Larger value with right associative tetration?
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0
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Given right associative tetration where:
$^{m}n =$ n^(n^(n^…))
And a situation such as:
$^{m}n = y$
$^{q}p = z$
What is a practical way to calculate which of $y$ and $z$ are larger?
I'm particularly looking at the case where:
$(n, m, p, q)$ are $> 1$
$p > n$
$m > q$
To simplify the question further, assume that $(n, m, p, q)$ are positive integers, they could for example be in the range 10 to 100.
Therefore $(y, z)$ are also positive integers.
exponentiation tetration
asked Nov 25 at 13:37
alan2here
516219
add a comment |
up vote
0
down vote
favorite
Given right associative tetration where:
$^{m}n =$ n^(n^(n^…))
And a situation such as:
$^{m}n = y$
$^{q}p = z$
What is a practical way to calculate which of $y$ and $z$ are larger?
I'm particularly looking at the case where:
$(n, m, p, q)$ are $> 1$
$p > n$
$m > q$
To simplify the question further, assume that $(n, m, p, q)$ are positive integers, they could for example be in the range 10 to 100.
Therefore $(y, z)$ are also positive integers.
exponentiation tetration
asked Nov 25 at 13:37
alan2here
516219
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given right associative tetration where:
$^{m}n =$ n^(n^(n^…))
And a situation such as:
$^{m}n = y$
$^{q}p = z$
What is a practical way to calculate which of $y$ and $z$ are larger?
I'm particularly looking at the case where:
$(n, m, p, q)$ are $> 1$
$p > n$
$m > q$
To simplify the question further, assume that $(n, m, p, q)$ are positive integers, they could for example be in the range 10 to 100.
Therefore $(y, z)$ are also positive integers.
exponentiation tetration
asked Nov 25 at 13:37
alan2here
516219
Given right associative tetration where:
$^{m}n =$ n^(n^(n^…))
And a situation such as:
$^{m}n = y$
$^{q}p = z$
What is a practical way to calculate which of $y$ and $z$ are larger?
I'm particularly looking at the case where:
$(n, m, p, q)$ are $> 1$
$p > n$
$m > q$
To simplify the question further, assume that $(n, m, p, q)$ are positive integers, they could for example be in the range 10 to 100.
Therefore $(y, z)$ are also positive integers.
exponentiation tetration
exponentiation tetration
asked Nov 25 at 13:37
alan2here
516219
asked Nov 25 at 13:37
alan2here
516219
edited Nov 25 at 23:44
asked Nov 25 at 13:37
alan2here
516219
asked Nov 25 at 13:37
alan2here
516219
asked Nov 25 at 13:37
alan2here
516219
516219
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Generally, the (exponent?, tetrand?)... the thing in the upper left, has more of an effect on the value than the base. For example the difference between $^43$ and $^37$ is astronomical (about 3.6 trillion orders of magnitude). This only gets more true as $m$ and $n$ increase.
As a rule of thumb, I would say whichever has the larger argument $m$ or $n$ is almost always larger. As for an exact answer, I don't know if there is one. Values resulting from tetration of large integers, say $^{987}4910$, tend do be so large that there is no way to calculate them exactly.
answered Nov 25 at 16:56
R. Burton
1417
I disagree with this answer. There is no quantification of formalization that a reasonable algorithm is impossible.
– AAAA
Nov 26 at 0:19
The quantification is that the number of digits in $^{987}4910$ exceeds the number of subatomic particles in the observable universe more than $10^{100000}$ times over. Even if you could devise an algorithm, it will still be physically impossible to write each digit of this number down, regardless of how small you make each digit. Furthermore, it would take more time than the universe will physically exist to run the algorithm. Even if you use scientific notation, you'll still be off by orders of magnitude.
– R. Burton
Nov 26 at 4:10
I do not think you understand the point. What you say is not a proof that two numbers represented in some way with $O(n)$ bits cannot be compared in $O(poly(n))$ time.
– AAAA
Nov 26 at 6:10
I suppose... but it doesn't make much difference if the time it takes just to calculate the time it will take to calculate the difference between two sufficiently large numbers exceeds the lifetime of the human species and $O(n)$ is larger than a universe's worth of storage space.
– R. Burton
Nov 26 at 12:23
add a comment |
up vote
1
down vote
There might be a way to use logarithms to simplify in a structured method. Look $a^{a^a}<b^{b^{b^b}}iff a^alog a<b^{b^b}log biff (alog a)+loglog a<(b^blog b)+loglog b$.
answered Nov 26 at 0:09
AAAA
1287
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Generally, the (exponent?, tetrand?)... the thing in the upper left, has more of an effect on the value than the base. For example the difference between $^43$ and $^37$ is astronomical (about 3.6 trillion orders of magnitude). This only gets more true as $m$ and $n$ increase.
As a rule of thumb, I would say whichever has the larger argument $m$ or $n$ is almost always larger. As for an exact answer, I don't know if there is one. Values resulting from tetration of large integers, say $^{987}4910$, tend do be so large that there is no way to calculate them exactly.
answered Nov 25 at 16:56
R. Burton
1417
I disagree with this answer. There is no quantification of formalization that a reasonable algorithm is impossible.
– AAAA
Nov 26 at 0:19
The quantification is that the number of digits in $^{987}4910$ exceeds the number of subatomic particles in the observable universe more than $10^{100000}$ times over. Even if you could devise an algorithm, it will still be physically impossible to write each digit of this number down, regardless of how small you make each digit. Furthermore, it would take more time than the universe will physically exist to run the algorithm. Even if you use scientific notation, you'll still be off by orders of magnitude.
– R. Burton
Nov 26 at 4:10
I do not think you understand the point. What you say is not a proof that two numbers represented in some way with $O(n)$ bits cannot be compared in $O(poly(n))$ time.
– AAAA
Nov 26 at 6:10
I suppose... but it doesn't make much difference if the time it takes just to calculate the time it will take to calculate the difference between two sufficiently large numbers exceeds the lifetime of the human species and $O(n)$ is larger than a universe's worth of storage space.
– R. Burton
Nov 26 at 12:23
add a comment |
up vote
1
down vote
accepted
Generally, the (exponent?, tetrand?)... the thing in the upper left, has more of an effect on the value than the base. For example the difference between $^43$ and $^37$ is astronomical (about 3.6 trillion orders of magnitude). This only gets more true as $m$ and $n$ increase.
As a rule of thumb, I would say whichever has the larger argument $m$ or $n$ is almost always larger. As for an exact answer, I don't know if there is one. Values resulting from tetration of large integers, say $^{987}4910$, tend do be so large that there is no way to calculate them exactly.
answered Nov 25 at 16:56
R. Burton
1417
I disagree with this answer. There is no quantification of formalization that a reasonable algorithm is impossible.
– AAAA
Nov 26 at 0:19
The quantification is that the number of digits in $^{987}4910$ exceeds the number of subatomic particles in the observable universe more than $10^{100000}$ times over. Even if you could devise an algorithm, it will still be physically impossible to write each digit of this number down, regardless of how small you make each digit. Furthermore, it would take more time than the universe will physically exist to run the algorithm. Even if you use scientific notation, you'll still be off by orders of magnitude.
– R. Burton
Nov 26 at 4:10
I do not think you understand the point. What you say is not a proof that two numbers represented in some way with $O(n)$ bits cannot be compared in $O(poly(n))$ time.
– AAAA
Nov 26 at 6:10
I suppose... but it doesn't make much difference if the time it takes just to calculate the time it will take to calculate the difference between two sufficiently large numbers exceeds the lifetime of the human species and $O(n)$ is larger than a universe's worth of storage space.
– R. Burton
Nov 26 at 12:23
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Generally, the (exponent?, tetrand?)... the thing in the upper left, has more of an effect on the value than the base. For example the difference between $^43$ and $^37$ is astronomical (about 3.6 trillion orders of magnitude). This only gets more true as $m$ and $n$ increase.
As a rule of thumb, I would say whichever has the larger argument $m$ or $n$ is almost always larger. As for an exact answer, I don't know if there is one. Values resulting from tetration of large integers, say $^{987}4910$, tend do be so large that there is no way to calculate them exactly.
answered Nov 25 at 16:56
R. Burton
1417
Generally, the (exponent?, tetrand?)... the thing in the upper left, has more of an effect on the value than the base. For example the difference between $^43$ and $^37$ is astronomical (about 3.6 trillion orders of magnitude). This only gets more true as $m$ and $n$ increase.
As a rule of thumb, I would say whichever has the larger argument $m$ or $n$ is almost always larger. As for an exact answer, I don't know if there is one. Values resulting from tetration of large integers, say $^{987}4910$, tend do be so large that there is no way to calculate them exactly.
answered Nov 25 at 16:56
R. Burton
1417
answered Nov 25 at 16:56
R. Burton
1417
answered Nov 25 at 16:56
R. Burton
1417
answered Nov 25 at 16:56
R. Burton
1417
1417
I disagree with this answer. There is no quantification of formalization that a reasonable algorithm is impossible.
– AAAA
Nov 26 at 0:19
The quantification is that the number of digits in $^{987}4910$ exceeds the number of subatomic particles in the observable universe more than $10^{100000}$ times over. Even if you could devise an algorithm, it will still be physically impossible to write each digit of this number down, regardless of how small you make each digit. Furthermore, it would take more time than the universe will physically exist to run the algorithm. Even if you use scientific notation, you'll still be off by orders of magnitude.
– R. Burton
Nov 26 at 4:10
I do not think you understand the point. What you say is not a proof that two numbers represented in some way with $O(n)$ bits cannot be compared in $O(poly(n))$ time.
– AAAA
Nov 26 at 6:10
I suppose... but it doesn't make much difference if the time it takes just to calculate the time it will take to calculate the difference between two sufficiently large numbers exceeds the lifetime of the human species and $O(n)$ is larger than a universe's worth of storage space.
– R. Burton
Nov 26 at 12:23
add a comment |
I disagree with this answer. There is no quantification of formalization that a reasonable algorithm is impossible.
– AAAA
Nov 26 at 0:19
The quantification is that the number of digits in $^{987}4910$ exceeds the number of subatomic particles in the observable universe more than $10^{100000}$ times over. Even if you could devise an algorithm, it will still be physically impossible to write each digit of this number down, regardless of how small you make each digit. Furthermore, it would take more time than the universe will physically exist to run the algorithm. Even if you use scientific notation, you'll still be off by orders of magnitude.
– R. Burton
Nov 26 at 4:10
I do not think you understand the point. What you say is not a proof that two numbers represented in some way with $O(n)$ bits cannot be compared in $O(poly(n))$ time.
– AAAA
Nov 26 at 6:10
I suppose... but it doesn't make much difference if the time it takes just to calculate the time it will take to calculate the difference between two sufficiently large numbers exceeds the lifetime of the human species and $O(n)$ is larger than a universe's worth of storage space.
– R. Burton
Nov 26 at 12:23
I disagree with this answer. There is no quantification of formalization that a reasonable algorithm is impossible.
– AAAA
Nov 26 at 0:19
I disagree with this answer. There is no quantification of formalization that a reasonable algorithm is impossible.
– AAAA
Nov 26 at 0:19
The quantification is that the number of digits in $^{987}4910$ exceeds the number of subatomic particles in the observable universe more than $10^{100000}$ times over. Even if you could devise an algorithm, it will still be physically impossible to write each digit of this number down, regardless of how small you make each digit. Furthermore, it would take more time than the universe will physically exist to run the algorithm. Even if you use scientific notation, you'll still be off by orders of magnitude.
– R. Burton
Nov 26 at 4:10
The quantification is that the number of digits in $^{987}4910$ exceeds the number of subatomic particles in the observable universe more than $10^{100000}$ times over. Even if you could devise an algorithm, it will still be physically impossible to write each digit of this number down, regardless of how small you make each digit. Furthermore, it would take more time than the universe will physically exist to run the algorithm. Even if you use scientific notation, you'll still be off by orders of magnitude.
– R. Burton
Nov 26 at 4:10
I do not think you understand the point. What you say is not a proof that two numbers represented in some way with $O(n)$ bits cannot be compared in $O(poly(n))$ time.
– AAAA
Nov 26 at 6:10
I do not think you understand the point. What you say is not a proof that two numbers represented in some way with $O(n)$ bits cannot be compared in $O(poly(n))$ time.
– AAAA
Nov 26 at 6:10
I suppose... but it doesn't make much difference if the time it takes just to calculate the time it will take to calculate the difference between two sufficiently large numbers exceeds the lifetime of the human species and $O(n)$ is larger than a universe's worth of storage space.
– R. Burton
Nov 26 at 12:23
I suppose... but it doesn't make much difference if the time it takes just to calculate the time it will take to calculate the difference between two sufficiently large numbers exceeds the lifetime of the human species and $O(n)$ is larger than a universe's worth of storage space.
– R. Burton
Nov 26 at 12:23
add a comment |
up vote
1
down vote
There might be a way to use logarithms to simplify in a structured method. Look $a^{a^a}<b^{b^{b^b}}iff a^alog a<b^{b^b}log biff (alog a)+loglog a<(b^blog b)+loglog b$.
answered Nov 26 at 0:09
AAAA
1287
add a comment |
up vote
1
down vote
There might be a way to use logarithms to simplify in a structured method. Look $a^{a^a}<b^{b^{b^b}}iff a^alog a<b^{b^b}log biff (alog a)+loglog a<(b^blog b)+loglog b$.
answered Nov 26 at 0:09
AAAA
1287
add a comment |
up vote
1
down vote
up vote
1
down vote
There might be a way to use logarithms to simplify in a structured method. Look $a^{a^a}<b^{b^{b^b}}iff a^alog a<b^{b^b}log biff (alog a)+loglog a<(b^blog b)+loglog b$.
answered Nov 26 at 0:09
AAAA
1287
There might be a way to use logarithms to simplify in a structured method. Look $a^{a^a}<b^{b^{b^b}}iff a^alog a<b^{b^b}log biff (alog a)+loglog a<(b^blog b)+loglog b$.
answered Nov 26 at 0:09
AAAA
1287
answered Nov 26 at 0:09
AAAA
1287
answered Nov 26 at 0:09
AAAA
1287
answered Nov 26 at 0:09
AAAA
1287
1287
add a comment |
add a comment |
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