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Lissajous Curve dense in a Volume
Clash Royale CLAN TAG#URR8PPP
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The Lissajous-type curve $left(sin(omega_1 t), sin(omega_2 t)right)$, with $t in mathrmR$, is dense in a certain region of the plane.
This can be seen for instance from excellent answers to a previous question of mine: Curve dense inside the unit circle.
Consider now the curve $left(sin(omega_1 t), sin(omega_2 t), sin(omega_3 t)right)$ with $t in mathrmR$, and assume that the three frequencies are incommensurable.
Is this curve dense in some 3-dimensional region of finite volume? Or is it also dense just inside a 2-dimensional surface?
$textbfADDED QUESTION$:
And how about a similar curve in four dimensions (four sines with incommensurate frequencies); would it be dense in a 4-dimensional region of space? If so, how to prove it?
real-analysis analysis fourier-analysis fourier-series recreational-mathematics
asked Aug 3 at 5:48
Jennifer
371313
add a comment |Â
up vote
0
down vote
favorite
The Lissajous-type curve $left(sin(omega_1 t), sin(omega_2 t)right)$, with $t in mathrmR$, is dense in a certain region of the plane.
This can be seen for instance from excellent answers to a previous question of mine: Curve dense inside the unit circle.
Consider now the curve $left(sin(omega_1 t), sin(omega_2 t), sin(omega_3 t)right)$ with $t in mathrmR$, and assume that the three frequencies are incommensurable.
Is this curve dense in some 3-dimensional region of finite volume? Or is it also dense just inside a 2-dimensional surface?
$textbfADDED QUESTION$:
And how about a similar curve in four dimensions (four sines with incommensurate frequencies); would it be dense in a 4-dimensional region of space? If so, how to prove it?
real-analysis analysis fourier-analysis fourier-series recreational-mathematics
asked Aug 3 at 5:48
Jennifer
371313
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The Lissajous-type curve $left(sin(omega_1 t), sin(omega_2 t)right)$, with $t in mathrmR$, is dense in a certain region of the plane.
This can be seen for instance from excellent answers to a previous question of mine: Curve dense inside the unit circle.
Consider now the curve $left(sin(omega_1 t), sin(omega_2 t), sin(omega_3 t)right)$ with $t in mathrmR$, and assume that the three frequencies are incommensurable.
Is this curve dense in some 3-dimensional region of finite volume? Or is it also dense just inside a 2-dimensional surface?
$textbfADDED QUESTION$:
And how about a similar curve in four dimensions (four sines with incommensurate frequencies); would it be dense in a 4-dimensional region of space? If so, how to prove it?
real-analysis analysis fourier-analysis fourier-series recreational-mathematics
asked Aug 3 at 5:48
Jennifer
371313
The Lissajous-type curve $left(sin(omega_1 t), sin(omega_2 t)right)$, with $t in mathrmR$, is dense in a certain region of the plane.
This can be seen for instance from excellent answers to a previous question of mine: Curve dense inside the unit circle.
Consider now the curve $left(sin(omega_1 t), sin(omega_2 t), sin(omega_3 t)right)$ with $t in mathrmR$, and assume that the three frequencies are incommensurable.
Is this curve dense in some 3-dimensional region of finite volume? Or is it also dense just inside a 2-dimensional surface?
$textbfADDED QUESTION$:
And how about a similar curve in four dimensions (four sines with incommensurate frequencies); would it be dense in a 4-dimensional region of space? If so, how to prove it?
real-analysis analysis fourier-analysis fourier-series recreational-mathematics
asked Aug 3 at 5:48
Jennifer
371313
edited Aug 3 at 14:42
asked Aug 3 at 5:48
Jennifer
371313
asked Aug 3 at 5:48
Jennifer
371313
asked Aug 3 at 5:48
Jennifer
371313
371313
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
"fill"... Since the curve is continuously differentiable its image must have enmpty interior. Regarding whether the image can be dense in a non-empty open set, (pairwise) incommensurable is not quite the right condition. In fact
The image is dense in $[-1,1]^3$ if and only if $omega_1$, $omega_2$, $omega_3$ are rationally independent.
Meaning that $r_1omega_1+r_2omega_2+r_3omega_3=0$, $r_j$ rational implies $r_1=r_2=r_3=0$.
If the $omega_j$ are not rationally independent there exist integers $n_j$ with $sum n_jomega_j=0$ and not all $n_j$ vanish. I think it's clear that this implies the curve is not dense, although I haven't written a formal proof.
Define $$gamma(t)=(e^iomega_1t,e^iomega_2t,e^iomega_3t).$$
Say $Bbb T=zinBbb C:$. It is clear that if the $omega_j$ are rationally dependent then $gamma$ is not dense in $Bbb T^3$.
On the other hand, if the $omega_j$ are rationally independent then a standard elegant argument using Fourier series shows that $gamma$ is dense in $Bbb T^3$ (hence your curve is dense in $[-1,1]^3$). Sketch:
Suppose the $omega_j$ are rationaly independent. Consider the equation $$left(frac12piint_0^2piright)^3f(e^it_1,e^it_2,e^it_3),dt_1dt_2dt_3=lim_Ttoinftyint_0^Tf(gamma(t)),dt,tag*$$which may or may not hold for a given $fin C(Bbb T^3)$. Show that $(*)$ holds if $f$ is a character, that is $$f(e^it_1,e^it_2,e^it_3)=e^i(n_1t_1+n_2t_2+n_3t_3)$$for some integers $n_j$. (Hint: If $n_1=n_2=n_3=0$ both sides equal $1$; if not, the fact that $sum n_jomega_jne0$ shows that both sides equal $0$.)
Hence $(*)$ holds if $f$ is a trigonometric polynomial (a linear combination of characters). Since the trigonometric polynomials are dense in $C(Bbb T^3)$ it follows that $(*)$ holds for every $fin C(Bbb T^3)$, hence $gamma$ must be dense in $Bbb T^3$.
answered Aug 3 at 17:04
David C. Ullrich
53.8k33481
add a comment |Â
up vote
1
down vote
Let $a=sqrt2$, $b=sqrt3$, $c=sqrt5$.
ParametricPlot3D[Sin[a t],Sin[b t],Sin[c t],t,-100,100]
The curve will not fill a surface or a volume. No differentiable curve can do that.
Answer to the edited question
There are cases where the image is not dense. The following follows this post in Wolfram Blog.
Let $phi$ be the golden ratio. Then the image of the curve $(sin t,sin(phi,t),sin(phi^2,t))$ is contained in the surface
$$
(2 - x^2 - y^2 - z^2)^2 = 4 (1 - x^2) (1 - y^2) (1 - z^2).
$$
answered Aug 3 at 9:47
Julián Aguirre
64.4k23894
Thank you!, but posting the image is not enough to prove that a curve is dense in a volume. It looks dense, but a proof is a different thing!
– Jennifer
Aug 3 at 10:53
When I wrote the answer, there was no mention to density, unless my memory fails.
– Julián Aguirre
Aug 3 at 13:24
You are right, I didn't word it properly. And your figure is helpful because it suggests the denseness.
– Jennifer
Aug 3 at 13:40
I'm actually particularly interested in the case where the frequencies are square roots of prime numbers. Numerically it seems that the image is dense in a volume.
– Jennifer
Aug 3 at 14:12
I suspect that the image will be dense "for almost all" choices.
– Julián Aguirre
Aug 3 at 14:19
 |Â
show 3 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
"fill"... Since the curve is continuously differentiable its image must have enmpty interior. Regarding whether the image can be dense in a non-empty open set, (pairwise) incommensurable is not quite the right condition. In fact
The image is dense in $[-1,1]^3$ if and only if $omega_1$, $omega_2$, $omega_3$ are rationally independent.
Meaning that $r_1omega_1+r_2omega_2+r_3omega_3=0$, $r_j$ rational implies $r_1=r_2=r_3=0$.
If the $omega_j$ are not rationally independent there exist integers $n_j$ with $sum n_jomega_j=0$ and not all $n_j$ vanish. I think it's clear that this implies the curve is not dense, although I haven't written a formal proof.
Define $$gamma(t)=(e^iomega_1t,e^iomega_2t,e^iomega_3t).$$
Say $Bbb T=zinBbb C:$. It is clear that if the $omega_j$ are rationally dependent then $gamma$ is not dense in $Bbb T^3$.
On the other hand, if the $omega_j$ are rationally independent then a standard elegant argument using Fourier series shows that $gamma$ is dense in $Bbb T^3$ (hence your curve is dense in $[-1,1]^3$). Sketch:
Suppose the $omega_j$ are rationaly independent. Consider the equation $$left(frac12piint_0^2piright)^3f(e^it_1,e^it_2,e^it_3),dt_1dt_2dt_3=lim_Ttoinftyint_0^Tf(gamma(t)),dt,tag*$$which may or may not hold for a given $fin C(Bbb T^3)$. Show that $(*)$ holds if $f$ is a character, that is $$f(e^it_1,e^it_2,e^it_3)=e^i(n_1t_1+n_2t_2+n_3t_3)$$for some integers $n_j$. (Hint: If $n_1=n_2=n_3=0$ both sides equal $1$; if not, the fact that $sum n_jomega_jne0$ shows that both sides equal $0$.)
Hence $(*)$ holds if $f$ is a trigonometric polynomial (a linear combination of characters). Since the trigonometric polynomials are dense in $C(Bbb T^3)$ it follows that $(*)$ holds for every $fin C(Bbb T^3)$, hence $gamma$ must be dense in $Bbb T^3$.
answered Aug 3 at 17:04
David C. Ullrich
53.8k33481
add a comment |Â
up vote
2
down vote
accepted
"fill"... Since the curve is continuously differentiable its image must have enmpty interior. Regarding whether the image can be dense in a non-empty open set, (pairwise) incommensurable is not quite the right condition. In fact
The image is dense in $[-1,1]^3$ if and only if $omega_1$, $omega_2$, $omega_3$ are rationally independent.
Meaning that $r_1omega_1+r_2omega_2+r_3omega_3=0$, $r_j$ rational implies $r_1=r_2=r_3=0$.
If the $omega_j$ are not rationally independent there exist integers $n_j$ with $sum n_jomega_j=0$ and not all $n_j$ vanish. I think it's clear that this implies the curve is not dense, although I haven't written a formal proof.
Define $$gamma(t)=(e^iomega_1t,e^iomega_2t,e^iomega_3t).$$
Say $Bbb T=zinBbb C:$. It is clear that if the $omega_j$ are rationally dependent then $gamma$ is not dense in $Bbb T^3$.
On the other hand, if the $omega_j$ are rationally independent then a standard elegant argument using Fourier series shows that $gamma$ is dense in $Bbb T^3$ (hence your curve is dense in $[-1,1]^3$). Sketch:
Suppose the $omega_j$ are rationaly independent. Consider the equation $$left(frac12piint_0^2piright)^3f(e^it_1,e^it_2,e^it_3),dt_1dt_2dt_3=lim_Ttoinftyint_0^Tf(gamma(t)),dt,tag*$$which may or may not hold for a given $fin C(Bbb T^3)$. Show that $(*)$ holds if $f$ is a character, that is $$f(e^it_1,e^it_2,e^it_3)=e^i(n_1t_1+n_2t_2+n_3t_3)$$for some integers $n_j$. (Hint: If $n_1=n_2=n_3=0$ both sides equal $1$; if not, the fact that $sum n_jomega_jne0$ shows that both sides equal $0$.)
Hence $(*)$ holds if $f$ is a trigonometric polynomial (a linear combination of characters). Since the trigonometric polynomials are dense in $C(Bbb T^3)$ it follows that $(*)$ holds for every $fin C(Bbb T^3)$, hence $gamma$ must be dense in $Bbb T^3$.
answered Aug 3 at 17:04
David C. Ullrich
53.8k33481
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
"fill"... Since the curve is continuously differentiable its image must have enmpty interior. Regarding whether the image can be dense in a non-empty open set, (pairwise) incommensurable is not quite the right condition. In fact
The image is dense in $[-1,1]^3$ if and only if $omega_1$, $omega_2$, $omega_3$ are rationally independent.
Meaning that $r_1omega_1+r_2omega_2+r_3omega_3=0$, $r_j$ rational implies $r_1=r_2=r_3=0$.
If the $omega_j$ are not rationally independent there exist integers $n_j$ with $sum n_jomega_j=0$ and not all $n_j$ vanish. I think it's clear that this implies the curve is not dense, although I haven't written a formal proof.
Define $$gamma(t)=(e^iomega_1t,e^iomega_2t,e^iomega_3t).$$
Say $Bbb T=zinBbb C:$. It is clear that if the $omega_j$ are rationally dependent then $gamma$ is not dense in $Bbb T^3$.
On the other hand, if the $omega_j$ are rationally independent then a standard elegant argument using Fourier series shows that $gamma$ is dense in $Bbb T^3$ (hence your curve is dense in $[-1,1]^3$). Sketch:
Suppose the $omega_j$ are rationaly independent. Consider the equation $$left(frac12piint_0^2piright)^3f(e^it_1,e^it_2,e^it_3),dt_1dt_2dt_3=lim_Ttoinftyint_0^Tf(gamma(t)),dt,tag*$$which may or may not hold for a given $fin C(Bbb T^3)$. Show that $(*)$ holds if $f$ is a character, that is $$f(e^it_1,e^it_2,e^it_3)=e^i(n_1t_1+n_2t_2+n_3t_3)$$for some integers $n_j$. (Hint: If $n_1=n_2=n_3=0$ both sides equal $1$; if not, the fact that $sum n_jomega_jne0$ shows that both sides equal $0$.)
Hence $(*)$ holds if $f$ is a trigonometric polynomial (a linear combination of characters). Since the trigonometric polynomials are dense in $C(Bbb T^3)$ it follows that $(*)$ holds for every $fin C(Bbb T^3)$, hence $gamma$ must be dense in $Bbb T^3$.
answered Aug 3 at 17:04
David C. Ullrich
53.8k33481
"fill"... Since the curve is continuously differentiable its image must have enmpty interior. Regarding whether the image can be dense in a non-empty open set, (pairwise) incommensurable is not quite the right condition. In fact
The image is dense in $[-1,1]^3$ if and only if $omega_1$, $omega_2$, $omega_3$ are rationally independent.
Meaning that $r_1omega_1+r_2omega_2+r_3omega_3=0$, $r_j$ rational implies $r_1=r_2=r_3=0$.
If the $omega_j$ are not rationally independent there exist integers $n_j$ with $sum n_jomega_j=0$ and not all $n_j$ vanish. I think it's clear that this implies the curve is not dense, although I haven't written a formal proof.
Define $$gamma(t)=(e^iomega_1t,e^iomega_2t,e^iomega_3t).$$
Say $Bbb T=zinBbb C:$. It is clear that if the $omega_j$ are rationally dependent then $gamma$ is not dense in $Bbb T^3$.
On the other hand, if the $omega_j$ are rationally independent then a standard elegant argument using Fourier series shows that $gamma$ is dense in $Bbb T^3$ (hence your curve is dense in $[-1,1]^3$). Sketch:
Suppose the $omega_j$ are rationaly independent. Consider the equation $$left(frac12piint_0^2piright)^3f(e^it_1,e^it_2,e^it_3),dt_1dt_2dt_3=lim_Ttoinftyint_0^Tf(gamma(t)),dt,tag*$$which may or may not hold for a given $fin C(Bbb T^3)$. Show that $(*)$ holds if $f$ is a character, that is $$f(e^it_1,e^it_2,e^it_3)=e^i(n_1t_1+n_2t_2+n_3t_3)$$for some integers $n_j$. (Hint: If $n_1=n_2=n_3=0$ both sides equal $1$; if not, the fact that $sum n_jomega_jne0$ shows that both sides equal $0$.)
Hence $(*)$ holds if $f$ is a trigonometric polynomial (a linear combination of characters). Since the trigonometric polynomials are dense in $C(Bbb T^3)$ it follows that $(*)$ holds for every $fin C(Bbb T^3)$, hence $gamma$ must be dense in $Bbb T^3$.
answered Aug 3 at 17:04
David C. Ullrich
53.8k33481
edited Aug 3 at 17:09
answered Aug 3 at 17:04
David C. Ullrich
53.8k33481
answered Aug 3 at 17:04
David C. Ullrich
53.8k33481
answered Aug 3 at 17:04
David C. Ullrich
53.8k33481
53.8k33481
add a comment |Â
add a comment |Â
up vote
1
down vote
Let $a=sqrt2$, $b=sqrt3$, $c=sqrt5$.
ParametricPlot3D[Sin[a t],Sin[b t],Sin[c t],t,-100,100]
The curve will not fill a surface or a volume. No differentiable curve can do that.
Answer to the edited question
There are cases where the image is not dense. The following follows this post in Wolfram Blog.
Let $phi$ be the golden ratio. Then the image of the curve $(sin t,sin(phi,t),sin(phi^2,t))$ is contained in the surface
$$
(2 - x^2 - y^2 - z^2)^2 = 4 (1 - x^2) (1 - y^2) (1 - z^2).
$$
answered Aug 3 at 9:47
Julián Aguirre
64.4k23894
Thank you!, but posting the image is not enough to prove that a curve is dense in a volume. It looks dense, but a proof is a different thing!
– Jennifer
Aug 3 at 10:53
When I wrote the answer, there was no mention to density, unless my memory fails.
– Julián Aguirre
Aug 3 at 13:24
You are right, I didn't word it properly. And your figure is helpful because it suggests the denseness.
– Jennifer
Aug 3 at 13:40
I'm actually particularly interested in the case where the frequencies are square roots of prime numbers. Numerically it seems that the image is dense in a volume.
– Jennifer
Aug 3 at 14:12
I suspect that the image will be dense "for almost all" choices.
– Julián Aguirre
Aug 3 at 14:19
 |Â
show 3 more comments
up vote
1
down vote
Let $a=sqrt2$, $b=sqrt3$, $c=sqrt5$.
ParametricPlot3D[Sin[a t],Sin[b t],Sin[c t],t,-100,100]
The curve will not fill a surface or a volume. No differentiable curve can do that.
Answer to the edited question
There are cases where the image is not dense. The following follows this post in Wolfram Blog.
Let $phi$ be the golden ratio. Then the image of the curve $(sin t,sin(phi,t),sin(phi^2,t))$ is contained in the surface
$$
(2 - x^2 - y^2 - z^2)^2 = 4 (1 - x^2) (1 - y^2) (1 - z^2).
$$
answered Aug 3 at 9:47
Julián Aguirre
64.4k23894
Thank you!, but posting the image is not enough to prove that a curve is dense in a volume. It looks dense, but a proof is a different thing!
– Jennifer
Aug 3 at 10:53
When I wrote the answer, there was no mention to density, unless my memory fails.
– Julián Aguirre
Aug 3 at 13:24
You are right, I didn't word it properly. And your figure is helpful because it suggests the denseness.
– Jennifer
Aug 3 at 13:40
I'm actually particularly interested in the case where the frequencies are square roots of prime numbers. Numerically it seems that the image is dense in a volume.
– Jennifer
Aug 3 at 14:12
I suspect that the image will be dense "for almost all" choices.
– Julián Aguirre
Aug 3 at 14:19
 |Â
show 3 more comments
up vote
1
down vote
up vote
1
down vote
Let $a=sqrt2$, $b=sqrt3$, $c=sqrt5$.
ParametricPlot3D[Sin[a t],Sin[b t],Sin[c t],t,-100,100]
The curve will not fill a surface or a volume. No differentiable curve can do that.
Answer to the edited question
There are cases where the image is not dense. The following follows this post in Wolfram Blog.
Let $phi$ be the golden ratio. Then the image of the curve $(sin t,sin(phi,t),sin(phi^2,t))$ is contained in the surface
$$
(2 - x^2 - y^2 - z^2)^2 = 4 (1 - x^2) (1 - y^2) (1 - z^2).
$$
answered Aug 3 at 9:47
Julián Aguirre
64.4k23894
Let $a=sqrt2$, $b=sqrt3$, $c=sqrt5$.
ParametricPlot3D[Sin[a t],Sin[b t],Sin[c t],t,-100,100]
The curve will not fill a surface or a volume. No differentiable curve can do that.
Answer to the edited question
There are cases where the image is not dense. The following follows this post in Wolfram Blog.
Let $phi$ be the golden ratio. Then the image of the curve $(sin t,sin(phi,t),sin(phi^2,t))$ is contained in the surface
$$
(2 - x^2 - y^2 - z^2)^2 = 4 (1 - x^2) (1 - y^2) (1 - z^2).
$$
answered Aug 3 at 9:47
Julián Aguirre
64.4k23894
edited Aug 3 at 13:57
answered Aug 3 at 9:47
Julián Aguirre
64.4k23894
answered Aug 3 at 9:47
Julián Aguirre
64.4k23894
answered Aug 3 at 9:47
Julián Aguirre
64.4k23894
64.4k23894
Thank you!, but posting the image is not enough to prove that a curve is dense in a volume. It looks dense, but a proof is a different thing!
– Jennifer
Aug 3 at 10:53
When I wrote the answer, there was no mention to density, unless my memory fails.
– Julián Aguirre
Aug 3 at 13:24
You are right, I didn't word it properly. And your figure is helpful because it suggests the denseness.
– Jennifer
Aug 3 at 13:40
I'm actually particularly interested in the case where the frequencies are square roots of prime numbers. Numerically it seems that the image is dense in a volume.
– Jennifer
Aug 3 at 14:12
I suspect that the image will be dense "for almost all" choices.
– Julián Aguirre
Aug 3 at 14:19
 |Â
show 3 more comments
Thank you!, but posting the image is not enough to prove that a curve is dense in a volume. It looks dense, but a proof is a different thing!
– Jennifer
Aug 3 at 10:53
When I wrote the answer, there was no mention to density, unless my memory fails.
– Julián Aguirre
Aug 3 at 13:24
You are right, I didn't word it properly. And your figure is helpful because it suggests the denseness.
– Jennifer
Aug 3 at 13:40
I'm actually particularly interested in the case where the frequencies are square roots of prime numbers. Numerically it seems that the image is dense in a volume.
– Jennifer
Aug 3 at 14:12
I suspect that the image will be dense "for almost all" choices.
– Julián Aguirre
Aug 3 at 14:19
Thank you!, but posting the image is not enough to prove that a curve is dense in a volume. It looks dense, but a proof is a different thing!
– Jennifer
Aug 3 at 10:53
Thank you!, but posting the image is not enough to prove that a curve is dense in a volume. It looks dense, but a proof is a different thing!
– Jennifer
Aug 3 at 10:53
When I wrote the answer, there was no mention to density, unless my memory fails.
– Julián Aguirre
Aug 3 at 13:24
When I wrote the answer, there was no mention to density, unless my memory fails.
– Julián Aguirre
Aug 3 at 13:24
You are right, I didn't word it properly. And your figure is helpful because it suggests the denseness.
– Jennifer
Aug 3 at 13:40
You are right, I didn't word it properly. And your figure is helpful because it suggests the denseness.
– Jennifer
Aug 3 at 13:40
I'm actually particularly interested in the case where the frequencies are square roots of prime numbers. Numerically it seems that the image is dense in a volume.
– Jennifer
Aug 3 at 14:12
I'm actually particularly interested in the case where the frequencies are square roots of prime numbers. Numerically it seems that the image is dense in a volume.
– Jennifer
Aug 3 at 14:12
I suspect that the image will be dense "for almost all" choices.
– Julián Aguirre
Aug 3 at 14:19
I suspect that the image will be dense "for almost all" choices.
– Julián Aguirre
Aug 3 at 14:19
 |Â
show 3 more comments
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