Convergence of a series involving cosine
3
$begingroup$
Let $x in (0, 2pi)$. Is the series $sum_{n=1}^{infty} frac{cos(n^2x)}{n}$ convergent? My guess is: YES and I would like to use Dirichlet test: however I have troubles proving that the partial sums $cos(x)+cos(4x)+...+cos(n^2x)$ are bounded due to the lack of the simple formula for this sum. Any ideas?
real-analysis sequences-and-series convergence
edited Dec 10 '18 at 0:28
Dando18
4,73241235
asked Dec 10 '18 at 0:12
truebarantruebaran
2,2352824
$endgroup$
add a comment |
3
$begingroup$
Let $x in (0, 2pi)$. Is the series $sum_{n=1}^{infty} frac{cos(n^2x)}{n}$ convergent? My guess is: YES and I would like to use Dirichlet test: however I have troubles proving that the partial sums $cos(x)+cos(4x)+...+cos(n^2x)$ are bounded due to the lack of the simple formula for this sum. Any ideas?
real-analysis sequences-and-series convergence
edited Dec 10 '18 at 0:28
Dando18
4,73241235
asked Dec 10 '18 at 0:12
truebarantruebaran
2,2352824
$endgroup$
$begingroup$
I'm not sure, but I don't think the partial sums are bounded. My guess is that there are some $x$ with annoying continued fraction expansion. However, I think the set of such $x$ is measure $0$.
$endgroup$
– mathworker21
Dec 10 '18 at 0:20
$begingroup$
You could look at the following question first: Is there an x such that $cos(n^2x)>epsilon$ for all n for some epsilon >0. If that is the case (which looks likely to me even though I can‘t imediatly see such an element) you can lower bound that sum by the harmonic sum which is divergent. Therefore for that x the sum would be divergent as well
$endgroup$
– Börge
Dec 10 '18 at 1:00
$begingroup$
At least sometimes it is convergent: take $x=pi$.
$endgroup$
– YiFan
Dec 10 '18 at 1:07
$begingroup$
And sometimes it is divergent: take $x=frac{pi}2$. For odd indeces $n$ the fraction vanishes, for even indeces $n$ it is $frac 1n$.
$endgroup$
– Pavel R.
Dec 10 '18 at 1:11
$begingroup$
It is the Fourier series of an $L^2$ function. Carleson's theorem implies that it converges a.e.
$endgroup$
– Julián Aguirre
Dec 10 '18 at 17:36
add a comment |
3
3
3
$begingroup$
Let $x in (0, 2pi)$. Is the series $sum_{n=1}^{infty} frac{cos(n^2x)}{n}$ convergent? My guess is: YES and I would like to use Dirichlet test: however I have troubles proving that the partial sums $cos(x)+cos(4x)+...+cos(n^2x)$ are bounded due to the lack of the simple formula for this sum. Any ideas?
real-analysis sequences-and-series convergence
edited Dec 10 '18 at 0:28
Dando18
4,73241235
asked Dec 10 '18 at 0:12
truebarantruebaran
2,2352824
$endgroup$
Let $x in (0, 2pi)$. Is the series $sum_{n=1}^{infty} frac{cos(n^2x)}{n}$ convergent? My guess is: YES and I would like to use Dirichlet test: however I have troubles proving that the partial sums $cos(x)+cos(4x)+...+cos(n^2x)$ are bounded due to the lack of the simple formula for this sum. Any ideas?
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
edited Dec 10 '18 at 0:28
Dando18
4,73241235
asked Dec 10 '18 at 0:12
truebarantruebaran
2,2352824
edited Dec 10 '18 at 0:28
Dando18
4,73241235
asked Dec 10 '18 at 0:12
truebarantruebaran
2,2352824
edited Dec 10 '18 at 0:28
Dando18
4,73241235
edited Dec 10 '18 at 0:28
Dando18
4,73241235
edited Dec 10 '18 at 0:28
Dando18
4,73241235
4,73241235
asked Dec 10 '18 at 0:12
truebarantruebaran
2,2352824
asked Dec 10 '18 at 0:12
truebarantruebaran
2,2352824
asked Dec 10 '18 at 0:12
truebarantruebaran
2,2352824
2,2352824
$begingroup$
I'm not sure, but I don't think the partial sums are bounded. My guess is that there are some $x$ with annoying continued fraction expansion. However, I think the set of such $x$ is measure $0$.
$endgroup$
– mathworker21
Dec 10 '18 at 0:20
$begingroup$
You could look at the following question first: Is there an x such that $cos(n^2x)>epsilon$ for all n for some epsilon >0. If that is the case (which looks likely to me even though I can‘t imediatly see such an element) you can lower bound that sum by the harmonic sum which is divergent. Therefore for that x the sum would be divergent as well
$endgroup$
– Börge
Dec 10 '18 at 1:00
$begingroup$
At least sometimes it is convergent: take $x=pi$.
$endgroup$
– YiFan
Dec 10 '18 at 1:07
$begingroup$
And sometimes it is divergent: take $x=frac{pi}2$. For odd indeces $n$ the fraction vanishes, for even indeces $n$ it is $frac 1n$.
$endgroup$
– Pavel R.
Dec 10 '18 at 1:11
$begingroup$
It is the Fourier series of an $L^2$ function. Carleson's theorem implies that it converges a.e.
$endgroup$
– Julián Aguirre
Dec 10 '18 at 17:36
add a comment |
$begingroup$
I'm not sure, but I don't think the partial sums are bounded. My guess is that there are some $x$ with annoying continued fraction expansion. However, I think the set of such $x$ is measure $0$.
$endgroup$
– mathworker21
Dec 10 '18 at 0:20
$begingroup$
You could look at the following question first: Is there an x such that $cos(n^2x)>epsilon$ for all n for some epsilon >0. If that is the case (which looks likely to me even though I can‘t imediatly see such an element) you can lower bound that sum by the harmonic sum which is divergent. Therefore for that x the sum would be divergent as well
$endgroup$
– Börge
Dec 10 '18 at 1:00
$begingroup$
At least sometimes it is convergent: take $x=pi$.
$endgroup$
– YiFan
Dec 10 '18 at 1:07
$begingroup$
And sometimes it is divergent: take $x=frac{pi}2$. For odd indeces $n$ the fraction vanishes, for even indeces $n$ it is $frac 1n$.
$endgroup$
– Pavel R.
Dec 10 '18 at 1:11
$begingroup$
It is the Fourier series of an $L^2$ function. Carleson's theorem implies that it converges a.e.
$endgroup$
– Julián Aguirre
Dec 10 '18 at 17:36
$begingroup$
I'm not sure, but I don't think the partial sums are bounded. My guess is that there are some $x$ with annoying continued fraction expansion. However, I think the set of such $x$ is measure $0$.
$endgroup$
– mathworker21
Dec 10 '18 at 0:20
$begingroup$
I'm not sure, but I don't think the partial sums are bounded. My guess is that there are some $x$ with annoying continued fraction expansion. However, I think the set of such $x$ is measure $0$.
$endgroup$
– mathworker21
Dec 10 '18 at 0:20
$begingroup$
You could look at the following question first: Is there an x such that $cos(n^2x)>epsilon$ for all n for some epsilon >0. If that is the case (which looks likely to me even though I can‘t imediatly see such an element) you can lower bound that sum by the harmonic sum which is divergent. Therefore for that x the sum would be divergent as well
$endgroup$
– Börge
Dec 10 '18 at 1:00
$begingroup$
You could look at the following question first: Is there an x such that $cos(n^2x)>epsilon$ for all n for some epsilon >0. If that is the case (which looks likely to me even though I can‘t imediatly see such an element) you can lower bound that sum by the harmonic sum which is divergent. Therefore for that x the sum would be divergent as well
$endgroup$
– Börge
Dec 10 '18 at 1:00
$begingroup$
At least sometimes it is convergent: take $x=pi$.
$endgroup$
– YiFan
Dec 10 '18 at 1:07
$begingroup$
At least sometimes it is convergent: take $x=pi$.
$endgroup$
– YiFan
Dec 10 '18 at 1:07
$begingroup$
And sometimes it is divergent: take $x=frac{pi}2$. For odd indeces $n$ the fraction vanishes, for even indeces $n$ it is $frac 1n$.
$endgroup$
– Pavel R.
Dec 10 '18 at 1:11
$begingroup$
And sometimes it is divergent: take $x=frac{pi}2$. For odd indeces $n$ the fraction vanishes, for even indeces $n$ it is $frac 1n$.
$endgroup$
– Pavel R.
Dec 10 '18 at 1:11
$begingroup$
It is the Fourier series of an $L^2$ function. Carleson's theorem implies that it converges a.e.
$endgroup$
– Julián Aguirre
Dec 10 '18 at 17:36
$begingroup$
It is the Fourier series of an $L^2$ function. Carleson's theorem implies that it converges a.e.
$endgroup$
– Julián Aguirre
Dec 10 '18 at 17:36
add a comment |
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$begingroup$
I'm not sure, but I don't think the partial sums are bounded. My guess is that there are some $x$ with annoying continued fraction expansion. However, I think the set of such $x$ is measure $0$.
$endgroup$
– mathworker21
Dec 10 '18 at 0:20
$begingroup$
You could look at the following question first: Is there an x such that $cos(n^2x)>epsilon$ for all n for some epsilon >0. If that is the case (which looks likely to me even though I can‘t imediatly see such an element) you can lower bound that sum by the harmonic sum which is divergent. Therefore for that x the sum would be divergent as well
$endgroup$
– Börge
Dec 10 '18 at 1:00
$begingroup$
At least sometimes it is convergent: take $x=pi$.
$endgroup$
– YiFan
Dec 10 '18 at 1:07
$begingroup$
And sometimes it is divergent: take $x=frac{pi}2$. For odd indeces $n$ the fraction vanishes, for even indeces $n$ it is $frac 1n$.
$endgroup$
– Pavel R.
Dec 10 '18 at 1:11
$begingroup$
It is the Fourier series of an $L^2$ function. Carleson's theorem implies that it converges a.e.
$endgroup$
– Julián Aguirre
Dec 10 '18 at 17:36