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Is it true that $forall epsilon>0, exists text infinitely many n in mathbbN, s.t. |sin(n) - 1| < epsilon$?
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I'm wondering if the following statement is true or not.
For me, it's quite 'intuitively' true, but I don't have any idea how to prove.
Statement:
$forall epsilon>0, exists text infinitely many n in mathbbN, s.t. |sin(n) - 1| < epsilon$
Is there anyone to help me out?
real-analysis trigonometry epsilon-delta
edited Jul 28 at 17:11
rtybase
8,86721433
asked Jul 21 at 3:43
moreblue
1738
add a comment |Â
up vote
3
down vote
favorite
I'm wondering if the following statement is true or not.
For me, it's quite 'intuitively' true, but I don't have any idea how to prove.
Statement:
$forall epsilon>0, exists text infinitely many n in mathbbN, s.t. |sin(n) - 1| < epsilon$
Is there anyone to help me out?
real-analysis trigonometry epsilon-delta
edited Jul 28 at 17:11
rtybase
8,86721433
asked Jul 21 at 3:43
moreblue
1738
Here's a question whose answer proves your statement true (and also proves a lot more): math.stackexchange.com/questions/1084945/…
– David K
Jul 21 at 23:28
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm wondering if the following statement is true or not.
For me, it's quite 'intuitively' true, but I don't have any idea how to prove.
Statement:
$forall epsilon>0, exists text infinitely many n in mathbbN, s.t. |sin(n) - 1| < epsilon$
Is there anyone to help me out?
real-analysis trigonometry epsilon-delta
edited Jul 28 at 17:11
rtybase
8,86721433
asked Jul 21 at 3:43
moreblue
1738
I'm wondering if the following statement is true or not.
For me, it's quite 'intuitively' true, but I don't have any idea how to prove.
Statement:
$forall epsilon>0, exists text infinitely many n in mathbbN, s.t. |sin(n) - 1| < epsilon$
Is there anyone to help me out?
real-analysis trigonometry epsilon-delta
edited Jul 28 at 17:11
rtybase
8,86721433
asked Jul 21 at 3:43
moreblue
1738
edited Jul 28 at 17:11
rtybase
8,86721433
edited Jul 28 at 17:11
rtybase
8,86721433
edited Jul 28 at 17:11
rtybase
8,86721433
8,86721433
asked Jul 21 at 3:43
moreblue
1738
asked Jul 21 at 3:43
moreblue
1738
asked Jul 21 at 3:43
moreblue
1738
1738
Here's a question whose answer proves your statement true (and also proves a lot more): math.stackexchange.com/questions/1084945/…
– David K
Jul 21 at 23:28
add a comment |Â
Here's a question whose answer proves your statement true (and also proves a lot more): math.stackexchange.com/questions/1084945/…
– David K
Jul 21 at 23:28
Here's a question whose answer proves your statement true (and also proves a lot more): math.stackexchange.com/questions/1084945/…
– David K
Jul 21 at 23:28
Here's a question whose answer proves your statement true (and also proves a lot more): math.stackexchange.com/questions/1084945/…
– David K
Jul 21 at 23:28
add a comment |Â
2 Answers
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up vote
2
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It follows from tree facts ...
Kronecker's approximation theorem (KAT) states that if $alpha$ is irrational then $leftnalpha+k mid n,k inmathbbZright$ is dense in $mathbbR$. It is not difficult to show that in fact
Proposition 1. $leftnalpha+k mid colorredninmathbbN,k inmathbbZright$ is dense in $mathbbR$.
I.e. from KAT, $forall x in mathbbR$ and $forall varepsilon>0, exists n_1,k_1 inmathbbZ: left|n_1alpha+k_1-xright|<fracvarepsilon2$. A consequence of Dirichlet's approximation theorem states that there will be infinitely many integers $q,p$ (in fact $qinmathbbN$) such that $|qalpha-p|<frac1q<fracvarepsilon2$, with large enough $q$. In fact, we can choose $q$ large enough to also have $q+n_1>0$. Then
$$left|colorred(q+n_1)alpha+colorred(k_1-p)-xright|=left|(qalpha -p)+(n_1alpha+k_1-x)right|leq \
left|qalpha -pright|+left|n_1alpha+k_1-xright|<fracvarepsilon2+fracvarepsilon2=varepsilon$$
and $n=q+n_1>0$, $k=k_1-p$. As a result
$$forall x in mathbbR text and forall varepsilon>0, exists n inmathbbN,k inmathbbZ: left|nalpha+k-xright|<varepsilon$$
and $leftnalpha+k mid ninmathbbN,k inmathbbZright$ is dense in $mathbbR$.
Because $frac12pi$ is irrational, from Proposition 1
Proposition 2. $leftfracn2pi+k mid n in mathbbN, k inmathbbZright$ is dense in $mathbbR$.
Because $f(x)=sin2pi x, f:mathbbRrightarrow[-1,1]$ is continuous, then
Proposition 3. $f(M)$ is dense in $[-1,1]$ for any set $M$ dense in $mathbbR$.
Altogether, $sinleft(2pileft(fracn2pi+kright)right)=sinn$, thus
$leftsinn mid n inmathbbNright$ is dense in $[-1,1]$.
As a result, there will be infinitely many $ninmathbbN$ such that $sinn$ is very close to $1$.
answered Jul 21 at 22:47
rtybase
8,86721433
add a comment |Â
up vote
1
down vote
It is true. For $n$ to satisfy this it needs to be close to an odd integer times $frac pi 2$. The equidistribution theorem promises you that the multiples of $frac pi 2$ are distributed in the unit interval, so some of them will be close to an integer. There are some loose ends to clean up-how close does $n$ have to be and why can't all the multiples of $frac pi 2$ that are close to integers be even multiples?
answered Jul 21 at 3:53
Ross Millikan
276k21186352
Those loose ends would be very useful to sort out this problem math.stackexchange.com/questions/2820234/… ... specifically, best rational approximations of $fracpi2$ with odd denominators, which seems to be an open problem yet.
– rtybase
Jul 21 at 14:43
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
It follows from tree facts ...
Kronecker's approximation theorem (KAT) states that if $alpha$ is irrational then $leftnalpha+k mid n,k inmathbbZright$ is dense in $mathbbR$. It is not difficult to show that in fact
Proposition 1. $leftnalpha+k mid colorredninmathbbN,k inmathbbZright$ is dense in $mathbbR$.
I.e. from KAT, $forall x in mathbbR$ and $forall varepsilon>0, exists n_1,k_1 inmathbbZ: left|n_1alpha+k_1-xright|<fracvarepsilon2$. A consequence of Dirichlet's approximation theorem states that there will be infinitely many integers $q,p$ (in fact $qinmathbbN$) such that $|qalpha-p|<frac1q<fracvarepsilon2$, with large enough $q$. In fact, we can choose $q$ large enough to also have $q+n_1>0$. Then
$$left|colorred(q+n_1)alpha+colorred(k_1-p)-xright|=left|(qalpha -p)+(n_1alpha+k_1-x)right|leq \
left|qalpha -pright|+left|n_1alpha+k_1-xright|<fracvarepsilon2+fracvarepsilon2=varepsilon$$
and $n=q+n_1>0$, $k=k_1-p$. As a result
$$forall x in mathbbR text and forall varepsilon>0, exists n inmathbbN,k inmathbbZ: left|nalpha+k-xright|<varepsilon$$
and $leftnalpha+k mid ninmathbbN,k inmathbbZright$ is dense in $mathbbR$.
Because $frac12pi$ is irrational, from Proposition 1
Proposition 2. $leftfracn2pi+k mid n in mathbbN, k inmathbbZright$ is dense in $mathbbR$.
Because $f(x)=sin2pi x, f:mathbbRrightarrow[-1,1]$ is continuous, then
Proposition 3. $f(M)$ is dense in $[-1,1]$ for any set $M$ dense in $mathbbR$.
Altogether, $sinleft(2pileft(fracn2pi+kright)right)=sinn$, thus
$leftsinn mid n inmathbbNright$ is dense in $[-1,1]$.
As a result, there will be infinitely many $ninmathbbN$ such that $sinn$ is very close to $1$.
answered Jul 21 at 22:47
rtybase
8,86721433
add a comment |Â
up vote
2
down vote
It follows from tree facts ...
Kronecker's approximation theorem (KAT) states that if $alpha$ is irrational then $leftnalpha+k mid n,k inmathbbZright$ is dense in $mathbbR$. It is not difficult to show that in fact
Proposition 1. $leftnalpha+k mid colorredninmathbbN,k inmathbbZright$ is dense in $mathbbR$.
I.e. from KAT, $forall x in mathbbR$ and $forall varepsilon>0, exists n_1,k_1 inmathbbZ: left|n_1alpha+k_1-xright|<fracvarepsilon2$. A consequence of Dirichlet's approximation theorem states that there will be infinitely many integers $q,p$ (in fact $qinmathbbN$) such that $|qalpha-p|<frac1q<fracvarepsilon2$, with large enough $q$. In fact, we can choose $q$ large enough to also have $q+n_1>0$. Then
$$left|colorred(q+n_1)alpha+colorred(k_1-p)-xright|=left|(qalpha -p)+(n_1alpha+k_1-x)right|leq \
left|qalpha -pright|+left|n_1alpha+k_1-xright|<fracvarepsilon2+fracvarepsilon2=varepsilon$$
and $n=q+n_1>0$, $k=k_1-p$. As a result
$$forall x in mathbbR text and forall varepsilon>0, exists n inmathbbN,k inmathbbZ: left|nalpha+k-xright|<varepsilon$$
and $leftnalpha+k mid ninmathbbN,k inmathbbZright$ is dense in $mathbbR$.
Because $frac12pi$ is irrational, from Proposition 1
Proposition 2. $leftfracn2pi+k mid n in mathbbN, k inmathbbZright$ is dense in $mathbbR$.
Because $f(x)=sin2pi x, f:mathbbRrightarrow[-1,1]$ is continuous, then
Proposition 3. $f(M)$ is dense in $[-1,1]$ for any set $M$ dense in $mathbbR$.
Altogether, $sinleft(2pileft(fracn2pi+kright)right)=sinn$, thus
$leftsinn mid n inmathbbNright$ is dense in $[-1,1]$.
As a result, there will be infinitely many $ninmathbbN$ such that $sinn$ is very close to $1$.
answered Jul 21 at 22:47
rtybase
8,86721433
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It follows from tree facts ...
Kronecker's approximation theorem (KAT) states that if $alpha$ is irrational then $leftnalpha+k mid n,k inmathbbZright$ is dense in $mathbbR$. It is not difficult to show that in fact
Proposition 1. $leftnalpha+k mid colorredninmathbbN,k inmathbbZright$ is dense in $mathbbR$.
I.e. from KAT, $forall x in mathbbR$ and $forall varepsilon>0, exists n_1,k_1 inmathbbZ: left|n_1alpha+k_1-xright|<fracvarepsilon2$. A consequence of Dirichlet's approximation theorem states that there will be infinitely many integers $q,p$ (in fact $qinmathbbN$) such that $|qalpha-p|<frac1q<fracvarepsilon2$, with large enough $q$. In fact, we can choose $q$ large enough to also have $q+n_1>0$. Then
$$left|colorred(q+n_1)alpha+colorred(k_1-p)-xright|=left|(qalpha -p)+(n_1alpha+k_1-x)right|leq \
left|qalpha -pright|+left|n_1alpha+k_1-xright|<fracvarepsilon2+fracvarepsilon2=varepsilon$$
and $n=q+n_1>0$, $k=k_1-p$. As a result
$$forall x in mathbbR text and forall varepsilon>0, exists n inmathbbN,k inmathbbZ: left|nalpha+k-xright|<varepsilon$$
and $leftnalpha+k mid ninmathbbN,k inmathbbZright$ is dense in $mathbbR$.
Because $frac12pi$ is irrational, from Proposition 1
Proposition 2. $leftfracn2pi+k mid n in mathbbN, k inmathbbZright$ is dense in $mathbbR$.
Because $f(x)=sin2pi x, f:mathbbRrightarrow[-1,1]$ is continuous, then
Proposition 3. $f(M)$ is dense in $[-1,1]$ for any set $M$ dense in $mathbbR$.
Altogether, $sinleft(2pileft(fracn2pi+kright)right)=sinn$, thus
$leftsinn mid n inmathbbNright$ is dense in $[-1,1]$.
As a result, there will be infinitely many $ninmathbbN$ such that $sinn$ is very close to $1$.
answered Jul 21 at 22:47
rtybase
8,86721433
It follows from tree facts ...
Kronecker's approximation theorem (KAT) states that if $alpha$ is irrational then $leftnalpha+k mid n,k inmathbbZright$ is dense in $mathbbR$. It is not difficult to show that in fact
Proposition 1. $leftnalpha+k mid colorredninmathbbN,k inmathbbZright$ is dense in $mathbbR$.
I.e. from KAT, $forall x in mathbbR$ and $forall varepsilon>0, exists n_1,k_1 inmathbbZ: left|n_1alpha+k_1-xright|<fracvarepsilon2$. A consequence of Dirichlet's approximation theorem states that there will be infinitely many integers $q,p$ (in fact $qinmathbbN$) such that $|qalpha-p|<frac1q<fracvarepsilon2$, with large enough $q$. In fact, we can choose $q$ large enough to also have $q+n_1>0$. Then
$$left|colorred(q+n_1)alpha+colorred(k_1-p)-xright|=left|(qalpha -p)+(n_1alpha+k_1-x)right|leq \
left|qalpha -pright|+left|n_1alpha+k_1-xright|<fracvarepsilon2+fracvarepsilon2=varepsilon$$
and $n=q+n_1>0$, $k=k_1-p$. As a result
$$forall x in mathbbR text and forall varepsilon>0, exists n inmathbbN,k inmathbbZ: left|nalpha+k-xright|<varepsilon$$
and $leftnalpha+k mid ninmathbbN,k inmathbbZright$ is dense in $mathbbR$.
Because $frac12pi$ is irrational, from Proposition 1
Proposition 2. $leftfracn2pi+k mid n in mathbbN, k inmathbbZright$ is dense in $mathbbR$.
Because $f(x)=sin2pi x, f:mathbbRrightarrow[-1,1]$ is continuous, then
Proposition 3. $f(M)$ is dense in $[-1,1]$ for any set $M$ dense in $mathbbR$.
Altogether, $sinleft(2pileft(fracn2pi+kright)right)=sinn$, thus
$leftsinn mid n inmathbbNright$ is dense in $[-1,1]$.
As a result, there will be infinitely many $ninmathbbN$ such that $sinn$ is very close to $1$.
answered Jul 21 at 22:47
rtybase
8,86721433
edited Jul 22 at 10:12
answered Jul 21 at 22:47
rtybase
8,86721433
answered Jul 21 at 22:47
rtybase
8,86721433
answered Jul 21 at 22:47
rtybase
8,86721433
8,86721433
add a comment |Â
add a comment |Â
up vote
1
down vote
It is true. For $n$ to satisfy this it needs to be close to an odd integer times $frac pi 2$. The equidistribution theorem promises you that the multiples of $frac pi 2$ are distributed in the unit interval, so some of them will be close to an integer. There are some loose ends to clean up-how close does $n$ have to be and why can't all the multiples of $frac pi 2$ that are close to integers be even multiples?
answered Jul 21 at 3:53
Ross Millikan
276k21186352
Those loose ends would be very useful to sort out this problem math.stackexchange.com/questions/2820234/… ... specifically, best rational approximations of $fracpi2$ with odd denominators, which seems to be an open problem yet.
– rtybase
Jul 21 at 14:43
add a comment |Â
up vote
1
down vote
It is true. For $n$ to satisfy this it needs to be close to an odd integer times $frac pi 2$. The equidistribution theorem promises you that the multiples of $frac pi 2$ are distributed in the unit interval, so some of them will be close to an integer. There are some loose ends to clean up-how close does $n$ have to be and why can't all the multiples of $frac pi 2$ that are close to integers be even multiples?
answered Jul 21 at 3:53
Ross Millikan
276k21186352
Those loose ends would be very useful to sort out this problem math.stackexchange.com/questions/2820234/… ... specifically, best rational approximations of $fracpi2$ with odd denominators, which seems to be an open problem yet.
– rtybase
Jul 21 at 14:43
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It is true. For $n$ to satisfy this it needs to be close to an odd integer times $frac pi 2$. The equidistribution theorem promises you that the multiples of $frac pi 2$ are distributed in the unit interval, so some of them will be close to an integer. There are some loose ends to clean up-how close does $n$ have to be and why can't all the multiples of $frac pi 2$ that are close to integers be even multiples?
answered Jul 21 at 3:53
Ross Millikan
276k21186352
It is true. For $n$ to satisfy this it needs to be close to an odd integer times $frac pi 2$. The equidistribution theorem promises you that the multiples of $frac pi 2$ are distributed in the unit interval, so some of them will be close to an integer. There are some loose ends to clean up-how close does $n$ have to be and why can't all the multiples of $frac pi 2$ that are close to integers be even multiples?
answered Jul 21 at 3:53
Ross Millikan
276k21186352
answered Jul 21 at 3:53
Ross Millikan
276k21186352
answered Jul 21 at 3:53
Ross Millikan
276k21186352
answered Jul 21 at 3:53
Ross Millikan
276k21186352
276k21186352
Those loose ends would be very useful to sort out this problem math.stackexchange.com/questions/2820234/… ... specifically, best rational approximations of $fracpi2$ with odd denominators, which seems to be an open problem yet.
– rtybase
Jul 21 at 14:43
add a comment |Â
Those loose ends would be very useful to sort out this problem math.stackexchange.com/questions/2820234/… ... specifically, best rational approximations of $fracpi2$ with odd denominators, which seems to be an open problem yet.
– rtybase
Jul 21 at 14:43
Those loose ends would be very useful to sort out this problem math.stackexchange.com/questions/2820234/… ... specifically, best rational approximations of $fracpi2$ with odd denominators, which seems to be an open problem yet.
– rtybase
Jul 21 at 14:43
Those loose ends would be very useful to sort out this problem math.stackexchange.com/questions/2820234/… ... specifically, best rational approximations of $fracpi2$ with odd denominators, which seems to be an open problem yet.
– rtybase
Jul 21 at 14:43
add a comment |Â
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Here's a question whose answer proves your statement true (and also proves a lot more): math.stackexchange.com/questions/1084945/…
– David K
Jul 21 at 23:28