Mixing
Any suggestions on how to compute $limsup |cos n|^n^2$?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
This problem has proven very difficult, does anyone have any suggestions on how to tackle it? Any little known theorems/identities that might help?
number-theory trigonometric-series diophantine-approximation
edited 2 days ago
Jyrki Lahtonen
104k12161355
asked 2 days ago
Sully Chen
866
add a comment |Â
up vote
3
down vote
favorite
This problem has proven very difficult, does anyone have any suggestions on how to tackle it? Any little known theorems/identities that might help?
number-theory trigonometric-series diophantine-approximation
edited 2 days ago
Jyrki Lahtonen
104k12161355
asked 2 days ago
Sully Chen
866
Would you like to share your thought on the problem? It is easy to show this limsup is non-zero.
– pisco
2 days ago
1
Based on the continued fraction approximations of $pi$ I'd expect it to be at least $exp(-pi^2/8)$.
– WimC
2 days ago
And the given problem is equivalent to finding the $limsup$ of the terms of the continued fraction of $pi$. This still is an open problem.
– Jack D'Aurizio♦
2 days ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This problem has proven very difficult, does anyone have any suggestions on how to tackle it? Any little known theorems/identities that might help?
number-theory trigonometric-series diophantine-approximation
edited 2 days ago
Jyrki Lahtonen
104k12161355
asked 2 days ago
Sully Chen
866
This problem has proven very difficult, does anyone have any suggestions on how to tackle it? Any little known theorems/identities that might help?
number-theory trigonometric-series diophantine-approximation
edited 2 days ago
Jyrki Lahtonen
104k12161355
asked 2 days ago
Sully Chen
866
edited 2 days ago
Jyrki Lahtonen
104k12161355
edited 2 days ago
Jyrki Lahtonen
104k12161355
edited 2 days ago
Jyrki Lahtonen
104k12161355
104k12161355
asked 2 days ago
Sully Chen
866
asked 2 days ago
Sully Chen
866
asked 2 days ago
Sully Chen
866
866
Would you like to share your thought on the problem? It is easy to show this limsup is non-zero.
– pisco
2 days ago
1
Based on the continued fraction approximations of $pi$ I'd expect it to be at least $exp(-pi^2/8)$.
– WimC
2 days ago
And the given problem is equivalent to finding the $limsup$ of the terms of the continued fraction of $pi$. This still is an open problem.
– Jack D'Aurizio♦
2 days ago
add a comment |Â
Would you like to share your thought on the problem? It is easy to show this limsup is non-zero.
– pisco
2 days ago
1
Based on the continued fraction approximations of $pi$ I'd expect it to be at least $exp(-pi^2/8)$.
– WimC
2 days ago
And the given problem is equivalent to finding the $limsup$ of the terms of the continued fraction of $pi$. This still is an open problem.
– Jack D'Aurizio♦
2 days ago
Would you like to share your thought on the problem? It is easy to show this limsup is non-zero.
– pisco
2 days ago
Would you like to share your thought on the problem? It is easy to show this limsup is non-zero.
– pisco
2 days ago
1
1
Based on the continued fraction approximations of $pi$ I'd expect it to be at least $exp(-pi^2/8)$.
– WimC
2 days ago
Based on the continued fraction approximations of $pi$ I'd expect it to be at least $exp(-pi^2/8)$.
– WimC
2 days ago
And the given problem is equivalent to finding the $limsup$ of the terms of the continued fraction of $pi$. This still is an open problem.
– Jack D'Aurizio♦
2 days ago
And the given problem is equivalent to finding the $limsup$ of the terms of the continued fraction of $pi$. This still is an open problem.
– Jack D'Aurizio♦
2 days ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Let us consider the continued fraction of $pi$:
$$ pi = [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14,ldots]$$
together with its convergents $fracp_1q_1=frac227,fracp_2q_2=frac333106,fracp_3q_3=frac355113$ etcetera.
The sequence $p_n_ngeq 1$ provides a set of natural numbers where the cosine function attains values estremely close to $pm 1$. By the properties of continued fractions we have $left|fracp_nq_n-piright|leq frac1q_n^2$, hence $left|p_n-pi q_nright|leq frac1q_napproxfracpip_n$ and by the Lipshitz-continuity of $sin$
$$left|cos(p_n)right|^p_n^2=left(1-sin^2 p_nright)^p_n^2/2 approx left(1-fracpi^2p_n^2right)^p_n^2/2to e^-pi^2/colorred2geq frac1140$$
so the wanted $limsup$ is at least as large as $frac1140$. On the other hand, if we assume that the terms of the continued fraction of $pi$ are unbounded, the constant $2$ above can be replaced by any larger constant and the wanted $limsup$ is just one. The pity is we do not actually know if the terms of the continued fraction of $pi$ are unbounded or not. For sure, if we consider $n=710$ the value of $left|cos nright|^n^2$ is very close to $1$, being equal to $approx 0.999084$. Additionally, since $pinotinmathbbQ(sqrt5)$, infinite terms of the continued fraction of $pi$ are $geq 2$, from which we have WimC's claim
$$ limsup_ninmathbbNleft|cos(n)right|^n^2geq e^-pi^2/8>frac2379.$$
Following the same lines we have
$$ limsup_ninmathbbNleft|cosfracpi neright|^n^2=1$$
since the terms of the continued fraction of $e$ are unbounded, due to
$$ e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,ldots].$$
answered 2 days ago
Jack D'Aurizio♦
268k30261623
At least terms $geq 2$ must occur infinitely often. That's where my denominator $8$ came from.
– WimC
2 days ago
@WimC: agreed (otherwise $pi$ would be algebraic with degree $2$ over $mathbbQ$, precisely of the form $a+bsqrt5$). If you like it, feel free to improve the answer.
– Jack D'Aurizio♦
2 days ago
Ah yes, I have proved the same result, but with a different method: by using the 2nd degree Taylor expansion of $cos$. Fascinating proof using with denominator 8 though!
– Sully Chen
2 days ago
@SullyChen: I would be curious to know how the Taylor series of $cos$ were used for describing the behaviour of $cos$ over the integers, but anyway I am pleased you found this useful.
– Jack D'Aurizio♦
2 days ago
1
Markov spectrum and transcendence of pi give us $left|fracp_nq_n-piright|leq frac13q_n^2$ (I don't have time to read the proof so I hope I didn't miss something) so we should get $e^-pi^2/18approx 0.5579$
– i9Fn
2 days ago
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Let us consider the continued fraction of $pi$:
$$ pi = [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14,ldots]$$
together with its convergents $fracp_1q_1=frac227,fracp_2q_2=frac333106,fracp_3q_3=frac355113$ etcetera.
The sequence $p_n_ngeq 1$ provides a set of natural numbers where the cosine function attains values estremely close to $pm 1$. By the properties of continued fractions we have $left|fracp_nq_n-piright|leq frac1q_n^2$, hence $left|p_n-pi q_nright|leq frac1q_napproxfracpip_n$ and by the Lipshitz-continuity of $sin$
$$left|cos(p_n)right|^p_n^2=left(1-sin^2 p_nright)^p_n^2/2 approx left(1-fracpi^2p_n^2right)^p_n^2/2to e^-pi^2/colorred2geq frac1140$$
so the wanted $limsup$ is at least as large as $frac1140$. On the other hand, if we assume that the terms of the continued fraction of $pi$ are unbounded, the constant $2$ above can be replaced by any larger constant and the wanted $limsup$ is just one. The pity is we do not actually know if the terms of the continued fraction of $pi$ are unbounded or not. For sure, if we consider $n=710$ the value of $left|cos nright|^n^2$ is very close to $1$, being equal to $approx 0.999084$. Additionally, since $pinotinmathbbQ(sqrt5)$, infinite terms of the continued fraction of $pi$ are $geq 2$, from which we have WimC's claim
$$ limsup_ninmathbbNleft|cos(n)right|^n^2geq e^-pi^2/8>frac2379.$$
Following the same lines we have
$$ limsup_ninmathbbNleft|cosfracpi neright|^n^2=1$$
since the terms of the continued fraction of $e$ are unbounded, due to
$$ e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,ldots].$$
answered 2 days ago
Jack D'Aurizio♦
268k30261623
At least terms $geq 2$ must occur infinitely often. That's where my denominator $8$ came from.
– WimC
2 days ago
@WimC: agreed (otherwise $pi$ would be algebraic with degree $2$ over $mathbbQ$, precisely of the form $a+bsqrt5$). If you like it, feel free to improve the answer.
– Jack D'Aurizio♦
2 days ago
Ah yes, I have proved the same result, but with a different method: by using the 2nd degree Taylor expansion of $cos$. Fascinating proof using with denominator 8 though!
– Sully Chen
2 days ago
@SullyChen: I would be curious to know how the Taylor series of $cos$ were used for describing the behaviour of $cos$ over the integers, but anyway I am pleased you found this useful.
– Jack D'Aurizio♦
2 days ago
1
Markov spectrum and transcendence of pi give us $left|fracp_nq_n-piright|leq frac13q_n^2$ (I don't have time to read the proof so I hope I didn't miss something) so we should get $e^-pi^2/18approx 0.5579$
– i9Fn
2 days ago
 |Â
show 5 more comments
up vote
4
down vote
accepted
Let us consider the continued fraction of $pi$:
$$ pi = [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14,ldots]$$
together with its convergents $fracp_1q_1=frac227,fracp_2q_2=frac333106,fracp_3q_3=frac355113$ etcetera.
The sequence $p_n_ngeq 1$ provides a set of natural numbers where the cosine function attains values estremely close to $pm 1$. By the properties of continued fractions we have $left|fracp_nq_n-piright|leq frac1q_n^2$, hence $left|p_n-pi q_nright|leq frac1q_napproxfracpip_n$ and by the Lipshitz-continuity of $sin$
$$left|cos(p_n)right|^p_n^2=left(1-sin^2 p_nright)^p_n^2/2 approx left(1-fracpi^2p_n^2right)^p_n^2/2to e^-pi^2/colorred2geq frac1140$$
so the wanted $limsup$ is at least as large as $frac1140$. On the other hand, if we assume that the terms of the continued fraction of $pi$ are unbounded, the constant $2$ above can be replaced by any larger constant and the wanted $limsup$ is just one. The pity is we do not actually know if the terms of the continued fraction of $pi$ are unbounded or not. For sure, if we consider $n=710$ the value of $left|cos nright|^n^2$ is very close to $1$, being equal to $approx 0.999084$. Additionally, since $pinotinmathbbQ(sqrt5)$, infinite terms of the continued fraction of $pi$ are $geq 2$, from which we have WimC's claim
$$ limsup_ninmathbbNleft|cos(n)right|^n^2geq e^-pi^2/8>frac2379.$$
Following the same lines we have
$$ limsup_ninmathbbNleft|cosfracpi neright|^n^2=1$$
since the terms of the continued fraction of $e$ are unbounded, due to
$$ e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,ldots].$$
answered 2 days ago
Jack D'Aurizio♦
268k30261623
At least terms $geq 2$ must occur infinitely often. That's where my denominator $8$ came from.
– WimC
2 days ago
@WimC: agreed (otherwise $pi$ would be algebraic with degree $2$ over $mathbbQ$, precisely of the form $a+bsqrt5$). If you like it, feel free to improve the answer.
– Jack D'Aurizio♦
2 days ago
Ah yes, I have proved the same result, but with a different method: by using the 2nd degree Taylor expansion of $cos$. Fascinating proof using with denominator 8 though!
– Sully Chen
2 days ago
@SullyChen: I would be curious to know how the Taylor series of $cos$ were used for describing the behaviour of $cos$ over the integers, but anyway I am pleased you found this useful.
– Jack D'Aurizio♦
2 days ago
1
Markov spectrum and transcendence of pi give us $left|fracp_nq_n-piright|leq frac13q_n^2$ (I don't have time to read the proof so I hope I didn't miss something) so we should get $e^-pi^2/18approx 0.5579$
– i9Fn
2 days ago
 |Â
show 5 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Let us consider the continued fraction of $pi$:
$$ pi = [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14,ldots]$$
together with its convergents $fracp_1q_1=frac227,fracp_2q_2=frac333106,fracp_3q_3=frac355113$ etcetera.
The sequence $p_n_ngeq 1$ provides a set of natural numbers where the cosine function attains values estremely close to $pm 1$. By the properties of continued fractions we have $left|fracp_nq_n-piright|leq frac1q_n^2$, hence $left|p_n-pi q_nright|leq frac1q_napproxfracpip_n$ and by the Lipshitz-continuity of $sin$
$$left|cos(p_n)right|^p_n^2=left(1-sin^2 p_nright)^p_n^2/2 approx left(1-fracpi^2p_n^2right)^p_n^2/2to e^-pi^2/colorred2geq frac1140$$
so the wanted $limsup$ is at least as large as $frac1140$. On the other hand, if we assume that the terms of the continued fraction of $pi$ are unbounded, the constant $2$ above can be replaced by any larger constant and the wanted $limsup$ is just one. The pity is we do not actually know if the terms of the continued fraction of $pi$ are unbounded or not. For sure, if we consider $n=710$ the value of $left|cos nright|^n^2$ is very close to $1$, being equal to $approx 0.999084$. Additionally, since $pinotinmathbbQ(sqrt5)$, infinite terms of the continued fraction of $pi$ are $geq 2$, from which we have WimC's claim
$$ limsup_ninmathbbNleft|cos(n)right|^n^2geq e^-pi^2/8>frac2379.$$
Following the same lines we have
$$ limsup_ninmathbbNleft|cosfracpi neright|^n^2=1$$
since the terms of the continued fraction of $e$ are unbounded, due to
$$ e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,ldots].$$
answered 2 days ago
Jack D'Aurizio♦
268k30261623
Let us consider the continued fraction of $pi$:
$$ pi = [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14,ldots]$$
together with its convergents $fracp_1q_1=frac227,fracp_2q_2=frac333106,fracp_3q_3=frac355113$ etcetera.
The sequence $p_n_ngeq 1$ provides a set of natural numbers where the cosine function attains values estremely close to $pm 1$. By the properties of continued fractions we have $left|fracp_nq_n-piright|leq frac1q_n^2$, hence $left|p_n-pi q_nright|leq frac1q_napproxfracpip_n$ and by the Lipshitz-continuity of $sin$
$$left|cos(p_n)right|^p_n^2=left(1-sin^2 p_nright)^p_n^2/2 approx left(1-fracpi^2p_n^2right)^p_n^2/2to e^-pi^2/colorred2geq frac1140$$
so the wanted $limsup$ is at least as large as $frac1140$. On the other hand, if we assume that the terms of the continued fraction of $pi$ are unbounded, the constant $2$ above can be replaced by any larger constant and the wanted $limsup$ is just one. The pity is we do not actually know if the terms of the continued fraction of $pi$ are unbounded or not. For sure, if we consider $n=710$ the value of $left|cos nright|^n^2$ is very close to $1$, being equal to $approx 0.999084$. Additionally, since $pinotinmathbbQ(sqrt5)$, infinite terms of the continued fraction of $pi$ are $geq 2$, from which we have WimC's claim
$$ limsup_ninmathbbNleft|cos(n)right|^n^2geq e^-pi^2/8>frac2379.$$
Following the same lines we have
$$ limsup_ninmathbbNleft|cosfracpi neright|^n^2=1$$
since the terms of the continued fraction of $e$ are unbounded, due to
$$ e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,ldots].$$
answered 2 days ago
Jack D'Aurizio♦
268k30261623
edited 2 days ago
answered 2 days ago
Jack D'Aurizio♦
268k30261623
answered 2 days ago
Jack D'Aurizio♦
268k30261623
answered 2 days ago
Jack D'Aurizio♦
268k30261623
268k30261623
At least terms $geq 2$ must occur infinitely often. That's where my denominator $8$ came from.
– WimC
2 days ago
@WimC: agreed (otherwise $pi$ would be algebraic with degree $2$ over $mathbbQ$, precisely of the form $a+bsqrt5$). If you like it, feel free to improve the answer.
– Jack D'Aurizio♦
2 days ago
Ah yes, I have proved the same result, but with a different method: by using the 2nd degree Taylor expansion of $cos$. Fascinating proof using with denominator 8 though!
– Sully Chen
2 days ago
@SullyChen: I would be curious to know how the Taylor series of $cos$ were used for describing the behaviour of $cos$ over the integers, but anyway I am pleased you found this useful.
– Jack D'Aurizio♦
2 days ago
1
Markov spectrum and transcendence of pi give us $left|fracp_nq_n-piright|leq frac13q_n^2$ (I don't have time to read the proof so I hope I didn't miss something) so we should get $e^-pi^2/18approx 0.5579$
– i9Fn
2 days ago
 |Â
show 5 more comments
At least terms $geq 2$ must occur infinitely often. That's where my denominator $8$ came from.
– WimC
2 days ago
@WimC: agreed (otherwise $pi$ would be algebraic with degree $2$ over $mathbbQ$, precisely of the form $a+bsqrt5$). If you like it, feel free to improve the answer.
– Jack D'Aurizio♦
2 days ago
Ah yes, I have proved the same result, but with a different method: by using the 2nd degree Taylor expansion of $cos$. Fascinating proof using with denominator 8 though!
– Sully Chen
2 days ago
@SullyChen: I would be curious to know how the Taylor series of $cos$ were used for describing the behaviour of $cos$ over the integers, but anyway I am pleased you found this useful.
– Jack D'Aurizio♦
2 days ago
1
Markov spectrum and transcendence of pi give us $left|fracp_nq_n-piright|leq frac13q_n^2$ (I don't have time to read the proof so I hope I didn't miss something) so we should get $e^-pi^2/18approx 0.5579$
– i9Fn
2 days ago
At least terms $geq 2$ must occur infinitely often. That's where my denominator $8$ came from.
– WimC
2 days ago
At least terms $geq 2$ must occur infinitely often. That's where my denominator $8$ came from.
– WimC
2 days ago
@WimC: agreed (otherwise $pi$ would be algebraic with degree $2$ over $mathbbQ$, precisely of the form $a+bsqrt5$). If you like it, feel free to improve the answer.
– Jack D'Aurizio♦
2 days ago
@WimC: agreed (otherwise $pi$ would be algebraic with degree $2$ over $mathbbQ$, precisely of the form $a+bsqrt5$). If you like it, feel free to improve the answer.
– Jack D'Aurizio♦
2 days ago
Ah yes, I have proved the same result, but with a different method: by using the 2nd degree Taylor expansion of $cos$. Fascinating proof using with denominator 8 though!
– Sully Chen
2 days ago
Ah yes, I have proved the same result, but with a different method: by using the 2nd degree Taylor expansion of $cos$. Fascinating proof using with denominator 8 though!
– Sully Chen
2 days ago
@SullyChen: I would be curious to know how the Taylor series of $cos$ were used for describing the behaviour of $cos$ over the integers, but anyway I am pleased you found this useful.
– Jack D'Aurizio♦
2 days ago
@SullyChen: I would be curious to know how the Taylor series of $cos$ were used for describing the behaviour of $cos$ over the integers, but anyway I am pleased you found this useful.
– Jack D'Aurizio♦
2 days ago
1
1
Markov spectrum and transcendence of pi give us $left|fracp_nq_n-piright|leq frac13q_n^2$ (I don't have time to read the proof so I hope I didn't miss something) so we should get $e^-pi^2/18approx 0.5579$
– i9Fn
2 days ago
Markov spectrum and transcendence of pi give us $left|fracp_nq_n-piright|leq frac13q_n^2$ (I don't have time to read the proof so I hope I didn't miss something) so we should get $e^-pi^2/18approx 0.5579$
– i9Fn
2 days ago
 |Â
show 5 more comments
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Would you like to share your thought on the problem? It is easy to show this limsup is non-zero.
– pisco
2 days ago
1
Based on the continued fraction approximations of $pi$ I'd expect it to be at least $exp(-pi^2/8)$.
– WimC
2 days ago
And the given problem is equivalent to finding the $limsup$ of the terms of the continued fraction of $pi$. This still is an open problem.
– Jack D'Aurizio♦
2 days ago