Does the sum $sumlimits^infty_k=1 fracsin(kx)k^alpha$ converge for $alpha > frac12$ and $x in [0,2 pi]$?
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Does the series
$$
sum^infty_k=1 fracsin(kx)k^alpha,
$$
converge for all $alpha > frac12$ and for all $x in [0,2 pi]$?
It is obvious that it does when $alpha > 1$, but I have no idea how to deal with the case
$$
frac12 < alpha le 1.
$$
I already appreciate your hints/ideas.
calculus real-analysis sequences-and-series convergence trigonometric-series
edited Aug 30 at 0:10
Robson
47320
asked Dec 28 '13 at 20:49
the8thone
1,803830
 |Â
show 2 more comments
up vote
5
down vote
favorite
Does the series
$$
sum^infty_k=1 fracsin(kx)k^alpha,
$$
converge for all $alpha > frac12$ and for all $x in [0,2 pi]$?
It is obvious that it does when $alpha > 1$, but I have no idea how to deal with the case
$$
frac12 < alpha le 1.
$$
I already appreciate your hints/ideas.
calculus real-analysis sequences-and-series convergence trigonometric-series
edited Aug 30 at 0:10
Robson
47320
asked Dec 28 '13 at 20:49
the8thone
1,803830
1
Did you try replacing $sin(kx)$ by $e^ikx$? This is probably some Fourier fact or something, since for $alpha > frac 12$ you have $sum_k ge 1 left( frac 1k^alpha right)^2 < infty$.
– Patrick Da Silva
Dec 28 '13 at 20:51
1
Dirichlet's test plus the fact the partial sums $sum_k=1^n sin(kx)$ are bounded is enough to show convergence for $alpha > 0$.
– achille hui
Dec 28 '13 at 20:58
1
@achille hui : How do you show the partial sums are bounded? I don't believe this statement.
– Patrick Da Silva
Dec 28 '13 at 21:00
2
@Roozbeh-unity : You're gonna have $$ sum_k=-infty^infty^* frack^alpha e^ikx $$ where $^*$ means you remove the term $k=0$. Do the sum of the absolute value of the squares of the coefficients give you a finite series? Then use Fourier.
– Patrick Da Silva
Dec 28 '13 at 21:01
1
@Roozbeh-unity: this will help you(believe):math.stackexchange.com/questions/108486/…
– Salech Alhasov
Dec 28 '13 at 21:04
 |Â
show 2 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Does the series
$$
sum^infty_k=1 fracsin(kx)k^alpha,
$$
converge for all $alpha > frac12$ and for all $x in [0,2 pi]$?
It is obvious that it does when $alpha > 1$, but I have no idea how to deal with the case
$$
frac12 < alpha le 1.
$$
I already appreciate your hints/ideas.
calculus real-analysis sequences-and-series convergence trigonometric-series
edited Aug 30 at 0:10
Robson
47320
asked Dec 28 '13 at 20:49
the8thone
1,803830
Does the series
$$
sum^infty_k=1 fracsin(kx)k^alpha,
$$
converge for all $alpha > frac12$ and for all $x in [0,2 pi]$?
It is obvious that it does when $alpha > 1$, but I have no idea how to deal with the case
$$
frac12 < alpha le 1.
$$
I already appreciate your hints/ideas.
calculus real-analysis sequences-and-series convergence trigonometric-series
calculus real-analysis sequences-and-series convergence trigonometric-series
edited Aug 30 at 0:10
Robson
47320
asked Dec 28 '13 at 20:49
the8thone
1,803830
edited Aug 30 at 0:10
Robson
47320
asked Dec 28 '13 at 20:49
the8thone
1,803830
edited Aug 30 at 0:10
Robson
47320
edited Aug 30 at 0:10
Robson
47320
edited Aug 30 at 0:10
Robson
47320
47320
asked Dec 28 '13 at 20:49
the8thone
1,803830
asked Dec 28 '13 at 20:49
the8thone
1,803830
asked Dec 28 '13 at 20:49
the8thone
1,803830
1,803830
1
Did you try replacing $sin(kx)$ by $e^ikx$? This is probably some Fourier fact or something, since for $alpha > frac 12$ you have $sum_k ge 1 left( frac 1k^alpha right)^2 < infty$.
– Patrick Da Silva
Dec 28 '13 at 20:51
1
Dirichlet's test plus the fact the partial sums $sum_k=1^n sin(kx)$ are bounded is enough to show convergence for $alpha > 0$.
– achille hui
Dec 28 '13 at 20:58
1
@achille hui : How do you show the partial sums are bounded? I don't believe this statement.
– Patrick Da Silva
Dec 28 '13 at 21:00
2
@Roozbeh-unity : You're gonna have $$ sum_k=-infty^infty^* frack^alpha e^ikx $$ where $^*$ means you remove the term $k=0$. Do the sum of the absolute value of the squares of the coefficients give you a finite series? Then use Fourier.
– Patrick Da Silva
Dec 28 '13 at 21:01
1
@Roozbeh-unity: this will help you(believe):math.stackexchange.com/questions/108486/…
– Salech Alhasov
Dec 28 '13 at 21:04
 |Â
show 2 more comments
1
Did you try replacing $sin(kx)$ by $e^ikx$? This is probably some Fourier fact or something, since for $alpha > frac 12$ you have $sum_k ge 1 left( frac 1k^alpha right)^2 < infty$.
– Patrick Da Silva
Dec 28 '13 at 20:51
1
Dirichlet's test plus the fact the partial sums $sum_k=1^n sin(kx)$ are bounded is enough to show convergence for $alpha > 0$.
– achille hui
Dec 28 '13 at 20:58
1
@achille hui : How do you show the partial sums are bounded? I don't believe this statement.
– Patrick Da Silva
Dec 28 '13 at 21:00
2
@Roozbeh-unity : You're gonna have $$ sum_k=-infty^infty^* frack^alpha e^ikx $$ where $^*$ means you remove the term $k=0$. Do the sum of the absolute value of the squares of the coefficients give you a finite series? Then use Fourier.
– Patrick Da Silva
Dec 28 '13 at 21:01
1
@Roozbeh-unity: this will help you(believe):math.stackexchange.com/questions/108486/…
– Salech Alhasov
Dec 28 '13 at 21:04
1
1
Did you try replacing $sin(kx)$ by $e^ikx$? This is probably some Fourier fact or something, since for $alpha > frac 12$ you have $sum_k ge 1 left( frac 1k^alpha right)^2 < infty$.
– Patrick Da Silva
Dec 28 '13 at 20:51
Did you try replacing $sin(kx)$ by $e^ikx$? This is probably some Fourier fact or something, since for $alpha > frac 12$ you have $sum_k ge 1 left( frac 1k^alpha right)^2 < infty$.
– Patrick Da Silva
Dec 28 '13 at 20:51
1
1
Dirichlet's test plus the fact the partial sums $sum_k=1^n sin(kx)$ are bounded is enough to show convergence for $alpha > 0$.
– achille hui
Dec 28 '13 at 20:58
Dirichlet's test plus the fact the partial sums $sum_k=1^n sin(kx)$ are bounded is enough to show convergence for $alpha > 0$.
– achille hui
Dec 28 '13 at 20:58
1
1
@achille hui : How do you show the partial sums are bounded? I don't believe this statement.
– Patrick Da Silva
Dec 28 '13 at 21:00
@achille hui : How do you show the partial sums are bounded? I don't believe this statement.
– Patrick Da Silva
Dec 28 '13 at 21:00
2
2
@Roozbeh-unity : You're gonna have $$ sum_k=-infty^infty^* frack^alpha e^ikx $$ where $^*$ means you remove the term $k=0$. Do the sum of the absolute value of the squares of the coefficients give you a finite series? Then use Fourier.
– Patrick Da Silva
Dec 28 '13 at 21:01
@Roozbeh-unity : You're gonna have $$ sum_k=-infty^infty^* frack^alpha e^ikx $$ where $^*$ means you remove the term $k=0$. Do the sum of the absolute value of the squares of the coefficients give you a finite series? Then use Fourier.
– Patrick Da Silva
Dec 28 '13 at 21:01
1
1
@Roozbeh-unity: this will help you(believe):math.stackexchange.com/questions/108486/…
– Salech Alhasov
Dec 28 '13 at 21:04
@Roozbeh-unity: this will help you(believe):math.stackexchange.com/questions/108486/…
– Salech Alhasov
Dec 28 '13 at 21:04
 |Â
show 2 more comments
2 Answers
2
active
oldest
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up vote
11
down vote
Actually, this sum converges for every $alpha>0$.
Step I. For every $xinmathbb R$, the sequence $s_n=sum_k=1^nsin kx$ is bounded.
Indeed, if $x=mpi$, then $s_n=0$. If $xne mpi$, then $sin(x/2)ne 0$, and
$$
s_n=sum_k=1^nsin kx=mathrmImleft(mathrme^xi+mathrme^2xi+cdotsmathrme^nxiright)=
mathrmImleft(mathrme^xifracmathrme^nxi-1mathrme^xi-1right)
$$
But
$$
left|mathrme^xifracmathrme^nxi-1mathrme^xi-1right|le frac2mathrme^xi-1=frac2=frac1.
$$
and hence $lvert s_nrvertle lvertsin(x/2)rvert^-1$.
Step II. Use Abel's summation method.
beginalign
sigma_n &=sum_k=1^nfracsin kxk^alpha=sum_k=1^nfracs_k-s_k-1k^alpha
=sum_k=1^nfracs_kk^alpha-sum_k=1^nfracs_k-1k^alpha \
&=sum_k=1^nfracs_kk^alpha-sum_k=0^n-1fracs_k(k+1)^alpha=
fracs_nn^a+sum_k=1^n-1s_kleft(frac1k^a-frac1(k+1)^aright).
endalign
But
$$
frac1k^alpha-frac1(k+1)^alphalefracak^1+alpha,
$$
and hence the series
$$
sum_n=1^inftys_nleft(frac1n^a-frac1(n+1)^aright),
$$
converges (indeed absolutely) due to the comparison test.
Note. This series converges conditionally and pointwise. It does not converge absolutely, but it does converge uniformly far from zero, i.e., in any interval
$[varepsilon,2pi-varepsilon]$.
answered Dec 28 '13 at 21:10
Yiorgos S. Smyrlis
60.3k1383161
$frac 1sin(x/2)$ is not bounded. What do you mean exactly?
– Patrick Da Silva
Dec 28 '13 at 21:13
@PatrickDaSilva: The sequence $s_n$ is bounded. I proved that the series converges for every $x$ - It does not converge uniformly.
– Yiorgos S. Smyrlis
Dec 28 '13 at 21:15
So you mean $s_n(x)$ is bounded for each $x$?
– Patrick Da Silva
Dec 28 '13 at 21:21
@PatrickDaSilva: Correct - What kind of convergence are you looking for?
– Yiorgos S. Smyrlis
Dec 28 '13 at 21:25
2
@YiorgosS.Smyrlis: you could simply refer to the Dirichlet Test, which you've essentially proven above.
– robjohn♦
Dec 28 '13 at 23:57
 |Â
show 2 more comments
up vote
3
down vote
The set $ , n in mathbb Z $ forms an Hilbert basis of $L^2([0,2pi])$ which means that for a function $f in L^2([0,2pi])$, we have
$$
f = sum_n in mathbb Z langle f, e_n rangle e_n
$$
with
$$
langle f,e_nrangle = fracn/2i n^alpha
$$
if $n neq 0$ and $0$ otherwise, if and only if
$$
sum_n in mathbb Z |langle f,e_n rangle|^2 < infty.
$$
Since $alpha > frac 12$, in our situation this is the case, therefore $sum_n in mathbb Z fracsin(nx)n^alpha$ is well-defined and converges pointwise almost everywhere to some function $f(x)$ which is in $L^2([0,2pi])$.
Hope that helps,
answered Dec 28 '13 at 21:11
Patrick Da Silva
31.7k352106
2
$L^2$ convergence of the Fourier series does not imply pointwise convergence in all points, if I don't misremember.
– Daniel Fischer♦
Dec 28 '13 at 21:15
@Daniel Fischer : It doesn't imply pointwise convergence to the considered function whose Fourier series we take (not here, but in the theory of "computing the Fourier series of a function", which is not what I do here), but it definitely converges pointwise to some function. $L^2$ convergence means $L^2$ convergence, so the series must converge to some element of $L^2$. That's all I need.
– Patrick Da Silva
Dec 28 '13 at 21:16
2
No, not in general. See e.g. Theorem 5.12 in Rudin's Real and Complex Analysis. The Fourier series of a continuous function is in general (meaning outside a meagre subspace of $C(T)$) unbounded in every point of a dense $G_delta$. Since $C(T) subset L^2(T)$, the existence of $L^2$ Fourier series that don't converge pointwise everywhere follows.
– Daniel Fischer♦
Dec 28 '13 at 21:24
@Daniel Fischer : Yeah you made me doubt. I mean, I wrote things down and then gave it some thought. My proof only tells us that for "almost all $x$" (in the Lebesgue-measure sense) the series converge. I edited my answer and left it there for fun, but Yiorgis's answer is better.
– Patrick Da Silva
Dec 28 '13 at 21:27
It's also because I thought "this was the proof" because of the $alpha > 1/2$ condition.
– Patrick Da Silva
Dec 28 '13 at 21:28
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
Actually, this sum converges for every $alpha>0$.
Step I. For every $xinmathbb R$, the sequence $s_n=sum_k=1^nsin kx$ is bounded.
Indeed, if $x=mpi$, then $s_n=0$. If $xne mpi$, then $sin(x/2)ne 0$, and
$$
s_n=sum_k=1^nsin kx=mathrmImleft(mathrme^xi+mathrme^2xi+cdotsmathrme^nxiright)=
mathrmImleft(mathrme^xifracmathrme^nxi-1mathrme^xi-1right)
$$
But
$$
left|mathrme^xifracmathrme^nxi-1mathrme^xi-1right|le frac2mathrme^xi-1=frac2=frac1.
$$
and hence $lvert s_nrvertle lvertsin(x/2)rvert^-1$.
Step II. Use Abel's summation method.
beginalign
sigma_n &=sum_k=1^nfracsin kxk^alpha=sum_k=1^nfracs_k-s_k-1k^alpha
=sum_k=1^nfracs_kk^alpha-sum_k=1^nfracs_k-1k^alpha \
&=sum_k=1^nfracs_kk^alpha-sum_k=0^n-1fracs_k(k+1)^alpha=
fracs_nn^a+sum_k=1^n-1s_kleft(frac1k^a-frac1(k+1)^aright).
endalign
But
$$
frac1k^alpha-frac1(k+1)^alphalefracak^1+alpha,
$$
and hence the series
$$
sum_n=1^inftys_nleft(frac1n^a-frac1(n+1)^aright),
$$
converges (indeed absolutely) due to the comparison test.
Note. This series converges conditionally and pointwise. It does not converge absolutely, but it does converge uniformly far from zero, i.e., in any interval
$[varepsilon,2pi-varepsilon]$.
answered Dec 28 '13 at 21:10
Yiorgos S. Smyrlis
60.3k1383161
$frac 1sin(x/2)$ is not bounded. What do you mean exactly?
– Patrick Da Silva
Dec 28 '13 at 21:13
@PatrickDaSilva: The sequence $s_n$ is bounded. I proved that the series converges for every $x$ - It does not converge uniformly.
– Yiorgos S. Smyrlis
Dec 28 '13 at 21:15
So you mean $s_n(x)$ is bounded for each $x$?
– Patrick Da Silva
Dec 28 '13 at 21:21
@PatrickDaSilva: Correct - What kind of convergence are you looking for?
– Yiorgos S. Smyrlis
Dec 28 '13 at 21:25
2
@YiorgosS.Smyrlis: you could simply refer to the Dirichlet Test, which you've essentially proven above.
– robjohn♦
Dec 28 '13 at 23:57
 |Â
show 2 more comments
up vote
11
down vote
Actually, this sum converges for every $alpha>0$.
Step I. For every $xinmathbb R$, the sequence $s_n=sum_k=1^nsin kx$ is bounded.
Indeed, if $x=mpi$, then $s_n=0$. If $xne mpi$, then $sin(x/2)ne 0$, and
$$
s_n=sum_k=1^nsin kx=mathrmImleft(mathrme^xi+mathrme^2xi+cdotsmathrme^nxiright)=
mathrmImleft(mathrme^xifracmathrme^nxi-1mathrme^xi-1right)
$$
But
$$
left|mathrme^xifracmathrme^nxi-1mathrme^xi-1right|le frac2mathrme^xi-1=frac2=frac1.
$$
and hence $lvert s_nrvertle lvertsin(x/2)rvert^-1$.
Step II. Use Abel's summation method.
beginalign
sigma_n &=sum_k=1^nfracsin kxk^alpha=sum_k=1^nfracs_k-s_k-1k^alpha
=sum_k=1^nfracs_kk^alpha-sum_k=1^nfracs_k-1k^alpha \
&=sum_k=1^nfracs_kk^alpha-sum_k=0^n-1fracs_k(k+1)^alpha=
fracs_nn^a+sum_k=1^n-1s_kleft(frac1k^a-frac1(k+1)^aright).
endalign
But
$$
frac1k^alpha-frac1(k+1)^alphalefracak^1+alpha,
$$
and hence the series
$$
sum_n=1^inftys_nleft(frac1n^a-frac1(n+1)^aright),
$$
converges (indeed absolutely) due to the comparison test.
Note. This series converges conditionally and pointwise. It does not converge absolutely, but it does converge uniformly far from zero, i.e., in any interval
$[varepsilon,2pi-varepsilon]$.
answered Dec 28 '13 at 21:10
Yiorgos S. Smyrlis
60.3k1383161
$frac 1sin(x/2)$ is not bounded. What do you mean exactly?
– Patrick Da Silva
Dec 28 '13 at 21:13
@PatrickDaSilva: The sequence $s_n$ is bounded. I proved that the series converges for every $x$ - It does not converge uniformly.
– Yiorgos S. Smyrlis
Dec 28 '13 at 21:15
So you mean $s_n(x)$ is bounded for each $x$?
– Patrick Da Silva
Dec 28 '13 at 21:21
@PatrickDaSilva: Correct - What kind of convergence are you looking for?
– Yiorgos S. Smyrlis
Dec 28 '13 at 21:25
2
@YiorgosS.Smyrlis: you could simply refer to the Dirichlet Test, which you've essentially proven above.
– robjohn♦
Dec 28 '13 at 23:57
 |Â
show 2 more comments
up vote
11
down vote
up vote
11
down vote
Actually, this sum converges for every $alpha>0$.
Step I. For every $xinmathbb R$, the sequence $s_n=sum_k=1^nsin kx$ is bounded.
Indeed, if $x=mpi$, then $s_n=0$. If $xne mpi$, then $sin(x/2)ne 0$, and
$$
s_n=sum_k=1^nsin kx=mathrmImleft(mathrme^xi+mathrme^2xi+cdotsmathrme^nxiright)=
mathrmImleft(mathrme^xifracmathrme^nxi-1mathrme^xi-1right)
$$
But
$$
left|mathrme^xifracmathrme^nxi-1mathrme^xi-1right|le frac2mathrme^xi-1=frac2=frac1.
$$
and hence $lvert s_nrvertle lvertsin(x/2)rvert^-1$.
Step II. Use Abel's summation method.
beginalign
sigma_n &=sum_k=1^nfracsin kxk^alpha=sum_k=1^nfracs_k-s_k-1k^alpha
=sum_k=1^nfracs_kk^alpha-sum_k=1^nfracs_k-1k^alpha \
&=sum_k=1^nfracs_kk^alpha-sum_k=0^n-1fracs_k(k+1)^alpha=
fracs_nn^a+sum_k=1^n-1s_kleft(frac1k^a-frac1(k+1)^aright).
endalign
But
$$
frac1k^alpha-frac1(k+1)^alphalefracak^1+alpha,
$$
and hence the series
$$
sum_n=1^inftys_nleft(frac1n^a-frac1(n+1)^aright),
$$
converges (indeed absolutely) due to the comparison test.
Note. This series converges conditionally and pointwise. It does not converge absolutely, but it does converge uniformly far from zero, i.e., in any interval
$[varepsilon,2pi-varepsilon]$.
answered Dec 28 '13 at 21:10
Yiorgos S. Smyrlis
60.3k1383161
Actually, this sum converges for every $alpha>0$.
Step I. For every $xinmathbb R$, the sequence $s_n=sum_k=1^nsin kx$ is bounded.
Indeed, if $x=mpi$, then $s_n=0$. If $xne mpi$, then $sin(x/2)ne 0$, and
$$
s_n=sum_k=1^nsin kx=mathrmImleft(mathrme^xi+mathrme^2xi+cdotsmathrme^nxiright)=
mathrmImleft(mathrme^xifracmathrme^nxi-1mathrme^xi-1right)
$$
But
$$
left|mathrme^xifracmathrme^nxi-1mathrme^xi-1right|le frac2mathrme^xi-1=frac2=frac1.
$$
and hence $lvert s_nrvertle lvertsin(x/2)rvert^-1$.
Step II. Use Abel's summation method.
beginalign
sigma_n &=sum_k=1^nfracsin kxk^alpha=sum_k=1^nfracs_k-s_k-1k^alpha
=sum_k=1^nfracs_kk^alpha-sum_k=1^nfracs_k-1k^alpha \
&=sum_k=1^nfracs_kk^alpha-sum_k=0^n-1fracs_k(k+1)^alpha=
fracs_nn^a+sum_k=1^n-1s_kleft(frac1k^a-frac1(k+1)^aright).
endalign
But
$$
frac1k^alpha-frac1(k+1)^alphalefracak^1+alpha,
$$
and hence the series
$$
sum_n=1^inftys_nleft(frac1n^a-frac1(n+1)^aright),
$$
converges (indeed absolutely) due to the comparison test.
Note. This series converges conditionally and pointwise. It does not converge absolutely, but it does converge uniformly far from zero, i.e., in any interval
$[varepsilon,2pi-varepsilon]$.
answered Dec 28 '13 at 21:10
Yiorgos S. Smyrlis
60.3k1383161
edited May 30 '15 at 18:11
answered Dec 28 '13 at 21:10
Yiorgos S. Smyrlis
60.3k1383161
answered Dec 28 '13 at 21:10
Yiorgos S. Smyrlis
60.3k1383161
answered Dec 28 '13 at 21:10
Yiorgos S. Smyrlis
60.3k1383161
60.3k1383161
$frac 1sin(x/2)$ is not bounded. What do you mean exactly?
– Patrick Da Silva
Dec 28 '13 at 21:13
@PatrickDaSilva: The sequence $s_n$ is bounded. I proved that the series converges for every $x$ - It does not converge uniformly.
– Yiorgos S. Smyrlis
Dec 28 '13 at 21:15
So you mean $s_n(x)$ is bounded for each $x$?
– Patrick Da Silva
Dec 28 '13 at 21:21
@PatrickDaSilva: Correct - What kind of convergence are you looking for?
– Yiorgos S. Smyrlis
Dec 28 '13 at 21:25
2
@YiorgosS.Smyrlis: you could simply refer to the Dirichlet Test, which you've essentially proven above.
– robjohn♦
Dec 28 '13 at 23:57
 |Â
show 2 more comments
$frac 1sin(x/2)$ is not bounded. What do you mean exactly?
– Patrick Da Silva
Dec 28 '13 at 21:13
@PatrickDaSilva: The sequence $s_n$ is bounded. I proved that the series converges for every $x$ - It does not converge uniformly.
– Yiorgos S. Smyrlis
Dec 28 '13 at 21:15
So you mean $s_n(x)$ is bounded for each $x$?
– Patrick Da Silva
Dec 28 '13 at 21:21
@PatrickDaSilva: Correct - What kind of convergence are you looking for?
– Yiorgos S. Smyrlis
Dec 28 '13 at 21:25
2
@YiorgosS.Smyrlis: you could simply refer to the Dirichlet Test, which you've essentially proven above.
– robjohn♦
Dec 28 '13 at 23:57
$frac 1sin(x/2)$ is not bounded. What do you mean exactly?
– Patrick Da Silva
Dec 28 '13 at 21:13
$frac 1sin(x/2)$ is not bounded. What do you mean exactly?
– Patrick Da Silva
Dec 28 '13 at 21:13
@PatrickDaSilva: The sequence $s_n$ is bounded. I proved that the series converges for every $x$ - It does not converge uniformly.
– Yiorgos S. Smyrlis
Dec 28 '13 at 21:15
@PatrickDaSilva: The sequence $s_n$ is bounded. I proved that the series converges for every $x$ - It does not converge uniformly.
– Yiorgos S. Smyrlis
Dec 28 '13 at 21:15
So you mean $s_n(x)$ is bounded for each $x$?
– Patrick Da Silva
Dec 28 '13 at 21:21
So you mean $s_n(x)$ is bounded for each $x$?
– Patrick Da Silva
Dec 28 '13 at 21:21
@PatrickDaSilva: Correct - What kind of convergence are you looking for?
– Yiorgos S. Smyrlis
Dec 28 '13 at 21:25
@PatrickDaSilva: Correct - What kind of convergence are you looking for?
– Yiorgos S. Smyrlis
Dec 28 '13 at 21:25
2
2
@YiorgosS.Smyrlis: you could simply refer to the Dirichlet Test, which you've essentially proven above.
– robjohn♦
Dec 28 '13 at 23:57
@YiorgosS.Smyrlis: you could simply refer to the Dirichlet Test, which you've essentially proven above.
– robjohn♦
Dec 28 '13 at 23:57
 |Â
show 2 more comments
up vote
3
down vote
The set $ , n in mathbb Z $ forms an Hilbert basis of $L^2([0,2pi])$ which means that for a function $f in L^2([0,2pi])$, we have
$$
f = sum_n in mathbb Z langle f, e_n rangle e_n
$$
with
$$
langle f,e_nrangle = fracn/2i n^alpha
$$
if $n neq 0$ and $0$ otherwise, if and only if
$$
sum_n in mathbb Z |langle f,e_n rangle|^2 < infty.
$$
Since $alpha > frac 12$, in our situation this is the case, therefore $sum_n in mathbb Z fracsin(nx)n^alpha$ is well-defined and converges pointwise almost everywhere to some function $f(x)$ which is in $L^2([0,2pi])$.
Hope that helps,
answered Dec 28 '13 at 21:11
Patrick Da Silva
31.7k352106
2
$L^2$ convergence of the Fourier series does not imply pointwise convergence in all points, if I don't misremember.
– Daniel Fischer♦
Dec 28 '13 at 21:15
@Daniel Fischer : It doesn't imply pointwise convergence to the considered function whose Fourier series we take (not here, but in the theory of "computing the Fourier series of a function", which is not what I do here), but it definitely converges pointwise to some function. $L^2$ convergence means $L^2$ convergence, so the series must converge to some element of $L^2$. That's all I need.
– Patrick Da Silva
Dec 28 '13 at 21:16
2
No, not in general. See e.g. Theorem 5.12 in Rudin's Real and Complex Analysis. The Fourier series of a continuous function is in general (meaning outside a meagre subspace of $C(T)$) unbounded in every point of a dense $G_delta$. Since $C(T) subset L^2(T)$, the existence of $L^2$ Fourier series that don't converge pointwise everywhere follows.
– Daniel Fischer♦
Dec 28 '13 at 21:24
@Daniel Fischer : Yeah you made me doubt. I mean, I wrote things down and then gave it some thought. My proof only tells us that for "almost all $x$" (in the Lebesgue-measure sense) the series converge. I edited my answer and left it there for fun, but Yiorgis's answer is better.
– Patrick Da Silva
Dec 28 '13 at 21:27
It's also because I thought "this was the proof" because of the $alpha > 1/2$ condition.
– Patrick Da Silva
Dec 28 '13 at 21:28
add a comment |Â
up vote
3
down vote
The set $ , n in mathbb Z $ forms an Hilbert basis of $L^2([0,2pi])$ which means that for a function $f in L^2([0,2pi])$, we have
$$
f = sum_n in mathbb Z langle f, e_n rangle e_n
$$
with
$$
langle f,e_nrangle = fracn/2i n^alpha
$$
if $n neq 0$ and $0$ otherwise, if and only if
$$
sum_n in mathbb Z |langle f,e_n rangle|^2 < infty.
$$
Since $alpha > frac 12$, in our situation this is the case, therefore $sum_n in mathbb Z fracsin(nx)n^alpha$ is well-defined and converges pointwise almost everywhere to some function $f(x)$ which is in $L^2([0,2pi])$.
Hope that helps,
answered Dec 28 '13 at 21:11
Patrick Da Silva
31.7k352106
2
$L^2$ convergence of the Fourier series does not imply pointwise convergence in all points, if I don't misremember.
– Daniel Fischer♦
Dec 28 '13 at 21:15
@Daniel Fischer : It doesn't imply pointwise convergence to the considered function whose Fourier series we take (not here, but in the theory of "computing the Fourier series of a function", which is not what I do here), but it definitely converges pointwise to some function. $L^2$ convergence means $L^2$ convergence, so the series must converge to some element of $L^2$. That's all I need.
– Patrick Da Silva
Dec 28 '13 at 21:16
2
No, not in general. See e.g. Theorem 5.12 in Rudin's Real and Complex Analysis. The Fourier series of a continuous function is in general (meaning outside a meagre subspace of $C(T)$) unbounded in every point of a dense $G_delta$. Since $C(T) subset L^2(T)$, the existence of $L^2$ Fourier series that don't converge pointwise everywhere follows.
– Daniel Fischer♦
Dec 28 '13 at 21:24
@Daniel Fischer : Yeah you made me doubt. I mean, I wrote things down and then gave it some thought. My proof only tells us that for "almost all $x$" (in the Lebesgue-measure sense) the series converge. I edited my answer and left it there for fun, but Yiorgis's answer is better.
– Patrick Da Silva
Dec 28 '13 at 21:27
It's also because I thought "this was the proof" because of the $alpha > 1/2$ condition.
– Patrick Da Silva
Dec 28 '13 at 21:28
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The set $ , n in mathbb Z $ forms an Hilbert basis of $L^2([0,2pi])$ which means that for a function $f in L^2([0,2pi])$, we have
$$
f = sum_n in mathbb Z langle f, e_n rangle e_n
$$
with
$$
langle f,e_nrangle = fracn/2i n^alpha
$$
if $n neq 0$ and $0$ otherwise, if and only if
$$
sum_n in mathbb Z |langle f,e_n rangle|^2 < infty.
$$
Since $alpha > frac 12$, in our situation this is the case, therefore $sum_n in mathbb Z fracsin(nx)n^alpha$ is well-defined and converges pointwise almost everywhere to some function $f(x)$ which is in $L^2([0,2pi])$.
Hope that helps,
answered Dec 28 '13 at 21:11
Patrick Da Silva
31.7k352106
The set $ , n in mathbb Z $ forms an Hilbert basis of $L^2([0,2pi])$ which means that for a function $f in L^2([0,2pi])$, we have
$$
f = sum_n in mathbb Z langle f, e_n rangle e_n
$$
with
$$
langle f,e_nrangle = fracn/2i n^alpha
$$
if $n neq 0$ and $0$ otherwise, if and only if
$$
sum_n in mathbb Z |langle f,e_n rangle|^2 < infty.
$$
Since $alpha > frac 12$, in our situation this is the case, therefore $sum_n in mathbb Z fracsin(nx)n^alpha$ is well-defined and converges pointwise almost everywhere to some function $f(x)$ which is in $L^2([0,2pi])$.
Hope that helps,
answered Dec 28 '13 at 21:11
Patrick Da Silva
31.7k352106
edited Dec 28 '13 at 21:26
answered Dec 28 '13 at 21:11
Patrick Da Silva
31.7k352106
answered Dec 28 '13 at 21:11
Patrick Da Silva
31.7k352106
answered Dec 28 '13 at 21:11
Patrick Da Silva
31.7k352106
31.7k352106
2
$L^2$ convergence of the Fourier series does not imply pointwise convergence in all points, if I don't misremember.
– Daniel Fischer♦
Dec 28 '13 at 21:15
@Daniel Fischer : It doesn't imply pointwise convergence to the considered function whose Fourier series we take (not here, but in the theory of "computing the Fourier series of a function", which is not what I do here), but it definitely converges pointwise to some function. $L^2$ convergence means $L^2$ convergence, so the series must converge to some element of $L^2$. That's all I need.
– Patrick Da Silva
Dec 28 '13 at 21:16
2
No, not in general. See e.g. Theorem 5.12 in Rudin's Real and Complex Analysis. The Fourier series of a continuous function is in general (meaning outside a meagre subspace of $C(T)$) unbounded in every point of a dense $G_delta$. Since $C(T) subset L^2(T)$, the existence of $L^2$ Fourier series that don't converge pointwise everywhere follows.
– Daniel Fischer♦
Dec 28 '13 at 21:24
@Daniel Fischer : Yeah you made me doubt. I mean, I wrote things down and then gave it some thought. My proof only tells us that for "almost all $x$" (in the Lebesgue-measure sense) the series converge. I edited my answer and left it there for fun, but Yiorgis's answer is better.
– Patrick Da Silva
Dec 28 '13 at 21:27
It's also because I thought "this was the proof" because of the $alpha > 1/2$ condition.
– Patrick Da Silva
Dec 28 '13 at 21:28
add a comment |Â
2
$L^2$ convergence of the Fourier series does not imply pointwise convergence in all points, if I don't misremember.
– Daniel Fischer♦
Dec 28 '13 at 21:15
@Daniel Fischer : It doesn't imply pointwise convergence to the considered function whose Fourier series we take (not here, but in the theory of "computing the Fourier series of a function", which is not what I do here), but it definitely converges pointwise to some function. $L^2$ convergence means $L^2$ convergence, so the series must converge to some element of $L^2$. That's all I need.
– Patrick Da Silva
Dec 28 '13 at 21:16
2
No, not in general. See e.g. Theorem 5.12 in Rudin's Real and Complex Analysis. The Fourier series of a continuous function is in general (meaning outside a meagre subspace of $C(T)$) unbounded in every point of a dense $G_delta$. Since $C(T) subset L^2(T)$, the existence of $L^2$ Fourier series that don't converge pointwise everywhere follows.
– Daniel Fischer♦
Dec 28 '13 at 21:24
@Daniel Fischer : Yeah you made me doubt. I mean, I wrote things down and then gave it some thought. My proof only tells us that for "almost all $x$" (in the Lebesgue-measure sense) the series converge. I edited my answer and left it there for fun, but Yiorgis's answer is better.
– Patrick Da Silva
Dec 28 '13 at 21:27
It's also because I thought "this was the proof" because of the $alpha > 1/2$ condition.
– Patrick Da Silva
Dec 28 '13 at 21:28
2
2
$L^2$ convergence of the Fourier series does not imply pointwise convergence in all points, if I don't misremember.
– Daniel Fischer♦
Dec 28 '13 at 21:15
$L^2$ convergence of the Fourier series does not imply pointwise convergence in all points, if I don't misremember.
– Daniel Fischer♦
Dec 28 '13 at 21:15
@Daniel Fischer : It doesn't imply pointwise convergence to the considered function whose Fourier series we take (not here, but in the theory of "computing the Fourier series of a function", which is not what I do here), but it definitely converges pointwise to some function. $L^2$ convergence means $L^2$ convergence, so the series must converge to some element of $L^2$. That's all I need.
– Patrick Da Silva
Dec 28 '13 at 21:16
@Daniel Fischer : It doesn't imply pointwise convergence to the considered function whose Fourier series we take (not here, but in the theory of "computing the Fourier series of a function", which is not what I do here), but it definitely converges pointwise to some function. $L^2$ convergence means $L^2$ convergence, so the series must converge to some element of $L^2$. That's all I need.
– Patrick Da Silva
Dec 28 '13 at 21:16
2
2
No, not in general. See e.g. Theorem 5.12 in Rudin's Real and Complex Analysis. The Fourier series of a continuous function is in general (meaning outside a meagre subspace of $C(T)$) unbounded in every point of a dense $G_delta$. Since $C(T) subset L^2(T)$, the existence of $L^2$ Fourier series that don't converge pointwise everywhere follows.
– Daniel Fischer♦
Dec 28 '13 at 21:24
No, not in general. See e.g. Theorem 5.12 in Rudin's Real and Complex Analysis. The Fourier series of a continuous function is in general (meaning outside a meagre subspace of $C(T)$) unbounded in every point of a dense $G_delta$. Since $C(T) subset L^2(T)$, the existence of $L^2$ Fourier series that don't converge pointwise everywhere follows.
– Daniel Fischer♦
Dec 28 '13 at 21:24
@Daniel Fischer : Yeah you made me doubt. I mean, I wrote things down and then gave it some thought. My proof only tells us that for "almost all $x$" (in the Lebesgue-measure sense) the series converge. I edited my answer and left it there for fun, but Yiorgis's answer is better.
– Patrick Da Silva
Dec 28 '13 at 21:27
@Daniel Fischer : Yeah you made me doubt. I mean, I wrote things down and then gave it some thought. My proof only tells us that for "almost all $x$" (in the Lebesgue-measure sense) the series converge. I edited my answer and left it there for fun, but Yiorgis's answer is better.
– Patrick Da Silva
Dec 28 '13 at 21:27
It's also because I thought "this was the proof" because of the $alpha > 1/2$ condition.
– Patrick Da Silva
Dec 28 '13 at 21:28
It's also because I thought "this was the proof" because of the $alpha > 1/2$ condition.
– Patrick Da Silva
Dec 28 '13 at 21:28
add a comment |Â
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1
Did you try replacing $sin(kx)$ by $e^ikx$? This is probably some Fourier fact or something, since for $alpha > frac 12$ you have $sum_k ge 1 left( frac 1k^alpha right)^2 < infty$.
– Patrick Da Silva
Dec 28 '13 at 20:51
1
Dirichlet's test plus the fact the partial sums $sum_k=1^n sin(kx)$ are bounded is enough to show convergence for $alpha > 0$.
– achille hui
Dec 28 '13 at 20:58
1
@achille hui : How do you show the partial sums are bounded? I don't believe this statement.
– Patrick Da Silva
Dec 28 '13 at 21:00
2
@Roozbeh-unity : You're gonna have $$ sum_k=-infty^infty^* frack^alpha e^ikx $$ where $^*$ means you remove the term $k=0$. Do the sum of the absolute value of the squares of the coefficients give you a finite series? Then use Fourier.
– Patrick Da Silva
Dec 28 '13 at 21:01
1
@Roozbeh-unity: this will help you(believe):math.stackexchange.com/questions/108486/…
– Salech Alhasov
Dec 28 '13 at 21:04