Mixing
A function that grows faster than any function in the sequence $e^x, e^{e^x}, e^{e^{e^x}}$…
$begingroup$
Is there a continuous function constructed by elementary functions, or by integral formula involved only elementary functions (like Gamma function) that grows faster than any $e^{e^{e...^x}}$ ($e$ appears $n$ times)?
I ask for the answer with a single formula. Gluing continuous function together is too trivial.
The function need not to be defined on whole $mathbb{R}$, the domain $(a, infty)$ is acceptable.
calculus functions exponential-function elementary-functions
asked Aug 15 '16 at 11:50
PluviophilePluviophile
346111
$endgroup$
|
show 13 more comments
$begingroup$
Is there a continuous function constructed by elementary functions, or by integral formula involved only elementary functions (like Gamma function) that grows faster than any $e^{e^{e...^x}}$ ($e$ appears $n$ times)?
I ask for the answer with a single formula. Gluing continuous function together is too trivial.
The function need not to be defined on whole $mathbb{R}$, the domain $(a, infty)$ is acceptable.
calculus functions exponential-function elementary-functions
asked Aug 15 '16 at 11:50
PluviophilePluviophile
346111
$endgroup$
1
$begingroup$
Do you accept iterative procedures? If so The Ackermann Function generalises the series addition, multiplication, exponentiation. Likewise Knuth's Arrow Notation.
$endgroup$
– AlphaNumeric
Aug 15 '16 at 11:54
1
$begingroup$
I suppose that applying any of the allowed operations (composition with elementasry function or integration) turns a function that is bounded by some e-tower into a function bounded by some (possibly higher) e-tower.
$endgroup$
– Hagen von Eitzen
Aug 15 '16 at 11:59
3
$begingroup$
Perhaps an annoying example, but for any $;a>e;,;;a^{a^{aldots^x}};$ grows faster
$endgroup$
– DonAntonio
Aug 15 '16 at 12:01
3
$begingroup$
I suppose you mean to find a function $f$ that eventually grows faster than each function in the set ${exp^n(x)mid ninBbb N}$, where $n$ in the power denotes iteration of function.
$endgroup$
– edm
Aug 15 '16 at 12:31
2
$begingroup$
I mean define $f(x) =$ $C_k$ + exp(exp(..(x)) (k iterations) on $[k, k+1]$, $C_k$ is chosen so that $f$ is continuous.
$endgroup$
– Pluviophile
Aug 15 '16 at 13:32
|
show 13 more comments
$begingroup$
Is there a continuous function constructed by elementary functions, or by integral formula involved only elementary functions (like Gamma function) that grows faster than any $e^{e^{e...^x}}$ ($e$ appears $n$ times)?
I ask for the answer with a single formula. Gluing continuous function together is too trivial.
The function need not to be defined on whole $mathbb{R}$, the domain $(a, infty)$ is acceptable.
calculus functions exponential-function elementary-functions
asked Aug 15 '16 at 11:50
PluviophilePluviophile
346111
$endgroup$
Is there a continuous function constructed by elementary functions, or by integral formula involved only elementary functions (like Gamma function) that grows faster than any $e^{e^{e...^x}}$ ($e$ appears $n$ times)?
I ask for the answer with a single formula. Gluing continuous function together is too trivial.
The function need not to be defined on whole $mathbb{R}$, the domain $(a, infty)$ is acceptable.
calculus functions exponential-function elementary-functions
calculus functions exponential-function elementary-functions
asked Aug 15 '16 at 11:50
PluviophilePluviophile
346111
asked Aug 15 '16 at 11:50
PluviophilePluviophile
346111
edited Aug 15 '16 at 13:08
Pluviophile
asked Aug 15 '16 at 11:50
PluviophilePluviophile
346111
asked Aug 15 '16 at 11:50
PluviophilePluviophile
346111
asked Aug 15 '16 at 11:50
PluviophilePluviophile
346111
346111
1
$begingroup$
Do you accept iterative procedures? If so The Ackermann Function generalises the series addition, multiplication, exponentiation. Likewise Knuth's Arrow Notation.
$endgroup$
– AlphaNumeric
Aug 15 '16 at 11:54
1
$begingroup$
I suppose that applying any of the allowed operations (composition with elementasry function or integration) turns a function that is bounded by some e-tower into a function bounded by some (possibly higher) e-tower.
$endgroup$
– Hagen von Eitzen
Aug 15 '16 at 11:59
3
$begingroup$
Perhaps an annoying example, but for any $;a>e;,;;a^{a^{aldots^x}};$ grows faster
$endgroup$
– DonAntonio
Aug 15 '16 at 12:01
3
$begingroup$
I suppose you mean to find a function $f$ that eventually grows faster than each function in the set ${exp^n(x)mid ninBbb N}$, where $n$ in the power denotes iteration of function.
$endgroup$
– edm
Aug 15 '16 at 12:31
2
$begingroup$
I mean define $f(x) =$ $C_k$ + exp(exp(..(x)) (k iterations) on $[k, k+1]$, $C_k$ is chosen so that $f$ is continuous.
$endgroup$
– Pluviophile
Aug 15 '16 at 13:32
|
show 13 more comments
1
$begingroup$
Do you accept iterative procedures? If so The Ackermann Function generalises the series addition, multiplication, exponentiation. Likewise Knuth's Arrow Notation.
$endgroup$
– AlphaNumeric
Aug 15 '16 at 11:54
1
$begingroup$
I suppose that applying any of the allowed operations (composition with elementasry function or integration) turns a function that is bounded by some e-tower into a function bounded by some (possibly higher) e-tower.
$endgroup$
– Hagen von Eitzen
Aug 15 '16 at 11:59
3
$begingroup$
Perhaps an annoying example, but for any $;a>e;,;;a^{a^{aldots^x}};$ grows faster
$endgroup$
– DonAntonio
Aug 15 '16 at 12:01
3
$begingroup$
I suppose you mean to find a function $f$ that eventually grows faster than each function in the set ${exp^n(x)mid ninBbb N}$, where $n$ in the power denotes iteration of function.
$endgroup$
– edm
Aug 15 '16 at 12:31
2
$begingroup$
I mean define $f(x) =$ $C_k$ + exp(exp(..(x)) (k iterations) on $[k, k+1]$, $C_k$ is chosen so that $f$ is continuous.
$endgroup$
– Pluviophile
Aug 15 '16 at 13:32
1
1
$begingroup$
Do you accept iterative procedures? If so The Ackermann Function generalises the series addition, multiplication, exponentiation. Likewise Knuth's Arrow Notation.
$endgroup$
– AlphaNumeric
Aug 15 '16 at 11:54
$begingroup$
Do you accept iterative procedures? If so The Ackermann Function generalises the series addition, multiplication, exponentiation. Likewise Knuth's Arrow Notation.
$endgroup$
– AlphaNumeric
Aug 15 '16 at 11:54
1
1
$begingroup$
I suppose that applying any of the allowed operations (composition with elementasry function or integration) turns a function that is bounded by some e-tower into a function bounded by some (possibly higher) e-tower.
$endgroup$
– Hagen von Eitzen
Aug 15 '16 at 11:59
$begingroup$
I suppose that applying any of the allowed operations (composition with elementasry function or integration) turns a function that is bounded by some e-tower into a function bounded by some (possibly higher) e-tower.
$endgroup$
– Hagen von Eitzen
Aug 15 '16 at 11:59
3
3
$begingroup$
Perhaps an annoying example, but for any $;a>e;,;;a^{a^{aldots^x}};$ grows faster
$endgroup$
– DonAntonio
Aug 15 '16 at 12:01
$begingroup$
Perhaps an annoying example, but for any $;a>e;,;;a^{a^{aldots^x}};$ grows faster
$endgroup$
– DonAntonio
Aug 15 '16 at 12:01
3
3
$begingroup$
I suppose you mean to find a function $f$ that eventually grows faster than each function in the set ${exp^n(x)mid ninBbb N}$, where $n$ in the power denotes iteration of function.
$endgroup$
– edm
Aug 15 '16 at 12:31
$begingroup$
I suppose you mean to find a function $f$ that eventually grows faster than each function in the set ${exp^n(x)mid ninBbb N}$, where $n$ in the power denotes iteration of function.
$endgroup$
– edm
Aug 15 '16 at 12:31
2
2
$begingroup$
I mean define $f(x) =$ $C_k$ + exp(exp(..(x)) (k iterations) on $[k, k+1]$, $C_k$ is chosen so that $f$ is continuous.
$endgroup$
– Pluviophile
Aug 15 '16 at 13:32
$begingroup$
I mean define $f(x) =$ $C_k$ + exp(exp(..(x)) (k iterations) on $[k, k+1]$, $C_k$ is chosen so that $f$ is continuous.
$endgroup$
– Pluviophile
Aug 15 '16 at 13:32
|
show 13 more comments
2 Answers
2
active
oldest
votes
$begingroup$
a negative answer
No, there is no such function constructed in a "natural" way, if we construe that word propertly.
Perhaps consult the literature on "transseries" ...
For example, transseries in the sense here:
G. A. Edgar, "Transseries for Beginners". Real Analysis Exchange 35 (2010) 253-310 .
The set of transseries includes real elementary functions, is closed under indefinite integration, composition, and many other operations.
But no transseries has growth rate beyond all $e^{dots e^x}$. There is an integer "exponentiality" associated with each (large, positive) transseries; for example Exercise 4.10 in:
J. van der Hoeven, Transseries and Real Differential Algebra (LNM 1888) (Springer 2006)
A (large, positive) transseries with exponentiality $n+1$ grows faster than any transseries with exponentiality ${} le n$.
And $e^{dots e^x}$ with $n$ exponentiations has exponentiality $n$.
answered Aug 15 '16 at 13:52
GEdgarGEdgar
62.4k267171
$endgroup$
1
$begingroup$
If $f$ is a power series that converges on $mathbb{R}$, do we still get a negative answer?
$endgroup$
– Pluviophile
Aug 15 '16 at 14:51
add a comment |
$begingroup$
Here is a proof that there is no such function und elementary function.
Let $A$ be a collection of functions which grow slower than the collection $e^x, e^{e^x} , dots $.
We know that the sum, product composition and integral of functions which grow slower than $e^x, e^{e^x} , dots $ also grows slower that $e^x, e^{e^x} , dots $.
Then the set of functions which are finite iterate sums, $dots$ of functions in $A$ also grows slower that $e^x, e^{e^x} , dots $
Now, depending on or definition of elementary function, if you take $A= {1, x, exp, log, x^a }$ you should be done.
answered Aug 15 '16 at 13:58
user60589user60589
2,2971815
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
a negative answer
No, there is no such function constructed in a "natural" way, if we construe that word propertly.
Perhaps consult the literature on "transseries" ...
For example, transseries in the sense here:
G. A. Edgar, "Transseries for Beginners". Real Analysis Exchange 35 (2010) 253-310 .
The set of transseries includes real elementary functions, is closed under indefinite integration, composition, and many other operations.
But no transseries has growth rate beyond all $e^{dots e^x}$. There is an integer "exponentiality" associated with each (large, positive) transseries; for example Exercise 4.10 in:
J. van der Hoeven, Transseries and Real Differential Algebra (LNM 1888) (Springer 2006)
A (large, positive) transseries with exponentiality $n+1$ grows faster than any transseries with exponentiality ${} le n$.
And $e^{dots e^x}$ with $n$ exponentiations has exponentiality $n$.
answered Aug 15 '16 at 13:52
GEdgarGEdgar
62.4k267171
$endgroup$
1
$begingroup$
If $f$ is a power series that converges on $mathbb{R}$, do we still get a negative answer?
$endgroup$
– Pluviophile
Aug 15 '16 at 14:51
add a comment |
$begingroup$
a negative answer
No, there is no such function constructed in a "natural" way, if we construe that word propertly.
Perhaps consult the literature on "transseries" ...
For example, transseries in the sense here:
G. A. Edgar, "Transseries for Beginners". Real Analysis Exchange 35 (2010) 253-310 .
The set of transseries includes real elementary functions, is closed under indefinite integration, composition, and many other operations.
But no transseries has growth rate beyond all $e^{dots e^x}$. There is an integer "exponentiality" associated with each (large, positive) transseries; for example Exercise 4.10 in:
J. van der Hoeven, Transseries and Real Differential Algebra (LNM 1888) (Springer 2006)
A (large, positive) transseries with exponentiality $n+1$ grows faster than any transseries with exponentiality ${} le n$.
And $e^{dots e^x}$ with $n$ exponentiations has exponentiality $n$.
answered Aug 15 '16 at 13:52
GEdgarGEdgar
62.4k267171
$endgroup$
1
$begingroup$
If $f$ is a power series that converges on $mathbb{R}$, do we still get a negative answer?
$endgroup$
– Pluviophile
Aug 15 '16 at 14:51
add a comment |
$begingroup$
a negative answer
No, there is no such function constructed in a "natural" way, if we construe that word propertly.
Perhaps consult the literature on "transseries" ...
For example, transseries in the sense here:
G. A. Edgar, "Transseries for Beginners". Real Analysis Exchange 35 (2010) 253-310 .
The set of transseries includes real elementary functions, is closed under indefinite integration, composition, and many other operations.
But no transseries has growth rate beyond all $e^{dots e^x}$. There is an integer "exponentiality" associated with each (large, positive) transseries; for example Exercise 4.10 in:
J. van der Hoeven, Transseries and Real Differential Algebra (LNM 1888) (Springer 2006)
A (large, positive) transseries with exponentiality $n+1$ grows faster than any transseries with exponentiality ${} le n$.
And $e^{dots e^x}$ with $n$ exponentiations has exponentiality $n$.
answered Aug 15 '16 at 13:52
GEdgarGEdgar
62.4k267171
$endgroup$
a negative answer
No, there is no such function constructed in a "natural" way, if we construe that word propertly.
Perhaps consult the literature on "transseries" ...
For example, transseries in the sense here:
G. A. Edgar, "Transseries for Beginners". Real Analysis Exchange 35 (2010) 253-310 .
The set of transseries includes real elementary functions, is closed under indefinite integration, composition, and many other operations.
But no transseries has growth rate beyond all $e^{dots e^x}$. There is an integer "exponentiality" associated with each (large, positive) transseries; for example Exercise 4.10 in:
J. van der Hoeven, Transseries and Real Differential Algebra (LNM 1888) (Springer 2006)
A (large, positive) transseries with exponentiality $n+1$ grows faster than any transseries with exponentiality ${} le n$.
And $e^{dots e^x}$ with $n$ exponentiations has exponentiality $n$.
answered Aug 15 '16 at 13:52
GEdgarGEdgar
62.4k267171
answered Aug 15 '16 at 13:52
GEdgarGEdgar
62.4k267171
answered Aug 15 '16 at 13:52
GEdgarGEdgar
62.4k267171
answered Aug 15 '16 at 13:52
GEdgarGEdgar
62.4k267171
62.4k267171
1
$begingroup$
If $f$ is a power series that converges on $mathbb{R}$, do we still get a negative answer?
$endgroup$
– Pluviophile
Aug 15 '16 at 14:51
add a comment |
1
$begingroup$
If $f$ is a power series that converges on $mathbb{R}$, do we still get a negative answer?
$endgroup$
– Pluviophile
Aug 15 '16 at 14:51
1
1
$begingroup$
If $f$ is a power series that converges on $mathbb{R}$, do we still get a negative answer?
$endgroup$
– Pluviophile
Aug 15 '16 at 14:51
$begingroup$
If $f$ is a power series that converges on $mathbb{R}$, do we still get a negative answer?
$endgroup$
– Pluviophile
Aug 15 '16 at 14:51
add a comment |
$begingroup$
Here is a proof that there is no such function und elementary function.
Let $A$ be a collection of functions which grow slower than the collection $e^x, e^{e^x} , dots $.
We know that the sum, product composition and integral of functions which grow slower than $e^x, e^{e^x} , dots $ also grows slower that $e^x, e^{e^x} , dots $.
Then the set of functions which are finite iterate sums, $dots$ of functions in $A$ also grows slower that $e^x, e^{e^x} , dots $
Now, depending on or definition of elementary function, if you take $A= {1, x, exp, log, x^a }$ you should be done.
answered Aug 15 '16 at 13:58
user60589user60589
2,2971815
$endgroup$
add a comment |
$begingroup$
Here is a proof that there is no such function und elementary function.
Let $A$ be a collection of functions which grow slower than the collection $e^x, e^{e^x} , dots $.
We know that the sum, product composition and integral of functions which grow slower than $e^x, e^{e^x} , dots $ also grows slower that $e^x, e^{e^x} , dots $.
Then the set of functions which are finite iterate sums, $dots$ of functions in $A$ also grows slower that $e^x, e^{e^x} , dots $
Now, depending on or definition of elementary function, if you take $A= {1, x, exp, log, x^a }$ you should be done.
answered Aug 15 '16 at 13:58
user60589user60589
2,2971815
$endgroup$
add a comment |
$begingroup$
Here is a proof that there is no such function und elementary function.
Let $A$ be a collection of functions which grow slower than the collection $e^x, e^{e^x} , dots $.
We know that the sum, product composition and integral of functions which grow slower than $e^x, e^{e^x} , dots $ also grows slower that $e^x, e^{e^x} , dots $.
Then the set of functions which are finite iterate sums, $dots$ of functions in $A$ also grows slower that $e^x, e^{e^x} , dots $
Now, depending on or definition of elementary function, if you take $A= {1, x, exp, log, x^a }$ you should be done.
answered Aug 15 '16 at 13:58
user60589user60589
2,2971815
$endgroup$
Here is a proof that there is no such function und elementary function.
Let $A$ be a collection of functions which grow slower than the collection $e^x, e^{e^x} , dots $.
We know that the sum, product composition and integral of functions which grow slower than $e^x, e^{e^x} , dots $ also grows slower that $e^x, e^{e^x} , dots $.
Then the set of functions which are finite iterate sums, $dots$ of functions in $A$ also grows slower that $e^x, e^{e^x} , dots $
Now, depending on or definition of elementary function, if you take $A= {1, x, exp, log, x^a }$ you should be done.
answered Aug 15 '16 at 13:58
user60589user60589
2,2971815
answered Aug 15 '16 at 13:58
user60589user60589
2,2971815
answered Aug 15 '16 at 13:58
user60589user60589
2,2971815
answered Aug 15 '16 at 13:58
user60589user60589
2,2971815
2,2971815
add a comment |
add a comment |
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1
$begingroup$
Do you accept iterative procedures? If so The Ackermann Function generalises the series addition, multiplication, exponentiation. Likewise Knuth's Arrow Notation.
$endgroup$
– AlphaNumeric
Aug 15 '16 at 11:54
1
$begingroup$
I suppose that applying any of the allowed operations (composition with elementasry function or integration) turns a function that is bounded by some e-tower into a function bounded by some (possibly higher) e-tower.
$endgroup$
– Hagen von Eitzen
Aug 15 '16 at 11:59
3
$begingroup$
Perhaps an annoying example, but for any $;a>e;,;;a^{a^{aldots^x}};$ grows faster
$endgroup$
– DonAntonio
Aug 15 '16 at 12:01
3
$begingroup$
I suppose you mean to find a function $f$ that eventually grows faster than each function in the set ${exp^n(x)mid ninBbb N}$, where $n$ in the power denotes iteration of function.
$endgroup$
– edm
Aug 15 '16 at 12:31
2
$begingroup$
I mean define $f(x) =$ $C_k$ + exp(exp(..(x)) (k iterations) on $[k, k+1]$, $C_k$ is chosen so that $f$ is continuous.
$endgroup$
– Pluviophile
Aug 15 '16 at 13:32