Series of $sin(nx)$ terms that sum up to $0$.
$begingroup$
Working in $mathbb{R}$, what sequences $a_n$ satisfy $sum_{n=0}^{infty} a_n sin(nx)=0$ for all $x$, pointwise ?
I've never thought about this and I'm not sure whether I'm not missing something trivial - I would have expected that $a_n = 0$ for all $n$ is the only solution, but I'm not sure I've seen this stated anywhere.
real-analysis fourier-analysis
edited Jan 5 at 22:52
Bernard
123k741117
asked Jan 5 at 22:42
John PJohn P
55727
$endgroup$
add a comment |
$begingroup$
Working in $mathbb{R}$, what sequences $a_n$ satisfy $sum_{n=0}^{infty} a_n sin(nx)=0$ for all $x$, pointwise ?
I've never thought about this and I'm not sure whether I'm not missing something trivial - I would have expected that $a_n = 0$ for all $n$ is the only solution, but I'm not sure I've seen this stated anywhere.
real-analysis fourier-analysis
edited Jan 5 at 22:52
Bernard
123k741117
asked Jan 5 at 22:42
John PJohn P
55727
$endgroup$
1
$begingroup$
Fourier series. They're linearly independent, even with infinitely many terms.
$endgroup$
– Matt Samuel
Jan 5 at 23:07
$begingroup$
But what justifies exchanging the integral and sum order?
$endgroup$
– John P
Jan 6 at 14:31
add a comment |
$begingroup$
Working in $mathbb{R}$, what sequences $a_n$ satisfy $sum_{n=0}^{infty} a_n sin(nx)=0$ for all $x$, pointwise ?
I've never thought about this and I'm not sure whether I'm not missing something trivial - I would have expected that $a_n = 0$ for all $n$ is the only solution, but I'm not sure I've seen this stated anywhere.
real-analysis fourier-analysis
edited Jan 5 at 22:52
Bernard
123k741117
asked Jan 5 at 22:42
John PJohn P
55727
$endgroup$
Working in $mathbb{R}$, what sequences $a_n$ satisfy $sum_{n=0}^{infty} a_n sin(nx)=0$ for all $x$, pointwise ?
I've never thought about this and I'm not sure whether I'm not missing something trivial - I would have expected that $a_n = 0$ for all $n$ is the only solution, but I'm not sure I've seen this stated anywhere.
real-analysis fourier-analysis
real-analysis fourier-analysis
edited Jan 5 at 22:52
Bernard
123k741117
asked Jan 5 at 22:42
John PJohn P
55727
edited Jan 5 at 22:52
Bernard
123k741117
asked Jan 5 at 22:42
John PJohn P
55727
edited Jan 5 at 22:52
Bernard
123k741117
edited Jan 5 at 22:52
Bernard
123k741117
edited Jan 5 at 22:52
Bernard
123k741117
123k741117
asked Jan 5 at 22:42
John PJohn P
55727
asked Jan 5 at 22:42
John PJohn P
55727
asked Jan 5 at 22:42
John PJohn P
55727
55727
1
$begingroup$
Fourier series. They're linearly independent, even with infinitely many terms.
$endgroup$
– Matt Samuel
Jan 5 at 23:07
$begingroup$
But what justifies exchanging the integral and sum order?
$endgroup$
– John P
Jan 6 at 14:31
add a comment |
1
$begingroup$
Fourier series. They're linearly independent, even with infinitely many terms.
$endgroup$
– Matt Samuel
Jan 5 at 23:07
$begingroup$
But what justifies exchanging the integral and sum order?
$endgroup$
– John P
Jan 6 at 14:31
1
1
$begingroup$
Fourier series. They're linearly independent, even with infinitely many terms.
$endgroup$
– Matt Samuel
Jan 5 at 23:07
$begingroup$
Fourier series. They're linearly independent, even with infinitely many terms.
$endgroup$
– Matt Samuel
Jan 5 at 23:07
$begingroup$
But what justifies exchanging the integral and sum order?
$endgroup$
– John P
Jan 6 at 14:31
$begingroup$
But what justifies exchanging the integral and sum order?
$endgroup$
– John P
Jan 6 at 14:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Riemann's uniqueness theorem says that if a trigonometric series converges pointwise to $0$, then its coefficients are all equal to $0$. This was later generalized by Cantor, leading him to the development of set theory.
answered Jan 6 at 17:54
Julián AguirreJulián Aguirre
69.4k24197
$endgroup$
add a comment |
$begingroup$
It may depend on the meaning of the summation (and, as a consequence, the speed of the decay of $a$).
To solve all convergence problems, we assume that the series converges normally on $mathbb{R}$, ie $a in ell^1$.
Let then $f(x)=sum_n{a_nsin(nx)}$. $f$ is continuous.
I suggest that you compute $int_0^{2pi}{f(x)sin(nx),dx}$ as a function of $a_n$ (using the normal convergence of the series).
answered Jan 5 at 23:08
MindlackMindlack
4,920211
$endgroup$
$begingroup$
I'm sorry but I have no idea what sort of approach to take here - exchanging the integral and sum signs would of course solve the problem, but I don't see how the use of this is justified. Uniform convergence would be enough, but I don't see how $sum a_n sin(nx)=0$ pointwise implies that this convergence is necessarily uniform. The general Lebesgue theorem I know concerning this requires some sort of bound on absolute value of the partial sums, or for example that $sum_n int |f_n| < infty$.
$endgroup$
– John P
Jan 6 at 14:29
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Riemann's uniqueness theorem says that if a trigonometric series converges pointwise to $0$, then its coefficients are all equal to $0$. This was later generalized by Cantor, leading him to the development of set theory.
answered Jan 6 at 17:54
Julián AguirreJulián Aguirre
69.4k24197
$endgroup$
add a comment |
$begingroup$
Riemann's uniqueness theorem says that if a trigonometric series converges pointwise to $0$, then its coefficients are all equal to $0$. This was later generalized by Cantor, leading him to the development of set theory.
answered Jan 6 at 17:54
Julián AguirreJulián Aguirre
69.4k24197
$endgroup$
add a comment |
$begingroup$
Riemann's uniqueness theorem says that if a trigonometric series converges pointwise to $0$, then its coefficients are all equal to $0$. This was later generalized by Cantor, leading him to the development of set theory.
answered Jan 6 at 17:54
Julián AguirreJulián Aguirre
69.4k24197
$endgroup$
Riemann's uniqueness theorem says that if a trigonometric series converges pointwise to $0$, then its coefficients are all equal to $0$. This was later generalized by Cantor, leading him to the development of set theory.
answered Jan 6 at 17:54
Julián AguirreJulián Aguirre
69.4k24197
answered Jan 6 at 17:54
Julián AguirreJulián Aguirre
69.4k24197
answered Jan 6 at 17:54
Julián AguirreJulián Aguirre
69.4k24197
answered Jan 6 at 17:54
Julián AguirreJulián Aguirre
69.4k24197
69.4k24197
add a comment |
add a comment |
$begingroup$
It may depend on the meaning of the summation (and, as a consequence, the speed of the decay of $a$).
To solve all convergence problems, we assume that the series converges normally on $mathbb{R}$, ie $a in ell^1$.
Let then $f(x)=sum_n{a_nsin(nx)}$. $f$ is continuous.
I suggest that you compute $int_0^{2pi}{f(x)sin(nx),dx}$ as a function of $a_n$ (using the normal convergence of the series).
answered Jan 5 at 23:08
MindlackMindlack
4,920211
$endgroup$
$begingroup$
I'm sorry but I have no idea what sort of approach to take here - exchanging the integral and sum signs would of course solve the problem, but I don't see how the use of this is justified. Uniform convergence would be enough, but I don't see how $sum a_n sin(nx)=0$ pointwise implies that this convergence is necessarily uniform. The general Lebesgue theorem I know concerning this requires some sort of bound on absolute value of the partial sums, or for example that $sum_n int |f_n| < infty$.
$endgroup$
– John P
Jan 6 at 14:29
add a comment |
$begingroup$
It may depend on the meaning of the summation (and, as a consequence, the speed of the decay of $a$).
To solve all convergence problems, we assume that the series converges normally on $mathbb{R}$, ie $a in ell^1$.
Let then $f(x)=sum_n{a_nsin(nx)}$. $f$ is continuous.
I suggest that you compute $int_0^{2pi}{f(x)sin(nx),dx}$ as a function of $a_n$ (using the normal convergence of the series).
answered Jan 5 at 23:08
MindlackMindlack
4,920211
$endgroup$
$begingroup$
I'm sorry but I have no idea what sort of approach to take here - exchanging the integral and sum signs would of course solve the problem, but I don't see how the use of this is justified. Uniform convergence would be enough, but I don't see how $sum a_n sin(nx)=0$ pointwise implies that this convergence is necessarily uniform. The general Lebesgue theorem I know concerning this requires some sort of bound on absolute value of the partial sums, or for example that $sum_n int |f_n| < infty$.
$endgroup$
– John P
Jan 6 at 14:29
add a comment |
$begingroup$
It may depend on the meaning of the summation (and, as a consequence, the speed of the decay of $a$).
To solve all convergence problems, we assume that the series converges normally on $mathbb{R}$, ie $a in ell^1$.
Let then $f(x)=sum_n{a_nsin(nx)}$. $f$ is continuous.
I suggest that you compute $int_0^{2pi}{f(x)sin(nx),dx}$ as a function of $a_n$ (using the normal convergence of the series).
answered Jan 5 at 23:08
MindlackMindlack
4,920211
$endgroup$
It may depend on the meaning of the summation (and, as a consequence, the speed of the decay of $a$).
To solve all convergence problems, we assume that the series converges normally on $mathbb{R}$, ie $a in ell^1$.
Let then $f(x)=sum_n{a_nsin(nx)}$. $f$ is continuous.
I suggest that you compute $int_0^{2pi}{f(x)sin(nx),dx}$ as a function of $a_n$ (using the normal convergence of the series).
answered Jan 5 at 23:08
MindlackMindlack
4,920211
answered Jan 5 at 23:08
MindlackMindlack
4,920211
answered Jan 5 at 23:08
MindlackMindlack
4,920211
answered Jan 5 at 23:08
MindlackMindlack
4,920211
4,920211
$begingroup$
I'm sorry but I have no idea what sort of approach to take here - exchanging the integral and sum signs would of course solve the problem, but I don't see how the use of this is justified. Uniform convergence would be enough, but I don't see how $sum a_n sin(nx)=0$ pointwise implies that this convergence is necessarily uniform. The general Lebesgue theorem I know concerning this requires some sort of bound on absolute value of the partial sums, or for example that $sum_n int |f_n| < infty$.
$endgroup$
– John P
Jan 6 at 14:29
add a comment |
$begingroup$
I'm sorry but I have no idea what sort of approach to take here - exchanging the integral and sum signs would of course solve the problem, but I don't see how the use of this is justified. Uniform convergence would be enough, but I don't see how $sum a_n sin(nx)=0$ pointwise implies that this convergence is necessarily uniform. The general Lebesgue theorem I know concerning this requires some sort of bound on absolute value of the partial sums, or for example that $sum_n int |f_n| < infty$.
$endgroup$
– John P
Jan 6 at 14:29
$begingroup$
I'm sorry but I have no idea what sort of approach to take here - exchanging the integral and sum signs would of course solve the problem, but I don't see how the use of this is justified. Uniform convergence would be enough, but I don't see how $sum a_n sin(nx)=0$ pointwise implies that this convergence is necessarily uniform. The general Lebesgue theorem I know concerning this requires some sort of bound on absolute value of the partial sums, or for example that $sum_n int |f_n| < infty$.
$endgroup$
– John P
Jan 6 at 14:29
$begingroup$
I'm sorry but I have no idea what sort of approach to take here - exchanging the integral and sum signs would of course solve the problem, but I don't see how the use of this is justified. Uniform convergence would be enough, but I don't see how $sum a_n sin(nx)=0$ pointwise implies that this convergence is necessarily uniform. The general Lebesgue theorem I know concerning this requires some sort of bound on absolute value of the partial sums, or for example that $sum_n int |f_n| < infty$.
$endgroup$
– John P
Jan 6 at 14:29
add a comment |
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1
$begingroup$
Fourier series. They're linearly independent, even with infinitely many terms.
$endgroup$
– Matt Samuel
Jan 5 at 23:07
$begingroup$
But what justifies exchanging the integral and sum order?
$endgroup$
– John P
Jan 6 at 14:31