Prove that trigonometric series is/isn't a Fourier series Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Example of a trigonometric series that is not fourier series?When convergence in mean implies uniform convergence?Trigonometric series as a Fourier series.Trigonometric series as a Fourier series of essentially bounded function.Fourier Series (Even and Odd Functions)Determine the trigonometric Fourier seriesConvergence of Fourier series in $L^2$ spaceFourier coefficients of a trigonometric series : identical to trigonometric coefficients?Trigonometric series - are there possible Fourier series applications for it?Prove uniform convergence of a Fourier series
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Prove that trigonometric series is/isn't a Fourier series
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Example of a trigonometric series that is not fourier series?When convergence in mean implies uniform convergence?Trigonometric series as a Fourier series.Trigonometric series as a Fourier series of essentially bounded function.Fourier Series (Even and Odd Functions)Determine the trigonometric Fourier seriesConvergence of Fourier series in $L^2$ spaceFourier coefficients of a trigonometric series : identical to trigonometric coefficients?Trigonometric series - are there possible Fourier series applications for it?Prove uniform convergence of a Fourier series
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Prove that trigonometric series $sum_n=2^infty fraccos(nx)ln(n)$ is a Fourier series and $sum_n=2^infty fracsin(nx)ln(n)$ is not.
I know a few tricks how to find out whether the trigonometric series are Fourier or not: check its uniform convergence, check the weights convergence to zero; check the Bessel's inequality for the weights.
But these series look so similar - I'm at a loss.
fourier-series trigonometric-series
asked Apr 1 at 14:26
SilverLightSilverLight
1317
$endgroup$
add a comment |
$begingroup$
Prove that trigonometric series $sum_n=2^infty fraccos(nx)ln(n)$ is a Fourier series and $sum_n=2^infty fracsin(nx)ln(n)$ is not.
I know a few tricks how to find out whether the trigonometric series are Fourier or not: check its uniform convergence, check the weights convergence to zero; check the Bessel's inequality for the weights.
But these series look so similar - I'm at a loss.
fourier-series trigonometric-series
asked Apr 1 at 14:26
SilverLightSilverLight
1317
$endgroup$
$begingroup$
Both series converge uniformly on every compact subset of $(0,2pi)$.
$endgroup$
– Mark Viola
Apr 1 at 14:38
add a comment |
$begingroup$
Prove that trigonometric series $sum_n=2^infty fraccos(nx)ln(n)$ is a Fourier series and $sum_n=2^infty fracsin(nx)ln(n)$ is not.
I know a few tricks how to find out whether the trigonometric series are Fourier or not: check its uniform convergence, check the weights convergence to zero; check the Bessel's inequality for the weights.
But these series look so similar - I'm at a loss.
fourier-series trigonometric-series
asked Apr 1 at 14:26
SilverLightSilverLight
1317
$endgroup$
Prove that trigonometric series $sum_n=2^infty fraccos(nx)ln(n)$ is a Fourier series and $sum_n=2^infty fracsin(nx)ln(n)$ is not.
I know a few tricks how to find out whether the trigonometric series are Fourier or not: check its uniform convergence, check the weights convergence to zero; check the Bessel's inequality for the weights.
But these series look so similar - I'm at a loss.
fourier-series trigonometric-series
fourier-series trigonometric-series
asked Apr 1 at 14:26
SilverLightSilverLight
1317
asked Apr 1 at 14:26
SilverLightSilverLight
1317
asked Apr 1 at 14:26
SilverLightSilverLight
1317
asked Apr 1 at 14:26
SilverLightSilverLight
1317
asked Apr 1 at 14:26
SilverLightSilverLight
1317
1317
$begingroup$
Both series converge uniformly on every compact subset of $(0,2pi)$.
$endgroup$
– Mark Viola
Apr 1 at 14:38
add a comment |
$begingroup$
Both series converge uniformly on every compact subset of $(0,2pi)$.
$endgroup$
– Mark Viola
Apr 1 at 14:38
$begingroup$
Both series converge uniformly on every compact subset of $(0,2pi)$.
$endgroup$
– Mark Viola
Apr 1 at 14:38
$begingroup$
Both series converge uniformly on every compact subset of $(0,2pi)$.
$endgroup$
– Mark Viola
Apr 1 at 14:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Theorem 4.2 of "Introduction to Harmonic Analysis" (Y. Katznelson) states that
Let $fin L^1(mathbbT)$ and assume that $hatf(|n|)=-hatf(-|n|)geq 0$. Then
$$sum_n>0frachatf(n)n<infty$$
As a corollary, if $a_n>0$ and $sum_n a_n/n$ is not finite, then the sine series $sum_na_nsin nt $ is not a Fourier series.
answered Apr 1 at 14:42
uniquesolutionuniquesolution
9,7241823
$endgroup$
$begingroup$
The expression $hatf(|n|) = -hatf(|n|)ge 0$ must contain a typo. Yua are literally saying $hatf(|n|) = 0$.
$endgroup$
– Martín-Blas Pérez Pinilla
Apr 1 at 15:19
$begingroup$
@Martín-BlasPérezPinilla. Yes of course, thank you.
$endgroup$
– uniquesolution
Apr 1 at 15:41
$begingroup$
@uniquesolution. Thanks, but I don't really understand the theorem. What do you mean by $L^1$? If it's a space of Lebesgue integrable function, i should admit we haven't studied Lebesgue Measure and Integration yet. I hope this could be solved without Lebesgue. Moreover, what does $hatf$ mean? Just some odd function?
$endgroup$
– SilverLight
Apr 1 at 19:30
$begingroup$
@SliverLight $hatf(n)$ is a standard notation for the $n$'th Fourier coefficient of $f$. In order to define the Fourier coefficient one usually assumes that $f$ is integrable (otherwise it is not clear why the Fourier coefficients exist at all). The notation $L^1(mathbbT)$ is a standard notation for the space of Lebesgue integrable functions on the unit circle. Frankly, I am not aware of any other method to prove that your example is not a Fourier series, other than the method presented in the textbook I have made reference to.
$endgroup$
– uniquesolution
Apr 1 at 20:46
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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active
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active
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$begingroup$
Theorem 4.2 of "Introduction to Harmonic Analysis" (Y. Katznelson) states that
Let $fin L^1(mathbbT)$ and assume that $hatf(|n|)=-hatf(-|n|)geq 0$. Then
$$sum_n>0frachatf(n)n<infty$$
As a corollary, if $a_n>0$ and $sum_n a_n/n$ is not finite, then the sine series $sum_na_nsin nt $ is not a Fourier series.
answered Apr 1 at 14:42
uniquesolutionuniquesolution
9,7241823
$endgroup$
$begingroup$
The expression $hatf(|n|) = -hatf(|n|)ge 0$ must contain a typo. Yua are literally saying $hatf(|n|) = 0$.
$endgroup$
– Martín-Blas Pérez Pinilla
Apr 1 at 15:19
$begingroup$
@Martín-BlasPérezPinilla. Yes of course, thank you.
$endgroup$
– uniquesolution
Apr 1 at 15:41
$begingroup$
@uniquesolution. Thanks, but I don't really understand the theorem. What do you mean by $L^1$? If it's a space of Lebesgue integrable function, i should admit we haven't studied Lebesgue Measure and Integration yet. I hope this could be solved without Lebesgue. Moreover, what does $hatf$ mean? Just some odd function?
$endgroup$
– SilverLight
Apr 1 at 19:30
$begingroup$
@SliverLight $hatf(n)$ is a standard notation for the $n$'th Fourier coefficient of $f$. In order to define the Fourier coefficient one usually assumes that $f$ is integrable (otherwise it is not clear why the Fourier coefficients exist at all). The notation $L^1(mathbbT)$ is a standard notation for the space of Lebesgue integrable functions on the unit circle. Frankly, I am not aware of any other method to prove that your example is not a Fourier series, other than the method presented in the textbook I have made reference to.
$endgroup$
– uniquesolution
Apr 1 at 20:46
add a comment |
$begingroup$
Theorem 4.2 of "Introduction to Harmonic Analysis" (Y. Katznelson) states that
Let $fin L^1(mathbbT)$ and assume that $hatf(|n|)=-hatf(-|n|)geq 0$. Then
$$sum_n>0frachatf(n)n<infty$$
As a corollary, if $a_n>0$ and $sum_n a_n/n$ is not finite, then the sine series $sum_na_nsin nt $ is not a Fourier series.
answered Apr 1 at 14:42
uniquesolutionuniquesolution
9,7241823
$endgroup$
$begingroup$
The expression $hatf(|n|) = -hatf(|n|)ge 0$ must contain a typo. Yua are literally saying $hatf(|n|) = 0$.
$endgroup$
– Martín-Blas Pérez Pinilla
Apr 1 at 15:19
$begingroup$
@Martín-BlasPérezPinilla. Yes of course, thank you.
$endgroup$
– uniquesolution
Apr 1 at 15:41
$begingroup$
@uniquesolution. Thanks, but I don't really understand the theorem. What do you mean by $L^1$? If it's a space of Lebesgue integrable function, i should admit we haven't studied Lebesgue Measure and Integration yet. I hope this could be solved without Lebesgue. Moreover, what does $hatf$ mean? Just some odd function?
$endgroup$
– SilverLight
Apr 1 at 19:30
$begingroup$
@SliverLight $hatf(n)$ is a standard notation for the $n$'th Fourier coefficient of $f$. In order to define the Fourier coefficient one usually assumes that $f$ is integrable (otherwise it is not clear why the Fourier coefficients exist at all). The notation $L^1(mathbbT)$ is a standard notation for the space of Lebesgue integrable functions on the unit circle. Frankly, I am not aware of any other method to prove that your example is not a Fourier series, other than the method presented in the textbook I have made reference to.
$endgroup$
– uniquesolution
Apr 1 at 20:46
add a comment |
$begingroup$
Theorem 4.2 of "Introduction to Harmonic Analysis" (Y. Katznelson) states that
Let $fin L^1(mathbbT)$ and assume that $hatf(|n|)=-hatf(-|n|)geq 0$. Then
$$sum_n>0frachatf(n)n<infty$$
As a corollary, if $a_n>0$ and $sum_n a_n/n$ is not finite, then the sine series $sum_na_nsin nt $ is not a Fourier series.
answered Apr 1 at 14:42
uniquesolutionuniquesolution
9,7241823
$endgroup$
Theorem 4.2 of "Introduction to Harmonic Analysis" (Y. Katznelson) states that
Let $fin L^1(mathbbT)$ and assume that $hatf(|n|)=-hatf(-|n|)geq 0$. Then
$$sum_n>0frachatf(n)n<infty$$
As a corollary, if $a_n>0$ and $sum_n a_n/n$ is not finite, then the sine series $sum_na_nsin nt $ is not a Fourier series.
answered Apr 1 at 14:42
uniquesolutionuniquesolution
9,7241823
edited Apr 1 at 15:41
answered Apr 1 at 14:42
uniquesolutionuniquesolution
9,7241823
answered Apr 1 at 14:42
uniquesolutionuniquesolution
9,7241823
answered Apr 1 at 14:42
uniquesolutionuniquesolution
9,7241823
9,7241823
$begingroup$
The expression $hatf(|n|) = -hatf(|n|)ge 0$ must contain a typo. Yua are literally saying $hatf(|n|) = 0$.
$endgroup$
– Martín-Blas Pérez Pinilla
Apr 1 at 15:19
$begingroup$
@Martín-BlasPérezPinilla. Yes of course, thank you.
$endgroup$
– uniquesolution
Apr 1 at 15:41
$begingroup$
@uniquesolution. Thanks, but I don't really understand the theorem. What do you mean by $L^1$? If it's a space of Lebesgue integrable function, i should admit we haven't studied Lebesgue Measure and Integration yet. I hope this could be solved without Lebesgue. Moreover, what does $hatf$ mean? Just some odd function?
$endgroup$
– SilverLight
Apr 1 at 19:30
$begingroup$
@SliverLight $hatf(n)$ is a standard notation for the $n$'th Fourier coefficient of $f$. In order to define the Fourier coefficient one usually assumes that $f$ is integrable (otherwise it is not clear why the Fourier coefficients exist at all). The notation $L^1(mathbbT)$ is a standard notation for the space of Lebesgue integrable functions on the unit circle. Frankly, I am not aware of any other method to prove that your example is not a Fourier series, other than the method presented in the textbook I have made reference to.
$endgroup$
– uniquesolution
Apr 1 at 20:46
add a comment |
$begingroup$
The expression $hatf(|n|) = -hatf(|n|)ge 0$ must contain a typo. Yua are literally saying $hatf(|n|) = 0$.
$endgroup$
– Martín-Blas Pérez Pinilla
Apr 1 at 15:19
$begingroup$
@Martín-BlasPérezPinilla. Yes of course, thank you.
$endgroup$
– uniquesolution
Apr 1 at 15:41
$begingroup$
@uniquesolution. Thanks, but I don't really understand the theorem. What do you mean by $L^1$? If it's a space of Lebesgue integrable function, i should admit we haven't studied Lebesgue Measure and Integration yet. I hope this could be solved without Lebesgue. Moreover, what does $hatf$ mean? Just some odd function?
$endgroup$
– SilverLight
Apr 1 at 19:30
$begingroup$
@SliverLight $hatf(n)$ is a standard notation for the $n$'th Fourier coefficient of $f$. In order to define the Fourier coefficient one usually assumes that $f$ is integrable (otherwise it is not clear why the Fourier coefficients exist at all). The notation $L^1(mathbbT)$ is a standard notation for the space of Lebesgue integrable functions on the unit circle. Frankly, I am not aware of any other method to prove that your example is not a Fourier series, other than the method presented in the textbook I have made reference to.
$endgroup$
– uniquesolution
Apr 1 at 20:46
$begingroup$
The expression $hatf(|n|) = -hatf(|n|)ge 0$ must contain a typo. Yua are literally saying $hatf(|n|) = 0$.
$endgroup$
– Martín-Blas Pérez Pinilla
Apr 1 at 15:19
$begingroup$
The expression $hatf(|n|) = -hatf(|n|)ge 0$ must contain a typo. Yua are literally saying $hatf(|n|) = 0$.
$endgroup$
– Martín-Blas Pérez Pinilla
Apr 1 at 15:19
$begingroup$
@Martín-BlasPérezPinilla. Yes of course, thank you.
$endgroup$
– uniquesolution
Apr 1 at 15:41
$begingroup$
@Martín-BlasPérezPinilla. Yes of course, thank you.
$endgroup$
– uniquesolution
Apr 1 at 15:41
$begingroup$
@uniquesolution. Thanks, but I don't really understand the theorem. What do you mean by $L^1$? If it's a space of Lebesgue integrable function, i should admit we haven't studied Lebesgue Measure and Integration yet. I hope this could be solved without Lebesgue. Moreover, what does $hatf$ mean? Just some odd function?
$endgroup$
– SilverLight
Apr 1 at 19:30
$begingroup$
@uniquesolution. Thanks, but I don't really understand the theorem. What do you mean by $L^1$? If it's a space of Lebesgue integrable function, i should admit we haven't studied Lebesgue Measure and Integration yet. I hope this could be solved without Lebesgue. Moreover, what does $hatf$ mean? Just some odd function?
$endgroup$
– SilverLight
Apr 1 at 19:30
$begingroup$
@SliverLight $hatf(n)$ is a standard notation for the $n$'th Fourier coefficient of $f$. In order to define the Fourier coefficient one usually assumes that $f$ is integrable (otherwise it is not clear why the Fourier coefficients exist at all). The notation $L^1(mathbbT)$ is a standard notation for the space of Lebesgue integrable functions on the unit circle. Frankly, I am not aware of any other method to prove that your example is not a Fourier series, other than the method presented in the textbook I have made reference to.
$endgroup$
– uniquesolution
Apr 1 at 20:46
$begingroup$
@SliverLight $hatf(n)$ is a standard notation for the $n$'th Fourier coefficient of $f$. In order to define the Fourier coefficient one usually assumes that $f$ is integrable (otherwise it is not clear why the Fourier coefficients exist at all). The notation $L^1(mathbbT)$ is a standard notation for the space of Lebesgue integrable functions on the unit circle. Frankly, I am not aware of any other method to prove that your example is not a Fourier series, other than the method presented in the textbook I have made reference to.
$endgroup$
– uniquesolution
Apr 1 at 20:46
add a comment |
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$begingroup$
Both series converge uniformly on every compact subset of $(0,2pi)$.
$endgroup$
– Mark Viola
Apr 1 at 14:38