How to solve this product recursion?
The recursion is given by
$$T(N)=2((log N)T(N^{3/8})T(N^{1/4}))^2$$
$$T(N)=1mbox{ if }M<1.$$
Is there a good upper bound?
analysis recurrence-relations asymptotics recursion upper-lower-bounds
asked Dec 26 at 13:56
Brout
2,1421428
|
show 1 more comment
The recursion is given by
$$T(N)=2((log N)T(N^{3/8})T(N^{1/4}))^2$$
$$T(N)=1mbox{ if }M<1.$$
Is there a good upper bound?
analysis recurrence-relations asymptotics recursion upper-lower-bounds
asked Dec 26 at 13:56
Brout
2,1421428
What is the domain of $T$?
– ajotatxe
Dec 26 at 14:04
$mathbb R_{geq0}$ and $Ninmathbb N$.
– Brout
Dec 26 at 14:05
But then we know absolutely nothing about, for example, $T(sqrt 2)$. It could be as big as you wish, and so it would be $T(4)$.
– ajotatxe
Dec 26 at 14:11
Ok how about large enough $N$? Also domain is integers?
– Brout
Dec 26 at 14:12
It is all the same. We don't have information about the values of $T(x)$ for irrational $x$. We don't have continuity, or any other good property of $T$. There's no possible bound. We only know that $T(1)=0$.
– ajotatxe
Dec 26 at 14:14
|
show 1 more comment
The recursion is given by
$$T(N)=2((log N)T(N^{3/8})T(N^{1/4}))^2$$
$$T(N)=1mbox{ if }M<1.$$
Is there a good upper bound?
analysis recurrence-relations asymptotics recursion upper-lower-bounds
asked Dec 26 at 13:56
Brout
2,1421428
The recursion is given by
$$T(N)=2((log N)T(N^{3/8})T(N^{1/4}))^2$$
$$T(N)=1mbox{ if }M<1.$$
Is there a good upper bound?
analysis recurrence-relations asymptotics recursion upper-lower-bounds
analysis recurrence-relations asymptotics recursion upper-lower-bounds
asked Dec 26 at 13:56
Brout
2,1421428
asked Dec 26 at 13:56
Brout
2,1421428
edited Dec 26 at 14:18
asked Dec 26 at 13:56
Brout
2,1421428
asked Dec 26 at 13:56
Brout
2,1421428
asked Dec 26 at 13:56
Brout
2,1421428
2,1421428
What is the domain of $T$?
– ajotatxe
Dec 26 at 14:04
$mathbb R_{geq0}$ and $Ninmathbb N$.
– Brout
Dec 26 at 14:05
But then we know absolutely nothing about, for example, $T(sqrt 2)$. It could be as big as you wish, and so it would be $T(4)$.
– ajotatxe
Dec 26 at 14:11
Ok how about large enough $N$? Also domain is integers?
– Brout
Dec 26 at 14:12
It is all the same. We don't have information about the values of $T(x)$ for irrational $x$. We don't have continuity, or any other good property of $T$. There's no possible bound. We only know that $T(1)=0$.
– ajotatxe
Dec 26 at 14:14
|
show 1 more comment
What is the domain of $T$?
– ajotatxe
Dec 26 at 14:04
$mathbb R_{geq0}$ and $Ninmathbb N$.
– Brout
Dec 26 at 14:05
But then we know absolutely nothing about, for example, $T(sqrt 2)$. It could be as big as you wish, and so it would be $T(4)$.
– ajotatxe
Dec 26 at 14:11
Ok how about large enough $N$? Also domain is integers?
– Brout
Dec 26 at 14:12
It is all the same. We don't have information about the values of $T(x)$ for irrational $x$. We don't have continuity, or any other good property of $T$. There's no possible bound. We only know that $T(1)=0$.
– ajotatxe
Dec 26 at 14:14
What is the domain of $T$?
– ajotatxe
Dec 26 at 14:04
What is the domain of $T$?
– ajotatxe
Dec 26 at 14:04
$mathbb R_{geq0}$ and $Ninmathbb N$.
– Brout
Dec 26 at 14:05
$mathbb R_{geq0}$ and $Ninmathbb N$.
– Brout
Dec 26 at 14:05
But then we know absolutely nothing about, for example, $T(sqrt 2)$. It could be as big as you wish, and so it would be $T(4)$.
– ajotatxe
Dec 26 at 14:11
But then we know absolutely nothing about, for example, $T(sqrt 2)$. It could be as big as you wish, and so it would be $T(4)$.
– ajotatxe
Dec 26 at 14:11
Ok how about large enough $N$? Also domain is integers?
– Brout
Dec 26 at 14:12
Ok how about large enough $N$? Also domain is integers?
– Brout
Dec 26 at 14:12
It is all the same. We don't have information about the values of $T(x)$ for irrational $x$. We don't have continuity, or any other good property of $T$. There's no possible bound. We only know that $T(1)=0$.
– ajotatxe
Dec 26 at 14:14
It is all the same. We don't have information about the values of $T(x)$ for irrational $x$. We don't have continuity, or any other good property of $T$. There's no possible bound. We only know that $T(1)=0$.
– ajotatxe
Dec 26 at 14:14
|
show 1 more comment
1 Answer
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No, there are no upper bounds on this function. In fact, either there are infinite families of functions satisfying this equation, or there are no solutions at all. Define the equivalence relation $sim$ on real numbers by $xsim y$ if there is exist $n,m in Bbb N$ such that $x^n = y^m$. The equivalence classes of this relation are countable, so there must be an uncountable number of such classes. And it is easy to verify that an infinite number of distinct equivalence classes contain natural numbers.
Yet $T$ can be defined independently on each equivalence class, as $N, N^{3/8}, N^{1/4}$ are all equivalent to each other. (In fact, there are certainly subclasses of these classes that can be defined independently - I just went with the easiest relation that worked, here.) For each independent set, you can choose an arbitrary $x$ for the set, choose an arbitrary value for $T(x)$ and figure out what values of $T(y)$ for $y$ in the same set are compatible with it (you will likely find other arbitrary choices are required to fully define $T$ even on that set).
As a result of being able to choose $T(x)$ arbitrarily for at least one $x$ in an infinite number of classes, $T$ does not satisfy any definite upper bound. You might be able to make your choices so that a bound exists, but you can also make choices that defy any bound.
answered Dec 27 at 2:03
Paul Sinclair
19.3k21441
add a comment |
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No, there are no upper bounds on this function. In fact, either there are infinite families of functions satisfying this equation, or there are no solutions at all. Define the equivalence relation $sim$ on real numbers by $xsim y$ if there is exist $n,m in Bbb N$ such that $x^n = y^m$. The equivalence classes of this relation are countable, so there must be an uncountable number of such classes. And it is easy to verify that an infinite number of distinct equivalence classes contain natural numbers.
Yet $T$ can be defined independently on each equivalence class, as $N, N^{3/8}, N^{1/4}$ are all equivalent to each other. (In fact, there are certainly subclasses of these classes that can be defined independently - I just went with the easiest relation that worked, here.) For each independent set, you can choose an arbitrary $x$ for the set, choose an arbitrary value for $T(x)$ and figure out what values of $T(y)$ for $y$ in the same set are compatible with it (you will likely find other arbitrary choices are required to fully define $T$ even on that set).
As a result of being able to choose $T(x)$ arbitrarily for at least one $x$ in an infinite number of classes, $T$ does not satisfy any definite upper bound. You might be able to make your choices so that a bound exists, but you can also make choices that defy any bound.
answered Dec 27 at 2:03
Paul Sinclair
19.3k21441
add a comment |
No, there are no upper bounds on this function. In fact, either there are infinite families of functions satisfying this equation, or there are no solutions at all. Define the equivalence relation $sim$ on real numbers by $xsim y$ if there is exist $n,m in Bbb N$ such that $x^n = y^m$. The equivalence classes of this relation are countable, so there must be an uncountable number of such classes. And it is easy to verify that an infinite number of distinct equivalence classes contain natural numbers.
Yet $T$ can be defined independently on each equivalence class, as $N, N^{3/8}, N^{1/4}$ are all equivalent to each other. (In fact, there are certainly subclasses of these classes that can be defined independently - I just went with the easiest relation that worked, here.) For each independent set, you can choose an arbitrary $x$ for the set, choose an arbitrary value for $T(x)$ and figure out what values of $T(y)$ for $y$ in the same set are compatible with it (you will likely find other arbitrary choices are required to fully define $T$ even on that set).
As a result of being able to choose $T(x)$ arbitrarily for at least one $x$ in an infinite number of classes, $T$ does not satisfy any definite upper bound. You might be able to make your choices so that a bound exists, but you can also make choices that defy any bound.
answered Dec 27 at 2:03
Paul Sinclair
19.3k21441
add a comment |
No, there are no upper bounds on this function. In fact, either there are infinite families of functions satisfying this equation, or there are no solutions at all. Define the equivalence relation $sim$ on real numbers by $xsim y$ if there is exist $n,m in Bbb N$ such that $x^n = y^m$. The equivalence classes of this relation are countable, so there must be an uncountable number of such classes. And it is easy to verify that an infinite number of distinct equivalence classes contain natural numbers.
Yet $T$ can be defined independently on each equivalence class, as $N, N^{3/8}, N^{1/4}$ are all equivalent to each other. (In fact, there are certainly subclasses of these classes that can be defined independently - I just went with the easiest relation that worked, here.) For each independent set, you can choose an arbitrary $x$ for the set, choose an arbitrary value for $T(x)$ and figure out what values of $T(y)$ for $y$ in the same set are compatible with it (you will likely find other arbitrary choices are required to fully define $T$ even on that set).
As a result of being able to choose $T(x)$ arbitrarily for at least one $x$ in an infinite number of classes, $T$ does not satisfy any definite upper bound. You might be able to make your choices so that a bound exists, but you can also make choices that defy any bound.
answered Dec 27 at 2:03
Paul Sinclair
19.3k21441
No, there are no upper bounds on this function. In fact, either there are infinite families of functions satisfying this equation, or there are no solutions at all. Define the equivalence relation $sim$ on real numbers by $xsim y$ if there is exist $n,m in Bbb N$ such that $x^n = y^m$. The equivalence classes of this relation are countable, so there must be an uncountable number of such classes. And it is easy to verify that an infinite number of distinct equivalence classes contain natural numbers.
Yet $T$ can be defined independently on each equivalence class, as $N, N^{3/8}, N^{1/4}$ are all equivalent to each other. (In fact, there are certainly subclasses of these classes that can be defined independently - I just went with the easiest relation that worked, here.) For each independent set, you can choose an arbitrary $x$ for the set, choose an arbitrary value for $T(x)$ and figure out what values of $T(y)$ for $y$ in the same set are compatible with it (you will likely find other arbitrary choices are required to fully define $T$ even on that set).
As a result of being able to choose $T(x)$ arbitrarily for at least one $x$ in an infinite number of classes, $T$ does not satisfy any definite upper bound. You might be able to make your choices so that a bound exists, but you can also make choices that defy any bound.
answered Dec 27 at 2:03
Paul Sinclair
19.3k21441
answered Dec 27 at 2:03
Paul Sinclair
19.3k21441
answered Dec 27 at 2:03
Paul Sinclair
19.3k21441
answered Dec 27 at 2:03
Paul Sinclair
19.3k21441
19.3k21441
add a comment |
add a comment |
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What is the domain of $T$?
– ajotatxe
Dec 26 at 14:04
$mathbb R_{geq0}$ and $Ninmathbb N$.
– Brout
Dec 26 at 14:05
But then we know absolutely nothing about, for example, $T(sqrt 2)$. It could be as big as you wish, and so it would be $T(4)$.
– ajotatxe
Dec 26 at 14:11
Ok how about large enough $N$? Also domain is integers?
– Brout
Dec 26 at 14:12
It is all the same. We don't have information about the values of $T(x)$ for irrational $x$. We don't have continuity, or any other good property of $T$. There's no possible bound. We only know that $T(1)=0$.
– ajotatxe
Dec 26 at 14:14