Is $(sum_{k=0}^n sin(k!))_{n in mathbb{N}} $ bounded?
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It is a relatively easy exercice to show that $(sum_{k=0}^n sin(k))_{n in mathbb{N}} $ is bounded by considering complex exponentials $e^{ik}$ (that form a geometric sum) and then taking the imaginary part. But what happens when we replace $k$ by $k!$ ?
I think that it is bounded by analogy. The terms somehow compensate one another... But I was unsucessfull at proving it.
One of my attempts : I tried to prove that $n!$ is an equidistributed sequence mod $2 pi$ (I'm not sure that's true), but even with that I struggle to conclude.
real-analysis sequences-and-series
asked Jan 9 at 20:24
C. LauverjatC. Lauverjat
262
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add a comment |
$begingroup$
It is a relatively easy exercice to show that $(sum_{k=0}^n sin(k))_{n in mathbb{N}} $ is bounded by considering complex exponentials $e^{ik}$ (that form a geometric sum) and then taking the imaginary part. But what happens when we replace $k$ by $k!$ ?
I think that it is bounded by analogy. The terms somehow compensate one another... But I was unsucessfull at proving it.
One of my attempts : I tried to prove that $n!$ is an equidistributed sequence mod $2 pi$ (I'm not sure that's true), but even with that I struggle to conclude.
real-analysis sequences-and-series
asked Jan 9 at 20:24
C. LauverjatC. Lauverjat
262
$endgroup$
$begingroup$
Well, $sum_{k=0}^nsin(n!pi e)$ is bounded, due to the Taylor expansion of $e^x$, but I doubt that'll help you here.
$endgroup$
– Simply Beautiful Art
Jan 9 at 20:45
4
$begingroup$
It should be noted that, for instance, the sequence $sum_{k=0}^nsin(k^2)$ are not bounded, even though squares are equidistributed modulo $2pi$. Your problem is probably even harder than that, since I doubt we know much about distribution of $k!$ mod $2pi$.
$endgroup$
– Wojowu
Jan 9 at 20:46
2
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Reference to my previous comment
$endgroup$
– Wojowu
Jan 9 at 20:47
2
$begingroup$
@Wojowu Why "even though" they are equidistributed? If $X_n$ were i.i.d. random variables uniformly distributed on $S^1$, I would expect $sum_{k=0}^n sin (X_k)$ to be almost surely unbounded since the standard deviation would be proportional to $sqrt{n+1}$.
$endgroup$
– Daniel Schepler
Jan 9 at 20:54
1
$begingroup$
@DanielSchepler "even though" because this has been OP's attempt (and that example implies this approach can't work).
$endgroup$
– Wojowu
Jan 9 at 20:56
add a comment |
$begingroup$
It is a relatively easy exercice to show that $(sum_{k=0}^n sin(k))_{n in mathbb{N}} $ is bounded by considering complex exponentials $e^{ik}$ (that form a geometric sum) and then taking the imaginary part. But what happens when we replace $k$ by $k!$ ?
I think that it is bounded by analogy. The terms somehow compensate one another... But I was unsucessfull at proving it.
One of my attempts : I tried to prove that $n!$ is an equidistributed sequence mod $2 pi$ (I'm not sure that's true), but even with that I struggle to conclude.
real-analysis sequences-and-series
asked Jan 9 at 20:24
C. LauverjatC. Lauverjat
262
$endgroup$
It is a relatively easy exercice to show that $(sum_{k=0}^n sin(k))_{n in mathbb{N}} $ is bounded by considering complex exponentials $e^{ik}$ (that form a geometric sum) and then taking the imaginary part. But what happens when we replace $k$ by $k!$ ?
I think that it is bounded by analogy. The terms somehow compensate one another... But I was unsucessfull at proving it.
One of my attempts : I tried to prove that $n!$ is an equidistributed sequence mod $2 pi$ (I'm not sure that's true), but even with that I struggle to conclude.
real-analysis sequences-and-series
real-analysis sequences-and-series
asked Jan 9 at 20:24
C. LauverjatC. Lauverjat
262
asked Jan 9 at 20:24
C. LauverjatC. Lauverjat
262
edited Jan 9 at 20:36
C. Lauverjat
asked Jan 9 at 20:24
C. LauverjatC. Lauverjat
262
asked Jan 9 at 20:24
C. LauverjatC. Lauverjat
262
asked Jan 9 at 20:24
C. LauverjatC. Lauverjat
262
262
$begingroup$
Well, $sum_{k=0}^nsin(n!pi e)$ is bounded, due to the Taylor expansion of $e^x$, but I doubt that'll help you here.
$endgroup$
– Simply Beautiful Art
Jan 9 at 20:45
4
$begingroup$
It should be noted that, for instance, the sequence $sum_{k=0}^nsin(k^2)$ are not bounded, even though squares are equidistributed modulo $2pi$. Your problem is probably even harder than that, since I doubt we know much about distribution of $k!$ mod $2pi$.
$endgroup$
– Wojowu
Jan 9 at 20:46
2
$begingroup$
Reference to my previous comment
$endgroup$
– Wojowu
Jan 9 at 20:47
2
$begingroup$
@Wojowu Why "even though" they are equidistributed? If $X_n$ were i.i.d. random variables uniformly distributed on $S^1$, I would expect $sum_{k=0}^n sin (X_k)$ to be almost surely unbounded since the standard deviation would be proportional to $sqrt{n+1}$.
$endgroup$
– Daniel Schepler
Jan 9 at 20:54
1
$begingroup$
@DanielSchepler "even though" because this has been OP's attempt (and that example implies this approach can't work).
$endgroup$
– Wojowu
Jan 9 at 20:56
add a comment |
$begingroup$
Well, $sum_{k=0}^nsin(n!pi e)$ is bounded, due to the Taylor expansion of $e^x$, but I doubt that'll help you here.
$endgroup$
– Simply Beautiful Art
Jan 9 at 20:45
4
$begingroup$
It should be noted that, for instance, the sequence $sum_{k=0}^nsin(k^2)$ are not bounded, even though squares are equidistributed modulo $2pi$. Your problem is probably even harder than that, since I doubt we know much about distribution of $k!$ mod $2pi$.
$endgroup$
– Wojowu
Jan 9 at 20:46
2
$begingroup$
Reference to my previous comment
$endgroup$
– Wojowu
Jan 9 at 20:47
2
$begingroup$
@Wojowu Why "even though" they are equidistributed? If $X_n$ were i.i.d. random variables uniformly distributed on $S^1$, I would expect $sum_{k=0}^n sin (X_k)$ to be almost surely unbounded since the standard deviation would be proportional to $sqrt{n+1}$.
$endgroup$
– Daniel Schepler
Jan 9 at 20:54
1
$begingroup$
@DanielSchepler "even though" because this has been OP's attempt (and that example implies this approach can't work).
$endgroup$
– Wojowu
Jan 9 at 20:56
$begingroup$
Well, $sum_{k=0}^nsin(n!pi e)$ is bounded, due to the Taylor expansion of $e^x$, but I doubt that'll help you here.
$endgroup$
– Simply Beautiful Art
Jan 9 at 20:45
$begingroup$
Well, $sum_{k=0}^nsin(n!pi e)$ is bounded, due to the Taylor expansion of $e^x$, but I doubt that'll help you here.
$endgroup$
– Simply Beautiful Art
Jan 9 at 20:45
4
4
$begingroup$
It should be noted that, for instance, the sequence $sum_{k=0}^nsin(k^2)$ are not bounded, even though squares are equidistributed modulo $2pi$. Your problem is probably even harder than that, since I doubt we know much about distribution of $k!$ mod $2pi$.
$endgroup$
– Wojowu
Jan 9 at 20:46
$begingroup$
It should be noted that, for instance, the sequence $sum_{k=0}^nsin(k^2)$ are not bounded, even though squares are equidistributed modulo $2pi$. Your problem is probably even harder than that, since I doubt we know much about distribution of $k!$ mod $2pi$.
$endgroup$
– Wojowu
Jan 9 at 20:46
2
2
$begingroup$
Reference to my previous comment
$endgroup$
– Wojowu
Jan 9 at 20:47
$begingroup$
Reference to my previous comment
$endgroup$
– Wojowu
Jan 9 at 20:47
2
2
$begingroup$
@Wojowu Why "even though" they are equidistributed? If $X_n$ were i.i.d. random variables uniformly distributed on $S^1$, I would expect $sum_{k=0}^n sin (X_k)$ to be almost surely unbounded since the standard deviation would be proportional to $sqrt{n+1}$.
$endgroup$
– Daniel Schepler
Jan 9 at 20:54
$begingroup$
@Wojowu Why "even though" they are equidistributed? If $X_n$ were i.i.d. random variables uniformly distributed on $S^1$, I would expect $sum_{k=0}^n sin (X_k)$ to be almost surely unbounded since the standard deviation would be proportional to $sqrt{n+1}$.
$endgroup$
– Daniel Schepler
Jan 9 at 20:54
1
1
$begingroup$
@DanielSchepler "even though" because this has been OP's attempt (and that example implies this approach can't work).
$endgroup$
– Wojowu
Jan 9 at 20:56
$begingroup$
@DanielSchepler "even though" because this has been OP's attempt (and that example implies this approach can't work).
$endgroup$
– Wojowu
Jan 9 at 20:56
add a comment |
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$begingroup$
Well, $sum_{k=0}^nsin(n!pi e)$ is bounded, due to the Taylor expansion of $e^x$, but I doubt that'll help you here.
$endgroup$
– Simply Beautiful Art
Jan 9 at 20:45
4
$begingroup$
It should be noted that, for instance, the sequence $sum_{k=0}^nsin(k^2)$ are not bounded, even though squares are equidistributed modulo $2pi$. Your problem is probably even harder than that, since I doubt we know much about distribution of $k!$ mod $2pi$.
$endgroup$
– Wojowu
Jan 9 at 20:46
2
$begingroup$
Reference to my previous comment
$endgroup$
– Wojowu
Jan 9 at 20:47
2
$begingroup$
@Wojowu Why "even though" they are equidistributed? If $X_n$ were i.i.d. random variables uniformly distributed on $S^1$, I would expect $sum_{k=0}^n sin (X_k)$ to be almost surely unbounded since the standard deviation would be proportional to $sqrt{n+1}$.
$endgroup$
– Daniel Schepler
Jan 9 at 20:54
1
$begingroup$
@DanielSchepler "even though" because this has been OP's attempt (and that example implies this approach can't work).
$endgroup$
– Wojowu
Jan 9 at 20:56