real $n.$ More common is that it applies for rational $n$ with a certain denominator. This is a result of Baker, finished by his student Liverpool.
$endgroup$
– Will Jagy
Feb 14 '14 at 11:48
2
$begingroup$
Your first link is broken.
$endgroup$
– Ruslan
Nov 21 '16 at 11:03
add a comment |
$begingroup$
Functional square roots are relatively easy to find through Functional Conjugacy specifications of generic iterated functions; find the general iterate, and set n=1/2 in the general expression, such as the ones available and listed there.
For example, the functional square root of the nth Chebyshev polynomial $T_n$ is evidently $cos (sqrt{n} arccos (x)) $ --- which, however, is not a polynomial, in general.
As Will Jagy already mentioned, in general, around a fixed point, say $f(0)=0$, the standard tried-and-true method is Schröder's equation, $psi(f(x))=f'(0) ~psi(x)$, a triumph of functional conjugacy, provided f ' there ≠1. Having found ψ, one has $f^{1/2}(x)=psi^{-1}left (sqrt{f'(0)} ~ psi(x)right )$.
If it is =1, at the fixed point, however, all is not lost, and conjugacy still delivers: Curtright, Jin, & Zachos, JouPhys A Math-Th 44.40 (2011): 405205 illustrates how $f^n circ g circ f^{-n}$ will improve an initial approximant g rapidly and dramatically for large n.
(This is illustrated there for intuitive functions such as $f=sin x$, blue, around the origin, where the half iterate is orange, the second iterate is red, and so on...)
answered May 26 '17 at 1:12
Cosmas ZachosCosmas Zachos
1,840522
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add a comment |
$begingroup$
Just some simple examples, allowing much obvious/natural versions of a fractional iterate:
Let $f(x)=x+a$ then $f(f(x))=x+2a$ , $f(f(f(x))) = x+3a$ and in general $f°^h(x)=x+hcdot a$. Then $f°^{0.5} (x) = x+0.5 a$
Let $f(x)=x cdot a$ then $f(f(x))=x cdot a^2$ , $f(f(f(x))) = x cdot a^3$ and in general $f°^h(x)=x cdot a^h $ . Then $f°^{0.5} (x) = x cdot a^{0.5} $
Let $f(x)=x ^ a$ then $f(f(x))=x ^{ a^2}$ , $f(f(f(x))) = x ^ { a^3}$ and in general $f°^h(x)=x ^{ a^h} $ . Then $f°^{0.5} (x) = x ^ {sqrt {a}} $
If a function has a power series without constant term , then you can find a formal power series for a fractional iterate. If $f(x) = ax + O(x^2)$ and $0<a<1$ then sometimes that series for fractional iterates might have a nonzero radius of convergence. If the power series has a constant term, one can sometimes use conjugacy to find a formal power series. More on this using the method of Carleman-matrices, see wikipedia:Carleman-matrix. For the specific case of $f(x) = a^x$ see wikipedia:tetration also there is something in wikipedia:hyperoperations .
Also note that a collection of Q&A about this subject is in MSE (tag::tetration, tag::hyperoperation) and in MO tag::fractional-iteration.
A very involved discussion can be found at tetration-forum: http://math.eretrandre.org/tetrationforum and many more links can be found using google-search. In mathoverflow you can find this question with answers: MO1 , MO2 , MO3
The problem is really not simple, the beginning of its consideration in a systematic way was only in the 19th century with N. Abel and E. Schroeder providing fairly general schemes - but only for functions of certain classes. Some singular gems might have been found earlier, say Euler's/Goldbach's finding of the range for the base $a$ in $f(x)=a^x$ where this converges even if infinitely iterated (but no fractional iterate has been discussed with this)
answered Jul 4 '17 at 8:41
Gottfried HelmsGottfried Helms
23.7k245101
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If you have a function g(x) with
g(g(x)) = f(x) , then you could formally set
g(x) = $f^{1/2}(x)$
answered Feb 14 '14 at 10:59
PeterPeter
49.3k1240138
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1
$begingroup$
But it will be difficult to find such a g(x) for a given f(x).
$endgroup$
– Peter
Feb 14 '14 at 11:01
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Thanks for your reply. But I think it is not the case. When I indicate $1/2$ I would like to mean that I am doing a partial composition. Suppose $n$ is the number of times of composition. Usually $n in N$. I would like to calculate when $n=1.2$ for example.
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– Erivelton Geraldo Nepomuceno
Feb 14 '14 at 11:33
add a comment |
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The fractional iterate is NON unique even with well behaved function : check the very nice book : Z.A Melzak : "Companion to concrete mathematics" Vol 1 : page 56 to 62
answered Feb 3 at 13:00
Jérôme JEAN-CHARLESJérôme JEAN-CHARLES
1114
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I put a number of relevant articles at http://zakuski.utsa.edu/~jagy/other.html I tell people to start with the obituary of Baker.
The short version is this:
- if your function has no fixpoints, you can mostly manage this by Kneser's method.
- If there is just one fixpoint,
- and the modulus of the derivative there is not $1,$ fairly easy by Schroder's equation.
- However, if the derivative there is $1,$ it is a big song and dance invented by Ecalle in the 1970's. I have only second-hand accounts, but enough to apply, see https://mathoverflow.net/questions/45608/formal-power-series-convergence including my own answer.
edited Jun 15 '17 at 8:15
Adam
1,1951919
answered Feb 14 '14 at 11:32
Will JagyWill Jagy
104k5103202
$endgroup$
$begingroup$
It applies for $n in R$?
$endgroup$
– Erivelton Geraldo Nepomuceno
Feb 14 '14 at 11:42
1
$begingroup$
Only under the most fortunate circumstances does it apply for all real $n.$ More common is that it applies for rational $n$ with a certain denominator. This is a result of Baker, finished by his student Liverpool.
$endgroup$
– Will Jagy
Feb 14 '14 at 11:48
2
$begingroup$
Your first link is broken.
$endgroup$
– Ruslan
Nov 21 '16 at 11:03
add a comment |
$begingroup$
I put a number of relevant articles at http://zakuski.utsa.edu/~jagy/other.html I tell people to start with the obituary of Baker.
The short version is this:
- if your function has no fixpoints, you can mostly manage this by Kneser's method.
- If there is just one fixpoint,
- and the modulus of the derivative there is not $1,$ fairly easy by Schroder's equation.
- However, if the derivative there is $1,$ it is a big song and dance invented by Ecalle in the 1970's. I have only second-hand accounts, but enough to apply, see https://mathoverflow.net/questions/45608/formal-power-series-convergence including my own answer.
edited Jun 15 '17 at 8:15
Adam
1,1951919
answered Feb 14 '14 at 11:32
Will JagyWill Jagy
104k5103202
$endgroup$
$begingroup$
It applies for $n in R$?
$endgroup$
– Erivelton Geraldo Nepomuceno
Feb 14 '14 at 11:42
1
$begingroup$
Only under the most fortunate circumstances does it apply for all real $n.$ More common is that it applies for rational $n$ with a certain denominator. This is a result of Baker, finished by his student Liverpool.
$endgroup$
– Will Jagy
Feb 14 '14 at 11:48
2
$begingroup$
Your first link is broken.
$endgroup$
– Ruslan
Nov 21 '16 at 11:03
add a comment |
$begingroup$
I put a number of relevant articles at http://zakuski.utsa.edu/~jagy/other.html I tell people to start with the obituary of Baker.
The short version is this:
- if your function has no fixpoints, you can mostly manage this by Kneser's method.
- If there is just one fixpoint,
- and the modulus of the derivative there is not $1,$ fairly easy by Schroder's equation.
- However, if the derivative there is $1,$ it is a big song and dance invented by Ecalle in the 1970's. I have only second-hand accounts, but enough to apply, see https://mathoverflow.net/questions/45608/formal-power-series-convergence including my own answer.
edited Jun 15 '17 at 8:15
Adam
1,1951919
answered Feb 14 '14 at 11:32
Will JagyWill Jagy
104k5103202
$endgroup$
I put a number of relevant articles at http://zakuski.utsa.edu/~jagy/other.html I tell people to start with the obituary of Baker.
The short version is this:
- if your function has no fixpoints, you can mostly manage this by Kneser's method.
- If there is just one fixpoint,
- and the modulus of the derivative there is not $1,$ fairly easy by Schroder's equation.
- However, if the derivative there is $1,$ it is a big song and dance invented by Ecalle in the 1970's. I have only second-hand accounts, but enough to apply, see https://mathoverflow.net/questions/45608/formal-power-series-convergence including my own answer.
edited Jun 15 '17 at 8:15
Adam
1,1951919
answered Feb 14 '14 at 11:32
Will JagyWill Jagy
104k5103202
edited Jun 15 '17 at 8:15
Adam
1,1951919
edited Jun 15 '17 at 8:15
Adam
1,1951919
edited Jun 15 '17 at 8:15
Adam
1,1951919
1,1951919
answered Feb 14 '14 at 11:32
Will JagyWill Jagy
104k5103202
answered Feb 14 '14 at 11:32
Will JagyWill Jagy
104k5103202
answered Feb 14 '14 at 11:32
Will JagyWill Jagy
104k5103202
104k5103202
$begingroup$
It applies for $n in R$?
$endgroup$
– Erivelton Geraldo Nepomuceno
Feb 14 '14 at 11:42
1
$begingroup$
Only under the most fortunate circumstances does it apply for all real $n.$ More common is that it applies for rational $n$ with a certain denominator. This is a result of Baker, finished by his student Liverpool.
$endgroup$
– Will Jagy
Feb 14 '14 at 11:48
2
$begingroup$
Your first link is broken.
$endgroup$
– Ruslan
Nov 21 '16 at 11:03
add a comment |
$begingroup$
It applies for $n in R$?
$endgroup$
– Erivelton Geraldo Nepomuceno
Feb 14 '14 at 11:42
1
$begingroup$
Only under the most fortunate circumstances does it apply for all real $n.$ More common is that it applies for rational $n$ with a certain denominator. This is a result of Baker, finished by his student Liverpool.
$endgroup$
– Will Jagy
Feb 14 '14 at 11:48
2
$begingroup$
Your first link is broken.
$endgroup$
– Ruslan
Nov 21 '16 at 11:03
$begingroup$
It applies for $n in R$?
$endgroup$
– Erivelton Geraldo Nepomuceno
Feb 14 '14 at 11:42
$begingroup$
It applies for $n in R$?
$endgroup$
– Erivelton Geraldo Nepomuceno
Feb 14 '14 at 11:42
1
1
$begingroup$
Only under the most fortunate circumstances does it apply for all real $n.$ More common is that it applies for rational $n$ with a certain denominator. This is a result of Baker, finished by his student Liverpool.
$endgroup$
– Will Jagy
Feb 14 '14 at 11:48
$begingroup$
Only under the most fortunate circumstances does it apply for all real $n.$ More common is that it applies for rational $n$ with a certain denominator. This is a result of Baker, finished by his student Liverpool.
$endgroup$
– Will Jagy
Feb 14 '14 at 11:48
2
2
$begingroup$
Your first link is broken.
$endgroup$
– Ruslan
Nov 21 '16 at 11:03
$begingroup$
Your first link is broken.
$endgroup$
– Ruslan
Nov 21 '16 at 11:03
add a comment |
$begingroup$
Functional square roots are relatively easy to find through Functional Conjugacy specifications of generic iterated functions; find the general iterate, and set n=1/2 in the general expression, such as the ones available and listed there.
For example, the functional square root of the nth Chebyshev polynomial $T_n$ is evidently $cos (sqrt{n} arccos (x)) $ --- which, however, is not a polynomial, in general.
As Will Jagy already mentioned, in general, around a fixed point, say $f(0)=0$, the standard tried-and-true method is Schröder's equation, $psi(f(x))=f'(0) ~psi(x)$, a triumph of functional conjugacy, provided f ' there ≠1. Having found ψ, one has $f^{1/2}(x)=psi^{-1}left (sqrt{f'(0)} ~ psi(x)right )$.
If it is =1, at the fixed point, however, all is not lost, and conjugacy still delivers: Curtright, Jin, & Zachos, JouPhys A Math-Th 44.40 (2011): 405205 illustrates how $f^n circ g circ f^{-n}$ will improve an initial approximant g rapidly and dramatically for large n.
(This is illustrated there for intuitive functions such as $f=sin x$, blue, around the origin, where the half iterate is orange, the second iterate is red, and so on...)
answered May 26 '17 at 1:12
Cosmas ZachosCosmas Zachos
1,840522
$endgroup$
add a comment |
$begingroup$
Functional square roots are relatively easy to find through Functional Conjugacy specifications of generic iterated functions; find the general iterate, and set n=1/2 in the general expression, such as the ones available and listed there.
For example, the functional square root of the nth Chebyshev polynomial $T_n$ is evidently $cos (sqrt{n} arccos (x)) $ --- which, however, is not a polynomial, in general.
As Will Jagy already mentioned, in general, around a fixed point, say $f(0)=0$, the standard tried-and-true method is Schröder's equation, $psi(f(x))=f'(0) ~psi(x)$, a triumph of functional conjugacy, provided f ' there ≠1. Having found ψ, one has $f^{1/2}(x)=psi^{-1}left (sqrt{f'(0)} ~ psi(x)right )$.
If it is =1, at the fixed point, however, all is not lost, and conjugacy still delivers: Curtright, Jin, & Zachos, JouPhys A Math-Th 44.40 (2011): 405205 illustrates how $f^n circ g circ f^{-n}$ will improve an initial approximant g rapidly and dramatically for large n.
(This is illustrated there for intuitive functions such as $f=sin x$, blue, around the origin, where the half iterate is orange, the second iterate is red, and so on...)
answered May 26 '17 at 1:12
Cosmas ZachosCosmas Zachos
1,840522
$endgroup$
add a comment |
$begingroup$
Functional square roots are relatively easy to find through Functional Conjugacy specifications of generic iterated functions; find the general iterate, and set n=1/2 in the general expression, such as the ones available and listed there.
For example, the functional square root of the nth Chebyshev polynomial $T_n$ is evidently $cos (sqrt{n} arccos (x)) $ --- which, however, is not a polynomial, in general.
As Will Jagy already mentioned, in general, around a fixed point, say $f(0)=0$, the standard tried-and-true method is Schröder's equation, $psi(f(x))=f'(0) ~psi(x)$, a triumph of functional conjugacy, provided f ' there ≠1. Having found ψ, one has $f^{1/2}(x)=psi^{-1}left (sqrt{f'(0)} ~ psi(x)right )$.
If it is =1, at the fixed point, however, all is not lost, and conjugacy still delivers: Curtright, Jin, & Zachos, JouPhys A Math-Th 44.40 (2011): 405205 illustrates how $f^n circ g circ f^{-n}$ will improve an initial approximant g rapidly and dramatically for large n.
(This is illustrated there for intuitive functions such as $f=sin x$, blue, around the origin, where the half iterate is orange, the second iterate is red, and so on...)
answered May 26 '17 at 1:12
Cosmas ZachosCosmas Zachos
1,840522
$endgroup$
Functional square roots are relatively easy to find through Functional Conjugacy specifications of generic iterated functions; find the general iterate, and set n=1/2 in the general expression, such as the ones available and listed there.
For example, the functional square root of the nth Chebyshev polynomial $T_n$ is evidently $cos (sqrt{n} arccos (x)) $ --- which, however, is not a polynomial, in general.
As Will Jagy already mentioned, in general, around a fixed point, say $f(0)=0$, the standard tried-and-true method is Schröder's equation, $psi(f(x))=f'(0) ~psi(x)$, a triumph of functional conjugacy, provided f ' there ≠1. Having found ψ, one has $f^{1/2}(x)=psi^{-1}left (sqrt{f'(0)} ~ psi(x)right )$.
If it is =1, at the fixed point, however, all is not lost, and conjugacy still delivers: Curtright, Jin, & Zachos, JouPhys A Math-Th 44.40 (2011): 405205 illustrates how $f^n circ g circ f^{-n}$ will improve an initial approximant g rapidly and dramatically for large n.
(This is illustrated there for intuitive functions such as $f=sin x$, blue, around the origin, where the half iterate is orange, the second iterate is red, and so on...)
answered May 26 '17 at 1:12
Cosmas ZachosCosmas Zachos
1,840522
edited Jul 10 '17 at 16:48
answered May 26 '17 at 1:12
Cosmas ZachosCosmas Zachos
1,840522
answered May 26 '17 at 1:12
Cosmas ZachosCosmas Zachos
1,840522
answered May 26 '17 at 1:12
Cosmas ZachosCosmas Zachos
1,840522
1,840522
add a comment |
add a comment |
$begingroup$
Just some simple examples, allowing much obvious/natural versions of a fractional iterate:
Let $f(x)=x+a$ then $f(f(x))=x+2a$ , $f(f(f(x))) = x+3a$ and in general $f°^h(x)=x+hcdot a$. Then $f°^{0.5} (x) = x+0.5 a$
Let $f(x)=x cdot a$ then $f(f(x))=x cdot a^2$ , $f(f(f(x))) = x cdot a^3$ and in general $f°^h(x)=x cdot a^h $ . Then $f°^{0.5} (x) = x cdot a^{0.5} $
Let $f(x)=x ^ a$ then $f(f(x))=x ^{ a^2}$ , $f(f(f(x))) = x ^ { a^3}$ and in general $f°^h(x)=x ^{ a^h} $ . Then $f°^{0.5} (x) = x ^ {sqrt {a}} $
If a function has a power series without constant term , then you can find a formal power series for a fractional iterate. If $f(x) = ax + O(x^2)$ and $0<a<1$ then sometimes that series for fractional iterates might have a nonzero radius of convergence. If the power series has a constant term, one can sometimes use conjugacy to find a formal power series. More on this using the method of Carleman-matrices, see wikipedia:Carleman-matrix. For the specific case of $f(x) = a^x$ see wikipedia:tetration also there is something in wikipedia:hyperoperations .
Also note that a collection of Q&A about this subject is in MSE (tag::tetration, tag::hyperoperation) and in MO tag::fractional-iteration.
A very involved discussion can be found at tetration-forum: http://math.eretrandre.org/tetrationforum and many more links can be found using google-search. In mathoverflow you can find this question with answers: MO1 , MO2 , MO3
The problem is really not simple, the beginning of its consideration in a systematic way was only in the 19th century with N. Abel and E. Schroeder providing fairly general schemes - but only for functions of certain classes. Some singular gems might have been found earlier, say Euler's/Goldbach's finding of the range for the base $a$ in $f(x)=a^x$ where this converges even if infinitely iterated (but no fractional iterate has been discussed with this)
answered Jul 4 '17 at 8:41
Gottfried HelmsGottfried Helms
23.7k245101
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Just some simple examples, allowing much obvious/natural versions of a fractional iterate:
Let $f(x)=x+a$ then $f(f(x))=x+2a$ , $f(f(f(x))) = x+3a$ and in general $f°^h(x)=x+hcdot a$. Then $f°^{0.5} (x) = x+0.5 a$
Let $f(x)=x cdot a$ then $f(f(x))=x cdot a^2$ , $f(f(f(x))) = x cdot a^3$ and in general $f°^h(x)=x cdot a^h $ . Then $f°^{0.5} (x) = x cdot a^{0.5} $
Let $f(x)=x ^ a$ then $f(f(x))=x ^{ a^2}$ , $f(f(f(x))) = x ^ { a^3}$ and in general $f°^h(x)=x ^{ a^h} $ . Then $f°^{0.5} (x) = x ^ {sqrt {a}} $
If a function has a power series without constant term , then you can find a formal power series for a fractional iterate. If $f(x) = ax + O(x^2)$ and $0<a<1$ then sometimes that series for fractional iterates might have a nonzero radius of convergence. If the power series has a constant term, one can sometimes use conjugacy to find a formal power series. More on this using the method of Carleman-matrices, see wikipedia:Carleman-matrix. For the specific case of $f(x) = a^x$ see wikipedia:tetration also there is something in wikipedia:hyperoperations .
Also note that a collection of Q&A about this subject is in MSE (tag::tetration, tag::hyperoperation) and in MO tag::fractional-iteration.
A very involved discussion can be found at tetration-forum: http://math.eretrandre.org/tetrationforum and many more links can be found using google-search. In mathoverflow you can find this question with answers: MO1 , MO2 , MO3
The problem is really not simple, the beginning of its consideration in a systematic way was only in the 19th century with N. Abel and E. Schroeder providing fairly general schemes - but only for functions of certain classes. Some singular gems might have been found earlier, say Euler's/Goldbach's finding of the range for the base $a$ in $f(x)=a^x$ where this converges even if infinitely iterated (but no fractional iterate has been discussed with this)
answered Jul 4 '17 at 8:41
Gottfried HelmsGottfried Helms
23.7k245101
$endgroup$
add a comment |
$begingroup$
Just some simple examples, allowing much obvious/natural versions of a fractional iterate:
Let $f(x)=x+a$ then $f(f(x))=x+2a$ , $f(f(f(x))) = x+3a$ and in general $f°^h(x)=x+hcdot a$. Then $f°^{0.5} (x) = x+0.5 a$
Let $f(x)=x cdot a$ then $f(f(x))=x cdot a^2$ , $f(f(f(x))) = x cdot a^3$ and in general $f°^h(x)=x cdot a^h $ . Then $f°^{0.5} (x) = x cdot a^{0.5} $
Let $f(x)=x ^ a$ then $f(f(x))=x ^{ a^2}$ , $f(f(f(x))) = x ^ { a^3}$ and in general $f°^h(x)=x ^{ a^h} $ . Then $f°^{0.5} (x) = x ^ {sqrt {a}} $
If a function has a power series without constant term , then you can find a formal power series for a fractional iterate. If $f(x) = ax + O(x^2)$ and $0<a<1$ then sometimes that series for fractional iterates might have a nonzero radius of convergence. If the power series has a constant term, one can sometimes use conjugacy to find a formal power series. More on this using the method of Carleman-matrices, see wikipedia:Carleman-matrix. For the specific case of $f(x) = a^x$ see wikipedia:tetration also there is something in wikipedia:hyperoperations .
Also note that a collection of Q&A about this subject is in MSE (tag::tetration, tag::hyperoperation) and in MO tag::fractional-iteration.
A very involved discussion can be found at tetration-forum: http://math.eretrandre.org/tetrationforum and many more links can be found using google-search. In mathoverflow you can find this question with answers: MO1 , MO2 , MO3
The problem is really not simple, the beginning of its consideration in a systematic way was only in the 19th century with N. Abel and E. Schroeder providing fairly general schemes - but only for functions of certain classes. Some singular gems might have been found earlier, say Euler's/Goldbach's finding of the range for the base $a$ in $f(x)=a^x$ where this converges even if infinitely iterated (but no fractional iterate has been discussed with this)
answered Jul 4 '17 at 8:41
Gottfried HelmsGottfried Helms
23.7k245101
$endgroup$
Just some simple examples, allowing much obvious/natural versions of a fractional iterate:
Let $f(x)=x+a$ then $f(f(x))=x+2a$ , $f(f(f(x))) = x+3a$ and in general $f°^h(x)=x+hcdot a$. Then $f°^{0.5} (x) = x+0.5 a$
Let $f(x)=x cdot a$ then $f(f(x))=x cdot a^2$ , $f(f(f(x))) = x cdot a^3$ and in general $f°^h(x)=x cdot a^h $ . Then $f°^{0.5} (x) = x cdot a^{0.5} $
Let $f(x)=x ^ a$ then $f(f(x))=x ^{ a^2}$ , $f(f(f(x))) = x ^ { a^3}$ and in general $f°^h(x)=x ^{ a^h} $ . Then $f°^{0.5} (x) = x ^ {sqrt {a}} $
If a function has a power series without constant term , then you can find a formal power series for a fractional iterate. If $f(x) = ax + O(x^2)$ and $0<a<1$ then sometimes that series for fractional iterates might have a nonzero radius of convergence. If the power series has a constant term, one can sometimes use conjugacy to find a formal power series. More on this using the method of Carleman-matrices, see wikipedia:Carleman-matrix. For the specific case of $f(x) = a^x$ see wikipedia:tetration also there is something in wikipedia:hyperoperations .
Also note that a collection of Q&A about this subject is in MSE (tag::tetration, tag::hyperoperation) and in MO tag::fractional-iteration.
A very involved discussion can be found at tetration-forum: http://math.eretrandre.org/tetrationforum and many more links can be found using google-search. In mathoverflow you can find this question with answers: MO1 , MO2 , MO3
The problem is really not simple, the beginning of its consideration in a systematic way was only in the 19th century with N. Abel and E. Schroeder providing fairly general schemes - but only for functions of certain classes. Some singular gems might have been found earlier, say Euler's/Goldbach's finding of the range for the base $a$ in $f(x)=a^x$ where this converges even if infinitely iterated (but no fractional iterate has been discussed with this)
answered Jul 4 '17 at 8:41
Gottfried HelmsGottfried Helms
23.7k245101
edited Jul 4 '17 at 23:08
answered Jul 4 '17 at 8:41
Gottfried HelmsGottfried Helms
23.7k245101
answered Jul 4 '17 at 8:41
Gottfried HelmsGottfried Helms
23.7k245101
answered Jul 4 '17 at 8:41
Gottfried HelmsGottfried Helms
23.7k245101
23.7k245101
add a comment |
add a comment |
$begingroup$
If you have a function g(x) with
g(g(x)) = f(x) , then you could formally set
g(x) = $f^{1/2}(x)$
answered Feb 14 '14 at 10:59
PeterPeter
49.3k1240138
$endgroup$
1
$begingroup$
But it will be difficult to find such a g(x) for a given f(x).
$endgroup$
– Peter
Feb 14 '14 at 11:01
$begingroup$
Thanks for your reply. But I think it is not the case. When I indicate $1/2$ I would like to mean that I am doing a partial composition. Suppose $n$ is the number of times of composition. Usually $n in N$. I would like to calculate when $n=1.2$ for example.
$endgroup$
– Erivelton Geraldo Nepomuceno
Feb 14 '14 at 11:33
add a comment |
$begingroup$
If you have a function g(x) with
g(g(x)) = f(x) , then you could formally set
g(x) = $f^{1/2}(x)$
answered Feb 14 '14 at 10:59
PeterPeter
49.3k1240138
$endgroup$
1
$begingroup$
But it will be difficult to find such a g(x) for a given f(x).
$endgroup$
– Peter
Feb 14 '14 at 11:01
$begingroup$
Thanks for your reply. But I think it is not the case. When I indicate $1/2$ I would like to mean that I am doing a partial composition. Suppose $n$ is the number of times of composition. Usually $n in N$. I would like to calculate when $n=1.2$ for example.
$endgroup$
– Erivelton Geraldo Nepomuceno
Feb 14 '14 at 11:33
add a comment |
$begingroup$
If you have a function g(x) with
g(g(x)) = f(x) , then you could formally set
g(x) = $f^{1/2}(x)$
answered Feb 14 '14 at 10:59
PeterPeter
49.3k1240138
$endgroup$
If you have a function g(x) with
g(g(x)) = f(x) , then you could formally set
g(x) = $f^{1/2}(x)$
answered Feb 14 '14 at 10:59
PeterPeter
49.3k1240138
answered Feb 14 '14 at 10:59
PeterPeter
49.3k1240138
answered Feb 14 '14 at 10:59
PeterPeter
49.3k1240138
answered Feb 14 '14 at 10:59
PeterPeter
49.3k1240138
49.3k1240138
1
$begingroup$
But it will be difficult to find such a g(x) for a given f(x).
$endgroup$
– Peter
Feb 14 '14 at 11:01
$begingroup$
Thanks for your reply. But I think it is not the case. When I indicate $1/2$ I would like to mean that I am doing a partial composition. Suppose $n$ is the number of times of composition. Usually $n in N$. I would like to calculate when $n=1.2$ for example.
$endgroup$
– Erivelton Geraldo Nepomuceno
Feb 14 '14 at 11:33
add a comment |
1
$begingroup$
But it will be difficult to find such a g(x) for a given f(x).
$endgroup$
– Peter
Feb 14 '14 at 11:01
$begingroup$
Thanks for your reply. But I think it is not the case. When I indicate $1/2$ I would like to mean that I am doing a partial composition. Suppose $n$ is the number of times of composition. Usually $n in N$. I would like to calculate when $n=1.2$ for example.
$endgroup$
– Erivelton Geraldo Nepomuceno
Feb 14 '14 at 11:33
1
1
$begingroup$
But it will be difficult to find such a g(x) for a given f(x).
$endgroup$
– Peter
Feb 14 '14 at 11:01
$begingroup$
But it will be difficult to find such a g(x) for a given f(x).
$endgroup$
– Peter
Feb 14 '14 at 11:01
$begingroup$
Thanks for your reply. But I think it is not the case. When I indicate $1/2$ I would like to mean that I am doing a partial composition. Suppose $n$ is the number of times of composition. Usually $n in N$. I would like to calculate when $n=1.2$ for example.
$endgroup$
– Erivelton Geraldo Nepomuceno
Feb 14 '14 at 11:33
$begingroup$
Thanks for your reply. But I think it is not the case. When I indicate $1/2$ I would like to mean that I am doing a partial composition. Suppose $n$ is the number of times of composition. Usually $n in N$. I would like to calculate when $n=1.2$ for example.
$endgroup$
– Erivelton Geraldo Nepomuceno
Feb 14 '14 at 11:33
add a comment |
$begingroup$
The fractional iterate is NON unique even with well behaved function : check the very nice book : Z.A Melzak : "Companion to concrete mathematics" Vol 1 : page 56 to 62
answered Feb 3 at 13:00
Jérôme JEAN-CHARLESJérôme JEAN-CHARLES
1114
$endgroup$
add a comment |
$begingroup$
The fractional iterate is NON unique even with well behaved function : check the very nice book : Z.A Melzak : "Companion to concrete mathematics" Vol 1 : page 56 to 62
answered Feb 3 at 13:00
Jérôme JEAN-CHARLESJérôme JEAN-CHARLES
1114
$endgroup$
add a comment |
$begingroup$
The fractional iterate is NON unique even with well behaved function : check the very nice book : Z.A Melzak : "Companion to concrete mathematics" Vol 1 : page 56 to 62
answered Feb 3 at 13:00
Jérôme JEAN-CHARLESJérôme JEAN-CHARLES
1114
$endgroup$
The fractional iterate is NON unique even with well behaved function : check the very nice book : Z.A Melzak : "Companion to concrete mathematics" Vol 1 : page 56 to 62
answered Feb 3 at 13:00
Jérôme JEAN-CHARLESJérôme JEAN-CHARLES
1114
answered Feb 3 at 13:00
Jérôme JEAN-CHARLESJérôme JEAN-CHARLES
1114
answered Feb 3 at 13:00
Jérôme JEAN-CHARLESJérôme JEAN-CHARLES
1114
answered Feb 3 at 13:00
Jérôme JEAN-CHARLESJérôme JEAN-CHARLES
1114
1114
add a comment |
add a comment |
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One can look at a collection of Q&A in mathoverflow: mathoverflow.net/questions/tagged/fractional-iteration or here: math.stackexchange.com/questions/tagged/tetration
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– Gottfried Helms
Jul 3 '17 at 18:52