Bounds of fractional tetration
$begingroup$
I know about Kneser, but if we take a simple recursion
$$^{1/d}b=c, a(0)=b, a(n)=b^{frac{1}{^{d-1}(a(n-1))}}$$
so
$$limlimits_{ntoinfty}a(n)=c$$
and we can quickly find $c$ for positive integer $d>1$ and positive $b$. But $b$ has bounds, ex.
$$d=2, left[(frac{1}{e})^{frac{1}{e}}; e^eright]$$
$$d=3, [0; approx2,421]$$
$$d=4, [approx0,593; approx1,904]$$
$$d=5, [0; approx1,762]$$
$$d=6, [approx0,543; approx1,689]$$
$$d=7, [0; approx1,632]$$
$$d=8, [approx0,511; approx1,592]$$
What is the law of those bounds? Why is lower bound for odd $d$ is $0$?
tetration hyperoperation
edited Jan 13 at 23:20
Simply Beautiful Art
50.9k580186
asked Oct 28 '18 at 11:08
user514787user514787
749410
$endgroup$
add a comment |
$begingroup$
I know about Kneser, but if we take a simple recursion
$$^{1/d}b=c, a(0)=b, a(n)=b^{frac{1}{^{d-1}(a(n-1))}}$$
so
$$limlimits_{ntoinfty}a(n)=c$$
and we can quickly find $c$ for positive integer $d>1$ and positive $b$. But $b$ has bounds, ex.
$$d=2, left[(frac{1}{e})^{frac{1}{e}}; e^eright]$$
$$d=3, [0; approx2,421]$$
$$d=4, [approx0,593; approx1,904]$$
$$d=5, [0; approx1,762]$$
$$d=6, [approx0,543; approx1,689]$$
$$d=7, [0; approx1,632]$$
$$d=8, [approx0,511; approx1,592]$$
What is the law of those bounds? Why is lower bound for odd $d$ is $0$?
tetration hyperoperation
edited Jan 13 at 23:20
Simply Beautiful Art
50.9k580186
asked Oct 28 '18 at 11:08
user514787user514787
749410
$endgroup$
1
$begingroup$
How do you define $^xb$ for non-integer $x$?
$endgroup$
– Simply Beautiful Art
Jan 13 at 23:19
$begingroup$
@SimplyBeautifulArt, thank you for comment! If $^xb=c$, so $^{1/x}c=b$.
$endgroup$
– user514787
Jan 14 at 0:17
$begingroup$
That doesn't define it though. And what's to guarantee it's uniqueness?
$endgroup$
– Simply Beautiful Art
Jan 14 at 0:19
add a comment |
$begingroup$
I know about Kneser, but if we take a simple recursion
$$^{1/d}b=c, a(0)=b, a(n)=b^{frac{1}{^{d-1}(a(n-1))}}$$
so
$$limlimits_{ntoinfty}a(n)=c$$
and we can quickly find $c$ for positive integer $d>1$ and positive $b$. But $b$ has bounds, ex.
$$d=2, left[(frac{1}{e})^{frac{1}{e}}; e^eright]$$
$$d=3, [0; approx2,421]$$
$$d=4, [approx0,593; approx1,904]$$
$$d=5, [0; approx1,762]$$
$$d=6, [approx0,543; approx1,689]$$
$$d=7, [0; approx1,632]$$
$$d=8, [approx0,511; approx1,592]$$
What is the law of those bounds? Why is lower bound for odd $d$ is $0$?
tetration hyperoperation
edited Jan 13 at 23:20
Simply Beautiful Art
50.9k580186
asked Oct 28 '18 at 11:08
user514787user514787
749410
$endgroup$
I know about Kneser, but if we take a simple recursion
$$^{1/d}b=c, a(0)=b, a(n)=b^{frac{1}{^{d-1}(a(n-1))}}$$
so
$$limlimits_{ntoinfty}a(n)=c$$
and we can quickly find $c$ for positive integer $d>1$ and positive $b$. But $b$ has bounds, ex.
$$d=2, left[(frac{1}{e})^{frac{1}{e}}; e^eright]$$
$$d=3, [0; approx2,421]$$
$$d=4, [approx0,593; approx1,904]$$
$$d=5, [0; approx1,762]$$
$$d=6, [approx0,543; approx1,689]$$
$$d=7, [0; approx1,632]$$
$$d=8, [approx0,511; approx1,592]$$
What is the law of those bounds? Why is lower bound for odd $d$ is $0$?
tetration hyperoperation
tetration hyperoperation
edited Jan 13 at 23:20
Simply Beautiful Art
50.9k580186
asked Oct 28 '18 at 11:08
user514787user514787
749410
edited Jan 13 at 23:20
Simply Beautiful Art
50.9k580186
asked Oct 28 '18 at 11:08
user514787user514787
749410
edited Jan 13 at 23:20
Simply Beautiful Art
50.9k580186
edited Jan 13 at 23:20
Simply Beautiful Art
50.9k580186
edited Jan 13 at 23:20
Simply Beautiful Art
50.9k580186
50.9k580186
asked Oct 28 '18 at 11:08
user514787user514787
749410
asked Oct 28 '18 at 11:08
user514787user514787
749410
asked Oct 28 '18 at 11:08
user514787user514787
749410
749410
1
$begingroup$
How do you define $^xb$ for non-integer $x$?
$endgroup$
– Simply Beautiful Art
Jan 13 at 23:19
$begingroup$
@SimplyBeautifulArt, thank you for comment! If $^xb=c$, so $^{1/x}c=b$.
$endgroup$
– user514787
Jan 14 at 0:17
$begingroup$
That doesn't define it though. And what's to guarantee it's uniqueness?
$endgroup$
– Simply Beautiful Art
Jan 14 at 0:19
add a comment |
1
$begingroup$
How do you define $^xb$ for non-integer $x$?
$endgroup$
– Simply Beautiful Art
Jan 13 at 23:19
$begingroup$
@SimplyBeautifulArt, thank you for comment! If $^xb=c$, so $^{1/x}c=b$.
$endgroup$
– user514787
Jan 14 at 0:17
$begingroup$
That doesn't define it though. And what's to guarantee it's uniqueness?
$endgroup$
– Simply Beautiful Art
Jan 14 at 0:19
1
1
$begingroup$
How do you define $^xb$ for non-integer $x$?
$endgroup$
– Simply Beautiful Art
Jan 13 at 23:19
$begingroup$
How do you define $^xb$ for non-integer $x$?
$endgroup$
– Simply Beautiful Art
Jan 13 at 23:19
$begingroup$
@SimplyBeautifulArt, thank you for comment! If $^xb=c$, so $^{1/x}c=b$.
$endgroup$
– user514787
Jan 14 at 0:17
$begingroup$
@SimplyBeautifulArt, thank you for comment! If $^xb=c$, so $^{1/x}c=b$.
$endgroup$
– user514787
Jan 14 at 0:17
$begingroup$
That doesn't define it though. And what's to guarantee it's uniqueness?
$endgroup$
– Simply Beautiful Art
Jan 14 at 0:19
$begingroup$
That doesn't define it though. And what's to guarantee it's uniqueness?
$endgroup$
– Simply Beautiful Art
Jan 14 at 0:19
add a comment |
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$begingroup$
How do you define $^xb$ for non-integer $x$?
$endgroup$
– Simply Beautiful Art
Jan 13 at 23:19
$begingroup$
@SimplyBeautifulArt, thank you for comment! If $^xb=c$, so $^{1/x}c=b$.
$endgroup$
– user514787
Jan 14 at 0:17
$begingroup$
That doesn't define it though. And what's to guarantee it's uniqueness?
$endgroup$
– Simply Beautiful Art
Jan 14 at 0:19