Mixing
$Q(x)=R(R(ldots(R(x))))$
Clash Royale CLAN TAG#URR8PPP
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Let $R(x) = fracF(x)G(x)$ be a rational function with $F(x), G(x) in mathbbZ[x]$ and
$F(x), G(x)$ have no common root modulo $p$ for all primes $p$.
Consider the rational function $Q(x)=underbraceR(R(ldots(R(x))))_text$n$ times$ where $n in mathbbN$.
Prove that if there is an integer $k$ such that $Q(k)=k$, then $R(R(k))=k$.
My thought,
in case $G(x) = 1$, $R(x)$ will be polynomial.
Let $a_1$ be the value such that $Q(a_1)=a_1$ and $R(a_i)=a_i+1, forall i= 1, 2, ldots, n-1$.
We have $a_2-a_1 mid a_3-a_2 mid ldots mid a_1-a_n mid a_2-a_1$, so $mid a_i+1-a_i mid$ is constant.
Then $mid R(a_i)-R(a_j)mid = mid a_i-a_j mid, forall i, j$
Since $mid a_i+1-a_i mid = c, forall i$ which will be true when orbit = $1$ or $2$,
so $R(R(k))=k$ or $R(k)=k$, we can conclude that $R(R(k))=k$
Please suggest how to proceed. Thank you.
polynomials algebraic-number-theory
asked Jul 30 at 6:25
Dan
1349
 |Â
show 2 more comments
up vote
2
down vote
favorite
Let $R(x) = fracF(x)G(x)$ be a rational function with $F(x), G(x) in mathbbZ[x]$ and
$F(x), G(x)$ have no common root modulo $p$ for all primes $p$.
Consider the rational function $Q(x)=underbraceR(R(ldots(R(x))))_text$n$ times$ where $n in mathbbN$.
Prove that if there is an integer $k$ such that $Q(k)=k$, then $R(R(k))=k$.
My thought,
in case $G(x) = 1$, $R(x)$ will be polynomial.
Let $a_1$ be the value such that $Q(a_1)=a_1$ and $R(a_i)=a_i+1, forall i= 1, 2, ldots, n-1$.
We have $a_2-a_1 mid a_3-a_2 mid ldots mid a_1-a_n mid a_2-a_1$, so $mid a_i+1-a_i mid$ is constant.
Then $mid R(a_i)-R(a_j)mid = mid a_i-a_j mid, forall i, j$
Since $mid a_i+1-a_i mid = c, forall i$ which will be true when orbit = $1$ or $2$,
so $R(R(k))=k$ or $R(k)=k$, we can conclude that $R(R(k))=k$
Please suggest how to proceed. Thank you.
polynomials algebraic-number-theory
asked Jul 30 at 6:25
Dan
1349
A few questions: 1. What is the $P(a_i)$ mentioned in your attempts? 2. Should there be some condition on n? Clearly, $ngeqslant 3$ for the question to be nontrivial but if $n$ is odd, $Q(k)=k$ and $R^2(k)=k$ would imply $R(k)=k$. (This would be a strictly stronger implication.)
– daruma
Jul 30 at 8:30
@daruma. Thank you, typo edited. There is no other condition on $n$.
– Dan
Jul 30 at 9:27
If $Q(k)=R^n(k)=k$ and $R^2(k)=k$, then we would get that $R(k)=k$ if $n$ is odd. Are you sure the question is correct?
– daruma
Jul 30 at 9:30
On a different point, why is $|P(a_i)-P(a_j)|=|a_i-a_j|$? In general, polynomials are not isometries of $mathbbR$
– daruma
Jul 30 at 9:32
@daruma. The question is correct. $mid R(a_i)-R(a_j)mid = mid a_i-a_j mid $ because $mid a_i+1-a_i mid$ is constant. Sorry, I don't understand "isometry".
– Dan
Jul 30 at 12:54
 |Â
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $R(x) = fracF(x)G(x)$ be a rational function with $F(x), G(x) in mathbbZ[x]$ and
$F(x), G(x)$ have no common root modulo $p$ for all primes $p$.
Consider the rational function $Q(x)=underbraceR(R(ldots(R(x))))_text$n$ times$ where $n in mathbbN$.
Prove that if there is an integer $k$ such that $Q(k)=k$, then $R(R(k))=k$.
My thought,
in case $G(x) = 1$, $R(x)$ will be polynomial.
Let $a_1$ be the value such that $Q(a_1)=a_1$ and $R(a_i)=a_i+1, forall i= 1, 2, ldots, n-1$.
We have $a_2-a_1 mid a_3-a_2 mid ldots mid a_1-a_n mid a_2-a_1$, so $mid a_i+1-a_i mid$ is constant.
Then $mid R(a_i)-R(a_j)mid = mid a_i-a_j mid, forall i, j$
Since $mid a_i+1-a_i mid = c, forall i$ which will be true when orbit = $1$ or $2$,
so $R(R(k))=k$ or $R(k)=k$, we can conclude that $R(R(k))=k$
Please suggest how to proceed. Thank you.
polynomials algebraic-number-theory
asked Jul 30 at 6:25
Dan
1349
Let $R(x) = fracF(x)G(x)$ be a rational function with $F(x), G(x) in mathbbZ[x]$ and
$F(x), G(x)$ have no common root modulo $p$ for all primes $p$.
Consider the rational function $Q(x)=underbraceR(R(ldots(R(x))))_text$n$ times$ where $n in mathbbN$.
Prove that if there is an integer $k$ such that $Q(k)=k$, then $R(R(k))=k$.
My thought,
in case $G(x) = 1$, $R(x)$ will be polynomial.
Let $a_1$ be the value such that $Q(a_1)=a_1$ and $R(a_i)=a_i+1, forall i= 1, 2, ldots, n-1$.
We have $a_2-a_1 mid a_3-a_2 mid ldots mid a_1-a_n mid a_2-a_1$, so $mid a_i+1-a_i mid$ is constant.
Then $mid R(a_i)-R(a_j)mid = mid a_i-a_j mid, forall i, j$
Since $mid a_i+1-a_i mid = c, forall i$ which will be true when orbit = $1$ or $2$,
so $R(R(k))=k$ or $R(k)=k$, we can conclude that $R(R(k))=k$
Please suggest how to proceed. Thank you.
polynomials algebraic-number-theory
asked Jul 30 at 6:25
Dan
1349
edited Jul 30 at 9:25
asked Jul 30 at 6:25
Dan
1349
asked Jul 30 at 6:25
Dan
1349
asked Jul 30 at 6:25
Dan
1349
1349
A few questions: 1. What is the $P(a_i)$ mentioned in your attempts? 2. Should there be some condition on n? Clearly, $ngeqslant 3$ for the question to be nontrivial but if $n$ is odd, $Q(k)=k$ and $R^2(k)=k$ would imply $R(k)=k$. (This would be a strictly stronger implication.)
– daruma
Jul 30 at 8:30
@daruma. Thank you, typo edited. There is no other condition on $n$.
– Dan
Jul 30 at 9:27
If $Q(k)=R^n(k)=k$ and $R^2(k)=k$, then we would get that $R(k)=k$ if $n$ is odd. Are you sure the question is correct?
– daruma
Jul 30 at 9:30
On a different point, why is $|P(a_i)-P(a_j)|=|a_i-a_j|$? In general, polynomials are not isometries of $mathbbR$
– daruma
Jul 30 at 9:32
@daruma. The question is correct. $mid R(a_i)-R(a_j)mid = mid a_i-a_j mid $ because $mid a_i+1-a_i mid$ is constant. Sorry, I don't understand "isometry".
– Dan
Jul 30 at 12:54
 |Â
show 2 more comments
A few questions: 1. What is the $P(a_i)$ mentioned in your attempts? 2. Should there be some condition on n? Clearly, $ngeqslant 3$ for the question to be nontrivial but if $n$ is odd, $Q(k)=k$ and $R^2(k)=k$ would imply $R(k)=k$. (This would be a strictly stronger implication.)
– daruma
Jul 30 at 8:30
@daruma. Thank you, typo edited. There is no other condition on $n$.
– Dan
Jul 30 at 9:27
If $Q(k)=R^n(k)=k$ and $R^2(k)=k$, then we would get that $R(k)=k$ if $n$ is odd. Are you sure the question is correct?
– daruma
Jul 30 at 9:30
On a different point, why is $|P(a_i)-P(a_j)|=|a_i-a_j|$? In general, polynomials are not isometries of $mathbbR$
– daruma
Jul 30 at 9:32
@daruma. The question is correct. $mid R(a_i)-R(a_j)mid = mid a_i-a_j mid $ because $mid a_i+1-a_i mid$ is constant. Sorry, I don't understand "isometry".
– Dan
Jul 30 at 12:54
A few questions: 1. What is the $P(a_i)$ mentioned in your attempts? 2. Should there be some condition on n? Clearly, $ngeqslant 3$ for the question to be nontrivial but if $n$ is odd, $Q(k)=k$ and $R^2(k)=k$ would imply $R(k)=k$. (This would be a strictly stronger implication.)
– daruma
Jul 30 at 8:30
A few questions: 1. What is the $P(a_i)$ mentioned in your attempts? 2. Should there be some condition on n? Clearly, $ngeqslant 3$ for the question to be nontrivial but if $n$ is odd, $Q(k)=k$ and $R^2(k)=k$ would imply $R(k)=k$. (This would be a strictly stronger implication.)
– daruma
Jul 30 at 8:30
@daruma. Thank you, typo edited. There is no other condition on $n$.
– Dan
Jul 30 at 9:27
@daruma. Thank you, typo edited. There is no other condition on $n$.
– Dan
Jul 30 at 9:27
If $Q(k)=R^n(k)=k$ and $R^2(k)=k$, then we would get that $R(k)=k$ if $n$ is odd. Are you sure the question is correct?
– daruma
Jul 30 at 9:30
If $Q(k)=R^n(k)=k$ and $R^2(k)=k$, then we would get that $R(k)=k$ if $n$ is odd. Are you sure the question is correct?
– daruma
Jul 30 at 9:30
On a different point, why is $|P(a_i)-P(a_j)|=|a_i-a_j|$? In general, polynomials are not isometries of $mathbbR$
– daruma
Jul 30 at 9:32
On a different point, why is $|P(a_i)-P(a_j)|=|a_i-a_j|$? In general, polynomials are not isometries of $mathbbR$
– daruma
Jul 30 at 9:32
@daruma. The question is correct. $mid R(a_i)-R(a_j)mid = mid a_i-a_j mid $ because $mid a_i+1-a_i mid$ is constant. Sorry, I don't understand "isometry".
– Dan
Jul 30 at 12:54
@daruma. The question is correct. $mid R(a_i)-R(a_j)mid = mid a_i-a_j mid $ because $mid a_i+1-a_i mid$ is constant. Sorry, I don't understand "isometry".
– Dan
Jul 30 at 12:54
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Unless there is something I don't understand, the assertion is false. An easy example is $F(x)=1$ and $G(x)=1-x$. Then $R(0)=1$, $R(1)=infty$ and $R(infty)=0$. So for $n=3$, we have $Q(0)=R^3(0)=0$, but $R^2(0)ne 0$.
answered Jul 30 at 16:40
xarles
83558
4
And of course you could have avoided the embarrassment of an infinity by starting with $k=2$ instead. The period of the transformation is three, so the counterexample kills the conjecture.
– Lubin
Jul 30 at 17:44
@Lubin, Thank you all. I got it now.
– Dan
Jul 31 at 14:11
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Unless there is something I don't understand, the assertion is false. An easy example is $F(x)=1$ and $G(x)=1-x$. Then $R(0)=1$, $R(1)=infty$ and $R(infty)=0$. So for $n=3$, we have $Q(0)=R^3(0)=0$, but $R^2(0)ne 0$.
answered Jul 30 at 16:40
xarles
83558
4
And of course you could have avoided the embarrassment of an infinity by starting with $k=2$ instead. The period of the transformation is three, so the counterexample kills the conjecture.
– Lubin
Jul 30 at 17:44
@Lubin, Thank you all. I got it now.
– Dan
Jul 31 at 14:11
add a comment |Â
up vote
4
down vote
accepted
Unless there is something I don't understand, the assertion is false. An easy example is $F(x)=1$ and $G(x)=1-x$. Then $R(0)=1$, $R(1)=infty$ and $R(infty)=0$. So for $n=3$, we have $Q(0)=R^3(0)=0$, but $R^2(0)ne 0$.
answered Jul 30 at 16:40
xarles
83558
4
And of course you could have avoided the embarrassment of an infinity by starting with $k=2$ instead. The period of the transformation is three, so the counterexample kills the conjecture.
– Lubin
Jul 30 at 17:44
@Lubin, Thank you all. I got it now.
– Dan
Jul 31 at 14:11
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Unless there is something I don't understand, the assertion is false. An easy example is $F(x)=1$ and $G(x)=1-x$. Then $R(0)=1$, $R(1)=infty$ and $R(infty)=0$. So for $n=3$, we have $Q(0)=R^3(0)=0$, but $R^2(0)ne 0$.
answered Jul 30 at 16:40
xarles
83558
Unless there is something I don't understand, the assertion is false. An easy example is $F(x)=1$ and $G(x)=1-x$. Then $R(0)=1$, $R(1)=infty$ and $R(infty)=0$. So for $n=3$, we have $Q(0)=R^3(0)=0$, but $R^2(0)ne 0$.
answered Jul 30 at 16:40
xarles
83558
answered Jul 30 at 16:40
xarles
83558
answered Jul 30 at 16:40
xarles
83558
answered Jul 30 at 16:40
xarles
83558
83558
4
And of course you could have avoided the embarrassment of an infinity by starting with $k=2$ instead. The period of the transformation is three, so the counterexample kills the conjecture.
– Lubin
Jul 30 at 17:44
@Lubin, Thank you all. I got it now.
– Dan
Jul 31 at 14:11
add a comment |Â
4
And of course you could have avoided the embarrassment of an infinity by starting with $k=2$ instead. The period of the transformation is three, so the counterexample kills the conjecture.
– Lubin
Jul 30 at 17:44
@Lubin, Thank you all. I got it now.
– Dan
Jul 31 at 14:11
4
4
And of course you could have avoided the embarrassment of an infinity by starting with $k=2$ instead. The period of the transformation is three, so the counterexample kills the conjecture.
– Lubin
Jul 30 at 17:44
And of course you could have avoided the embarrassment of an infinity by starting with $k=2$ instead. The period of the transformation is three, so the counterexample kills the conjecture.
– Lubin
Jul 30 at 17:44
@Lubin, Thank you all. I got it now.
– Dan
Jul 31 at 14:11
@Lubin, Thank you all. I got it now.
– Dan
Jul 31 at 14:11
add a comment |Â
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A few questions: 1. What is the $P(a_i)$ mentioned in your attempts? 2. Should there be some condition on n? Clearly, $ngeqslant 3$ for the question to be nontrivial but if $n$ is odd, $Q(k)=k$ and $R^2(k)=k$ would imply $R(k)=k$. (This would be a strictly stronger implication.)
– daruma
Jul 30 at 8:30
@daruma. Thank you, typo edited. There is no other condition on $n$.
– Dan
Jul 30 at 9:27
If $Q(k)=R^n(k)=k$ and $R^2(k)=k$, then we would get that $R(k)=k$ if $n$ is odd. Are you sure the question is correct?
– daruma
Jul 30 at 9:30
On a different point, why is $|P(a_i)-P(a_j)|=|a_i-a_j|$? In general, polynomials are not isometries of $mathbbR$
– daruma
Jul 30 at 9:32
@daruma. The question is correct. $mid R(a_i)-R(a_j)mid = mid a_i-a_j mid $ because $mid a_i+1-a_i mid$ is constant. Sorry, I don't understand "isometry".
– Dan
Jul 30 at 12:54