“Evenly” dense orbit?
$begingroup$
I want to prove the following: let $a$ be an irrational constant and $m$ an integer. Then
$$lim_{n toinfty} frac{1}{n}sum_{k=0}^{n-1}e^{2pi m i (x+ka)} = begin{cases} 0, & mnot=0 \ 1, & m=0 end{cases}$$
for any $x$. Intuitively I can understand this result: because $a$ is irrational, the points $x+ka$ mod $1$ are dense in $[0,1]$. So the sum on the left hand side is just the average of the values of $e^{2pi m i x}$ on the interval $[0,1]$. And this average is of course $0$ (both the real and imaginary parts are just sine waves), except when $m=0$ and thus $e^{2pi m i x} = 1$.
However, I am having difficulty formalizing the idea. In particular, here is something that troubles me: I know that the points $x+ka$ mod $1$ are dense in $[0,1]$, but how do I prove that they are "evenly distributed" across the interval, so that we get an unweighted average?
dynamical-systems irrational-numbers
edited Dec 14 '18 at 21:04
Adam
1,1951919
asked Dec 14 '14 at 3:20
AndreaAndrea
754823
$endgroup$
add a comment |
$begingroup$
I want to prove the following: let $a$ be an irrational constant and $m$ an integer. Then
$$lim_{n toinfty} frac{1}{n}sum_{k=0}^{n-1}e^{2pi m i (x+ka)} = begin{cases} 0, & mnot=0 \ 1, & m=0 end{cases}$$
for any $x$. Intuitively I can understand this result: because $a$ is irrational, the points $x+ka$ mod $1$ are dense in $[0,1]$. So the sum on the left hand side is just the average of the values of $e^{2pi m i x}$ on the interval $[0,1]$. And this average is of course $0$ (both the real and imaginary parts are just sine waves), except when $m=0$ and thus $e^{2pi m i x} = 1$.
However, I am having difficulty formalizing the idea. In particular, here is something that troubles me: I know that the points $x+ka$ mod $1$ are dense in $[0,1]$, but how do I prove that they are "evenly distributed" across the interval, so that we get an unweighted average?
dynamical-systems irrational-numbers
edited Dec 14 '18 at 21:04
Adam
1,1951919
asked Dec 14 '14 at 3:20
AndreaAndrea
754823
$endgroup$
$begingroup$
What is the source of this?
$endgroup$
– Will Jagy
Dec 14 '14 at 3:51
add a comment |
$begingroup$
I want to prove the following: let $a$ be an irrational constant and $m$ an integer. Then
$$lim_{n toinfty} frac{1}{n}sum_{k=0}^{n-1}e^{2pi m i (x+ka)} = begin{cases} 0, & mnot=0 \ 1, & m=0 end{cases}$$
for any $x$. Intuitively I can understand this result: because $a$ is irrational, the points $x+ka$ mod $1$ are dense in $[0,1]$. So the sum on the left hand side is just the average of the values of $e^{2pi m i x}$ on the interval $[0,1]$. And this average is of course $0$ (both the real and imaginary parts are just sine waves), except when $m=0$ and thus $e^{2pi m i x} = 1$.
However, I am having difficulty formalizing the idea. In particular, here is something that troubles me: I know that the points $x+ka$ mod $1$ are dense in $[0,1]$, but how do I prove that they are "evenly distributed" across the interval, so that we get an unweighted average?
dynamical-systems irrational-numbers
edited Dec 14 '18 at 21:04
Adam
1,1951919
asked Dec 14 '14 at 3:20
AndreaAndrea
754823
$endgroup$
I want to prove the following: let $a$ be an irrational constant and $m$ an integer. Then
$$lim_{n toinfty} frac{1}{n}sum_{k=0}^{n-1}e^{2pi m i (x+ka)} = begin{cases} 0, & mnot=0 \ 1, & m=0 end{cases}$$
for any $x$. Intuitively I can understand this result: because $a$ is irrational, the points $x+ka$ mod $1$ are dense in $[0,1]$. So the sum on the left hand side is just the average of the values of $e^{2pi m i x}$ on the interval $[0,1]$. And this average is of course $0$ (both the real and imaginary parts are just sine waves), except when $m=0$ and thus $e^{2pi m i x} = 1$.
However, I am having difficulty formalizing the idea. In particular, here is something that troubles me: I know that the points $x+ka$ mod $1$ are dense in $[0,1]$, but how do I prove that they are "evenly distributed" across the interval, so that we get an unweighted average?
dynamical-systems irrational-numbers
dynamical-systems irrational-numbers
edited Dec 14 '18 at 21:04
Adam
1,1951919
asked Dec 14 '14 at 3:20
AndreaAndrea
754823
edited Dec 14 '18 at 21:04
Adam
1,1951919
asked Dec 14 '14 at 3:20
AndreaAndrea
754823
edited Dec 14 '18 at 21:04
Adam
1,1951919
edited Dec 14 '18 at 21:04
Adam
1,1951919
edited Dec 14 '18 at 21:04
Adam
1,1951919
1,1951919
asked Dec 14 '14 at 3:20
AndreaAndrea
754823
asked Dec 14 '14 at 3:20
AndreaAndrea
754823
asked Dec 14 '14 at 3:20
AndreaAndrea
754823
754823
$begingroup$
What is the source of this?
$endgroup$
– Will Jagy
Dec 14 '14 at 3:51
add a comment |
$begingroup$
What is the source of this?
$endgroup$
– Will Jagy
Dec 14 '14 at 3:51
$begingroup$
What is the source of this?
$endgroup$
– Will Jagy
Dec 14 '14 at 3:51
$begingroup$
What is the source of this?
$endgroup$
– Will Jagy
Dec 14 '14 at 3:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A good way to do this would be to look for cancellations. In particular, we can use the property that we can find a $n$ such that $na$ is arbitrarily close to $frac{1}2$. By this and continuity, we could choose $n$ such that $|e^{2pi i na}+1|<varepsilon$ for any $varepsilon>0$. Then we could take the sum:
$$sum_{n=0}^{2k-1}e^{2pi na}$$
and rearrange to:
$$sum_{n=0}^{k-1}e^{2pi i na} + e^{2pi i(n+k)a} $$
$$sum_{n=0}^{k-1}e^{2pi ina} (1+e^{2pi ka}) $$
where each term has absolute value less than $varepsilon$, and so the sum must have absolute value less than $kvarepsilon$. If you generalize this argument a little, you can prove the desired result.
answered Dec 14 '14 at 4:07
Milo BrandtMilo Brandt
39.9k476140
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A good way to do this would be to look for cancellations. In particular, we can use the property that we can find a $n$ such that $na$ is arbitrarily close to $frac{1}2$. By this and continuity, we could choose $n$ such that $|e^{2pi i na}+1|<varepsilon$ for any $varepsilon>0$. Then we could take the sum:
$$sum_{n=0}^{2k-1}e^{2pi na}$$
and rearrange to:
$$sum_{n=0}^{k-1}e^{2pi i na} + e^{2pi i(n+k)a} $$
$$sum_{n=0}^{k-1}e^{2pi ina} (1+e^{2pi ka}) $$
where each term has absolute value less than $varepsilon$, and so the sum must have absolute value less than $kvarepsilon$. If you generalize this argument a little, you can prove the desired result.
answered Dec 14 '14 at 4:07
Milo BrandtMilo Brandt
39.9k476140
$endgroup$
add a comment |
$begingroup$
A good way to do this would be to look for cancellations. In particular, we can use the property that we can find a $n$ such that $na$ is arbitrarily close to $frac{1}2$. By this and continuity, we could choose $n$ such that $|e^{2pi i na}+1|<varepsilon$ for any $varepsilon>0$. Then we could take the sum:
$$sum_{n=0}^{2k-1}e^{2pi na}$$
and rearrange to:
$$sum_{n=0}^{k-1}e^{2pi i na} + e^{2pi i(n+k)a} $$
$$sum_{n=0}^{k-1}e^{2pi ina} (1+e^{2pi ka}) $$
where each term has absolute value less than $varepsilon$, and so the sum must have absolute value less than $kvarepsilon$. If you generalize this argument a little, you can prove the desired result.
answered Dec 14 '14 at 4:07
Milo BrandtMilo Brandt
39.9k476140
$endgroup$
add a comment |
$begingroup$
A good way to do this would be to look for cancellations. In particular, we can use the property that we can find a $n$ such that $na$ is arbitrarily close to $frac{1}2$. By this and continuity, we could choose $n$ such that $|e^{2pi i na}+1|<varepsilon$ for any $varepsilon>0$. Then we could take the sum:
$$sum_{n=0}^{2k-1}e^{2pi na}$$
and rearrange to:
$$sum_{n=0}^{k-1}e^{2pi i na} + e^{2pi i(n+k)a} $$
$$sum_{n=0}^{k-1}e^{2pi ina} (1+e^{2pi ka}) $$
where each term has absolute value less than $varepsilon$, and so the sum must have absolute value less than $kvarepsilon$. If you generalize this argument a little, you can prove the desired result.
answered Dec 14 '14 at 4:07
Milo BrandtMilo Brandt
39.9k476140
$endgroup$
A good way to do this would be to look for cancellations. In particular, we can use the property that we can find a $n$ such that $na$ is arbitrarily close to $frac{1}2$. By this and continuity, we could choose $n$ such that $|e^{2pi i na}+1|<varepsilon$ for any $varepsilon>0$. Then we could take the sum:
$$sum_{n=0}^{2k-1}e^{2pi na}$$
and rearrange to:
$$sum_{n=0}^{k-1}e^{2pi i na} + e^{2pi i(n+k)a} $$
$$sum_{n=0}^{k-1}e^{2pi ina} (1+e^{2pi ka}) $$
where each term has absolute value less than $varepsilon$, and so the sum must have absolute value less than $kvarepsilon$. If you generalize this argument a little, you can prove the desired result.
answered Dec 14 '14 at 4:07
Milo BrandtMilo Brandt
39.9k476140
answered Dec 14 '14 at 4:07
Milo BrandtMilo Brandt
39.9k476140
answered Dec 14 '14 at 4:07
Milo BrandtMilo Brandt
39.9k476140
answered Dec 14 '14 at 4:07
Milo BrandtMilo Brandt
39.9k476140
39.9k476140
add a comment |
add a comment |
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$begingroup$
What is the source of this?
$endgroup$
– Will Jagy
Dec 14 '14 at 3:51