Mixing
Bad approximation, alternative definiton
$begingroup$
I read Complex dynamics of Carleson and Gamelin. They state without a proof the following
$theta$ is called bad approximate if there exists $c>0$ and $mu<infty$ such that $$ Bigvert theta - frac{p}{q}Big vert > frac{c}{q^{mu}}$$
for all integer $p$ and $q$ ( $ q neq 0$). This is supposedly only occures if and only if:
$lambda = e^{2pi i theta}$ satisfying
$$ Big vert lambda^n -1 Big vert > c n^{1-mu} , quad forall ngeq 1$$
For some $c>0$ and $mu<infty$
I can not se any way to prove this statement, grateful for help.
number-theory analytic-number-theory diophantine-approximation
asked Jan 10 at 16:57
tomasfotomasfo
63
$endgroup$
add a comment |
$begingroup$
I read Complex dynamics of Carleson and Gamelin. They state without a proof the following
$theta$ is called bad approximate if there exists $c>0$ and $mu<infty$ such that $$ Bigvert theta - frac{p}{q}Big vert > frac{c}{q^{mu}}$$
for all integer $p$ and $q$ ( $ q neq 0$). This is supposedly only occures if and only if:
$lambda = e^{2pi i theta}$ satisfying
$$ Big vert lambda^n -1 Big vert > c n^{1-mu} , quad forall ngeq 1$$
For some $c>0$ and $mu<infty$
I can not se any way to prove this statement, grateful for help.
number-theory analytic-number-theory diophantine-approximation
asked Jan 10 at 16:57
tomasfotomasfo
63
$endgroup$
$begingroup$
Did you mean $forall n geq 1$? Also, is the thing you want to show equivalence to the statement that (for some given bad approximale $theta$) for any n ($geq 1 ?$) there exist $c$ and $mu$ such that $|lambda^n-1|>cn^{1-mu}$ ?
$endgroup$
– Cardioid_Ass_22
Jan 10 at 19:45
add a comment |
$begingroup$
I read Complex dynamics of Carleson and Gamelin. They state without a proof the following
$theta$ is called bad approximate if there exists $c>0$ and $mu<infty$ such that $$ Bigvert theta - frac{p}{q}Big vert > frac{c}{q^{mu}}$$
for all integer $p$ and $q$ ( $ q neq 0$). This is supposedly only occures if and only if:
$lambda = e^{2pi i theta}$ satisfying
$$ Big vert lambda^n -1 Big vert > c n^{1-mu} , quad forall ngeq 1$$
For some $c>0$ and $mu<infty$
I can not se any way to prove this statement, grateful for help.
number-theory analytic-number-theory diophantine-approximation
asked Jan 10 at 16:57
tomasfotomasfo
63
$endgroup$
I read Complex dynamics of Carleson and Gamelin. They state without a proof the following
$theta$ is called bad approximate if there exists $c>0$ and $mu<infty$ such that $$ Bigvert theta - frac{p}{q}Big vert > frac{c}{q^{mu}}$$
for all integer $p$ and $q$ ( $ q neq 0$). This is supposedly only occures if and only if:
$lambda = e^{2pi i theta}$ satisfying
$$ Big vert lambda^n -1 Big vert > c n^{1-mu} , quad forall ngeq 1$$
For some $c>0$ and $mu<infty$
I can not se any way to prove this statement, grateful for help.
number-theory analytic-number-theory diophantine-approximation
number-theory analytic-number-theory diophantine-approximation
asked Jan 10 at 16:57
tomasfotomasfo
63
asked Jan 10 at 16:57
tomasfotomasfo
63
edited Jan 12 at 23:44
tomasfo
asked Jan 10 at 16:57
tomasfotomasfo
63
asked Jan 10 at 16:57
tomasfotomasfo
63
asked Jan 10 at 16:57
tomasfotomasfo
63
63
$begingroup$
Did you mean $forall n geq 1$? Also, is the thing you want to show equivalence to the statement that (for some given bad approximale $theta$) for any n ($geq 1 ?$) there exist $c$ and $mu$ such that $|lambda^n-1|>cn^{1-mu}$ ?
$endgroup$
– Cardioid_Ass_22
Jan 10 at 19:45
add a comment |
$begingroup$
Did you mean $forall n geq 1$? Also, is the thing you want to show equivalence to the statement that (for some given bad approximale $theta$) for any n ($geq 1 ?$) there exist $c$ and $mu$ such that $|lambda^n-1|>cn^{1-mu}$ ?
$endgroup$
– Cardioid_Ass_22
Jan 10 at 19:45
$begingroup$
Did you mean $forall n geq 1$? Also, is the thing you want to show equivalence to the statement that (for some given bad approximale $theta$) for any n ($geq 1 ?$) there exist $c$ and $mu$ such that $|lambda^n-1|>cn^{1-mu}$ ?
$endgroup$
– Cardioid_Ass_22
Jan 10 at 19:45
$begingroup$
Did you mean $forall n geq 1$? Also, is the thing you want to show equivalence to the statement that (for some given bad approximale $theta$) for any n ($geq 1 ?$) there exist $c$ and $mu$ such that $|lambda^n-1|>cn^{1-mu}$ ?
$endgroup$
– Cardioid_Ass_22
Jan 10 at 19:45
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Not quite sure if this is along the lines of what you're looking for, but I thought I'd give it a shot anyway. We have the exponential map
begin{align*}
exp : mathbb{R} / mathbb{Z} &longrightarrow mathbb{T},\
left[x right] &longmapsto e^{2 pi i x}
end{align*}
from $mathbb{R}/mathbb{Z}$ to the (complex) unit circle. The topology on $mathbb{R}/mathbb{Z}$ is induced by the following metric: For $left[ x right], left[ y right] in mathbb{R}/mathbb{Z}$,
begin{align*}
d_1 ( left[ x right], left[ y right] ) = min_{k in mathbb{Z}} |x-kcdot y|.
end{align*}
The topology on $mathbb{T}$ is induced by the metric
begin{align*}
d_2 ( e^{2 pi i x}, e^{2 pi i y} ) = text{length of the shortest path on $mathbb{T}$ from $e^{2 pi i x}$ to $e^{2 pi i y}$}.
end{align*}
These metrics are related in the following way: $2 pi cdot d_1 (left[x right], left[ y right]) = d_2 (exp (x), exp (y))$.
Now, if there is $c > 0$ and $mu geqslant 1$ such that $| theta - p/q | > c/q^{mu}$ for all $p/q in mathbb{Q}$, then for any fixed integer $p$, $|n theta - p | > c/n^{mu - 1}$ for all $n geqslant 1$. This implies that
begin{align*}
d_1 (left[ p right], left[n theta right]) > frac{c}{n^{mu - 1}},
end{align*}
which implies that
begin{align*}
d_2 ( exp (p), exp(n theta)) = d_2 (1, e^{2 pi i n theta}) > frac{2 pi c}{n^{mu - 1}}.
end{align*}
So the shortest arc from $1$ to $e^{2 pi i n theta}$ has length at least $2 pi c / n^{mu - 1}$. But since
begin{align*}
text{chordal length} geqslant frac{2}{pi} cdot text{arc length},
end{align*}
we have
begin{align*}
|e^{2 pi i n theta} - 1| > frac{2}{pi} cdot frac{2 pi c}{n^{mu - 1}} = frac{4c}{n^{mu - 1}}.
end{align*}
It is also possible to prove the other direction this way, although possibly with a smaller $c$. But that shouldn't matter since being badly approximable just means that there is some $c > 0$ such that the relevant inequality is satisfied.
answered Jan 11 at 13:41
Teddan the TerranTeddan the Terran
1,206210
$endgroup$
$begingroup$
I have edited the question, but i think this should to it! thank you
$endgroup$
– tomasfo
Jan 13 at 9:12
add a comment |
$begingroup$
I have thought a bit and think i have solved one direction myself. The proof is based on proving following inequality.
$$ vert lambda^n -1 vert=vert e^{2pi i n theta}-1vert < 2 pi vert theta n -pvert,quad forall p,n in mathbb{Z}$$.
We have
$$
begin{align}
vert e^{2 pi i n theta}-1 vert& = vert e^{pi i ntheta}vert vert e^{2 pi i n theta}-1 vert\
& = vert e^{ pi i n theta}- e^{-pi i n theta} vert \
& =2 vert frac{e^{ pi i n theta}- e^{-pi i n theta}}{2i} vert \
& = 2 vertsin{pi ntheta}vert \
& = {vert sin{x}vert text{ is } pitext{-periodic}}\
& = 2 vert sin{pi n theta - pi p}vert\
& = { text{ Use that } vert sin{x}vert leq vert x vert quad forall x in mathbb{R} }\
& = 2vert pi (ntheta -p)vert\
& = 2pivert n theta -pvert
end{align}
$$
So given this inequality we can first prove the first direction. Assume $theta$ is not bad approximal. Then for all $c>0$ and $ mu < infty$ we have some $pin mathbb{Z},qin mathbb{Z}^*$ such that
$$vert theta - frac{p}{q}vert < frac{c}{q^{mu}}$$
But then we have that
$$
begin{align}
vert lambda^n-1 vert & < 2 pi vert n theta -pvert\
& = 2pi nvert theta -frac{p}{n}vert\
& < 2 pi n frac{c}{n^{mu }}\
& = 2 pi cn^{1-mu}
end{align}
$$
For the converse i can not see if one can apply the same metod.
answered Jan 13 at 9:09
tomasfotomasfo
63
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068907%2fbad-approximation-alternative-definiton%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Not quite sure if this is along the lines of what you're looking for, but I thought I'd give it a shot anyway. We have the exponential map
begin{align*}
exp : mathbb{R} / mathbb{Z} &longrightarrow mathbb{T},\
left[x right] &longmapsto e^{2 pi i x}
end{align*}
from $mathbb{R}/mathbb{Z}$ to the (complex) unit circle. The topology on $mathbb{R}/mathbb{Z}$ is induced by the following metric: For $left[ x right], left[ y right] in mathbb{R}/mathbb{Z}$,
begin{align*}
d_1 ( left[ x right], left[ y right] ) = min_{k in mathbb{Z}} |x-kcdot y|.
end{align*}
The topology on $mathbb{T}$ is induced by the metric
begin{align*}
d_2 ( e^{2 pi i x}, e^{2 pi i y} ) = text{length of the shortest path on $mathbb{T}$ from $e^{2 pi i x}$ to $e^{2 pi i y}$}.
end{align*}
These metrics are related in the following way: $2 pi cdot d_1 (left[x right], left[ y right]) = d_2 (exp (x), exp (y))$.
Now, if there is $c > 0$ and $mu geqslant 1$ such that $| theta - p/q | > c/q^{mu}$ for all $p/q in mathbb{Q}$, then for any fixed integer $p$, $|n theta - p | > c/n^{mu - 1}$ for all $n geqslant 1$. This implies that
begin{align*}
d_1 (left[ p right], left[n theta right]) > frac{c}{n^{mu - 1}},
end{align*}
which implies that
begin{align*}
d_2 ( exp (p), exp(n theta)) = d_2 (1, e^{2 pi i n theta}) > frac{2 pi c}{n^{mu - 1}}.
end{align*}
So the shortest arc from $1$ to $e^{2 pi i n theta}$ has length at least $2 pi c / n^{mu - 1}$. But since
begin{align*}
text{chordal length} geqslant frac{2}{pi} cdot text{arc length},
end{align*}
we have
begin{align*}
|e^{2 pi i n theta} - 1| > frac{2}{pi} cdot frac{2 pi c}{n^{mu - 1}} = frac{4c}{n^{mu - 1}}.
end{align*}
It is also possible to prove the other direction this way, although possibly with a smaller $c$. But that shouldn't matter since being badly approximable just means that there is some $c > 0$ such that the relevant inequality is satisfied.
answered Jan 11 at 13:41
Teddan the TerranTeddan the Terran
1,206210
$endgroup$
$begingroup$
I have edited the question, but i think this should to it! thank you
$endgroup$
– tomasfo
Jan 13 at 9:12
add a comment |
$begingroup$
Not quite sure if this is along the lines of what you're looking for, but I thought I'd give it a shot anyway. We have the exponential map
begin{align*}
exp : mathbb{R} / mathbb{Z} &longrightarrow mathbb{T},\
left[x right] &longmapsto e^{2 pi i x}
end{align*}
from $mathbb{R}/mathbb{Z}$ to the (complex) unit circle. The topology on $mathbb{R}/mathbb{Z}$ is induced by the following metric: For $left[ x right], left[ y right] in mathbb{R}/mathbb{Z}$,
begin{align*}
d_1 ( left[ x right], left[ y right] ) = min_{k in mathbb{Z}} |x-kcdot y|.
end{align*}
The topology on $mathbb{T}$ is induced by the metric
begin{align*}
d_2 ( e^{2 pi i x}, e^{2 pi i y} ) = text{length of the shortest path on $mathbb{T}$ from $e^{2 pi i x}$ to $e^{2 pi i y}$}.
end{align*}
These metrics are related in the following way: $2 pi cdot d_1 (left[x right], left[ y right]) = d_2 (exp (x), exp (y))$.
Now, if there is $c > 0$ and $mu geqslant 1$ such that $| theta - p/q | > c/q^{mu}$ for all $p/q in mathbb{Q}$, then for any fixed integer $p$, $|n theta - p | > c/n^{mu - 1}$ for all $n geqslant 1$. This implies that
begin{align*}
d_1 (left[ p right], left[n theta right]) > frac{c}{n^{mu - 1}},
end{align*}
which implies that
begin{align*}
d_2 ( exp (p), exp(n theta)) = d_2 (1, e^{2 pi i n theta}) > frac{2 pi c}{n^{mu - 1}}.
end{align*}
So the shortest arc from $1$ to $e^{2 pi i n theta}$ has length at least $2 pi c / n^{mu - 1}$. But since
begin{align*}
text{chordal length} geqslant frac{2}{pi} cdot text{arc length},
end{align*}
we have
begin{align*}
|e^{2 pi i n theta} - 1| > frac{2}{pi} cdot frac{2 pi c}{n^{mu - 1}} = frac{4c}{n^{mu - 1}}.
end{align*}
It is also possible to prove the other direction this way, although possibly with a smaller $c$. But that shouldn't matter since being badly approximable just means that there is some $c > 0$ such that the relevant inequality is satisfied.
answered Jan 11 at 13:41
Teddan the TerranTeddan the Terran
1,206210
$endgroup$
$begingroup$
I have edited the question, but i think this should to it! thank you
$endgroup$
– tomasfo
Jan 13 at 9:12
add a comment |
$begingroup$
Not quite sure if this is along the lines of what you're looking for, but I thought I'd give it a shot anyway. We have the exponential map
begin{align*}
exp : mathbb{R} / mathbb{Z} &longrightarrow mathbb{T},\
left[x right] &longmapsto e^{2 pi i x}
end{align*}
from $mathbb{R}/mathbb{Z}$ to the (complex) unit circle. The topology on $mathbb{R}/mathbb{Z}$ is induced by the following metric: For $left[ x right], left[ y right] in mathbb{R}/mathbb{Z}$,
begin{align*}
d_1 ( left[ x right], left[ y right] ) = min_{k in mathbb{Z}} |x-kcdot y|.
end{align*}
The topology on $mathbb{T}$ is induced by the metric
begin{align*}
d_2 ( e^{2 pi i x}, e^{2 pi i y} ) = text{length of the shortest path on $mathbb{T}$ from $e^{2 pi i x}$ to $e^{2 pi i y}$}.
end{align*}
These metrics are related in the following way: $2 pi cdot d_1 (left[x right], left[ y right]) = d_2 (exp (x), exp (y))$.
Now, if there is $c > 0$ and $mu geqslant 1$ such that $| theta - p/q | > c/q^{mu}$ for all $p/q in mathbb{Q}$, then for any fixed integer $p$, $|n theta - p | > c/n^{mu - 1}$ for all $n geqslant 1$. This implies that
begin{align*}
d_1 (left[ p right], left[n theta right]) > frac{c}{n^{mu - 1}},
end{align*}
which implies that
begin{align*}
d_2 ( exp (p), exp(n theta)) = d_2 (1, e^{2 pi i n theta}) > frac{2 pi c}{n^{mu - 1}}.
end{align*}
So the shortest arc from $1$ to $e^{2 pi i n theta}$ has length at least $2 pi c / n^{mu - 1}$. But since
begin{align*}
text{chordal length} geqslant frac{2}{pi} cdot text{arc length},
end{align*}
we have
begin{align*}
|e^{2 pi i n theta} - 1| > frac{2}{pi} cdot frac{2 pi c}{n^{mu - 1}} = frac{4c}{n^{mu - 1}}.
end{align*}
It is also possible to prove the other direction this way, although possibly with a smaller $c$. But that shouldn't matter since being badly approximable just means that there is some $c > 0$ such that the relevant inequality is satisfied.
answered Jan 11 at 13:41
Teddan the TerranTeddan the Terran
1,206210
$endgroup$
Not quite sure if this is along the lines of what you're looking for, but I thought I'd give it a shot anyway. We have the exponential map
begin{align*}
exp : mathbb{R} / mathbb{Z} &longrightarrow mathbb{T},\
left[x right] &longmapsto e^{2 pi i x}
end{align*}
from $mathbb{R}/mathbb{Z}$ to the (complex) unit circle. The topology on $mathbb{R}/mathbb{Z}$ is induced by the following metric: For $left[ x right], left[ y right] in mathbb{R}/mathbb{Z}$,
begin{align*}
d_1 ( left[ x right], left[ y right] ) = min_{k in mathbb{Z}} |x-kcdot y|.
end{align*}
The topology on $mathbb{T}$ is induced by the metric
begin{align*}
d_2 ( e^{2 pi i x}, e^{2 pi i y} ) = text{length of the shortest path on $mathbb{T}$ from $e^{2 pi i x}$ to $e^{2 pi i y}$}.
end{align*}
These metrics are related in the following way: $2 pi cdot d_1 (left[x right], left[ y right]) = d_2 (exp (x), exp (y))$.
Now, if there is $c > 0$ and $mu geqslant 1$ such that $| theta - p/q | > c/q^{mu}$ for all $p/q in mathbb{Q}$, then for any fixed integer $p$, $|n theta - p | > c/n^{mu - 1}$ for all $n geqslant 1$. This implies that
begin{align*}
d_1 (left[ p right], left[n theta right]) > frac{c}{n^{mu - 1}},
end{align*}
which implies that
begin{align*}
d_2 ( exp (p), exp(n theta)) = d_2 (1, e^{2 pi i n theta}) > frac{2 pi c}{n^{mu - 1}}.
end{align*}
So the shortest arc from $1$ to $e^{2 pi i n theta}$ has length at least $2 pi c / n^{mu - 1}$. But since
begin{align*}
text{chordal length} geqslant frac{2}{pi} cdot text{arc length},
end{align*}
we have
begin{align*}
|e^{2 pi i n theta} - 1| > frac{2}{pi} cdot frac{2 pi c}{n^{mu - 1}} = frac{4c}{n^{mu - 1}}.
end{align*}
It is also possible to prove the other direction this way, although possibly with a smaller $c$. But that shouldn't matter since being badly approximable just means that there is some $c > 0$ such that the relevant inequality is satisfied.
answered Jan 11 at 13:41
Teddan the TerranTeddan the Terran
1,206210
answered Jan 11 at 13:41
Teddan the TerranTeddan the Terran
1,206210
answered Jan 11 at 13:41
Teddan the TerranTeddan the Terran
1,206210
answered Jan 11 at 13:41
Teddan the TerranTeddan the Terran
1,206210
1,206210
$begingroup$
I have edited the question, but i think this should to it! thank you
$endgroup$
– tomasfo
Jan 13 at 9:12
add a comment |
$begingroup$
I have edited the question, but i think this should to it! thank you
$endgroup$
– tomasfo
Jan 13 at 9:12
$begingroup$
I have edited the question, but i think this should to it! thank you
$endgroup$
– tomasfo
Jan 13 at 9:12
$begingroup$
I have edited the question, but i think this should to it! thank you
$endgroup$
– tomasfo
Jan 13 at 9:12
add a comment |
$begingroup$
I have thought a bit and think i have solved one direction myself. The proof is based on proving following inequality.
$$ vert lambda^n -1 vert=vert e^{2pi i n theta}-1vert < 2 pi vert theta n -pvert,quad forall p,n in mathbb{Z}$$.
We have
$$
begin{align}
vert e^{2 pi i n theta}-1 vert& = vert e^{pi i ntheta}vert vert e^{2 pi i n theta}-1 vert\
& = vert e^{ pi i n theta}- e^{-pi i n theta} vert \
& =2 vert frac{e^{ pi i n theta}- e^{-pi i n theta}}{2i} vert \
& = 2 vertsin{pi ntheta}vert \
& = {vert sin{x}vert text{ is } pitext{-periodic}}\
& = 2 vert sin{pi n theta - pi p}vert\
& = { text{ Use that } vert sin{x}vert leq vert x vert quad forall x in mathbb{R} }\
& = 2vert pi (ntheta -p)vert\
& = 2pivert n theta -pvert
end{align}
$$
So given this inequality we can first prove the first direction. Assume $theta$ is not bad approximal. Then for all $c>0$ and $ mu < infty$ we have some $pin mathbb{Z},qin mathbb{Z}^*$ such that
$$vert theta - frac{p}{q}vert < frac{c}{q^{mu}}$$
But then we have that
$$
begin{align}
vert lambda^n-1 vert & < 2 pi vert n theta -pvert\
& = 2pi nvert theta -frac{p}{n}vert\
& < 2 pi n frac{c}{n^{mu }}\
& = 2 pi cn^{1-mu}
end{align}
$$
For the converse i can not see if one can apply the same metod.
answered Jan 13 at 9:09
tomasfotomasfo
63
$endgroup$
add a comment |
$begingroup$
I have thought a bit and think i have solved one direction myself. The proof is based on proving following inequality.
$$ vert lambda^n -1 vert=vert e^{2pi i n theta}-1vert < 2 pi vert theta n -pvert,quad forall p,n in mathbb{Z}$$.
We have
$$
begin{align}
vert e^{2 pi i n theta}-1 vert& = vert e^{pi i ntheta}vert vert e^{2 pi i n theta}-1 vert\
& = vert e^{ pi i n theta}- e^{-pi i n theta} vert \
& =2 vert frac{e^{ pi i n theta}- e^{-pi i n theta}}{2i} vert \
& = 2 vertsin{pi ntheta}vert \
& = {vert sin{x}vert text{ is } pitext{-periodic}}\
& = 2 vert sin{pi n theta - pi p}vert\
& = { text{ Use that } vert sin{x}vert leq vert x vert quad forall x in mathbb{R} }\
& = 2vert pi (ntheta -p)vert\
& = 2pivert n theta -pvert
end{align}
$$
So given this inequality we can first prove the first direction. Assume $theta$ is not bad approximal. Then for all $c>0$ and $ mu < infty$ we have some $pin mathbb{Z},qin mathbb{Z}^*$ such that
$$vert theta - frac{p}{q}vert < frac{c}{q^{mu}}$$
But then we have that
$$
begin{align}
vert lambda^n-1 vert & < 2 pi vert n theta -pvert\
& = 2pi nvert theta -frac{p}{n}vert\
& < 2 pi n frac{c}{n^{mu }}\
& = 2 pi cn^{1-mu}
end{align}
$$
For the converse i can not see if one can apply the same metod.
answered Jan 13 at 9:09
tomasfotomasfo
63
$endgroup$
add a comment |
$begingroup$
I have thought a bit and think i have solved one direction myself. The proof is based on proving following inequality.
$$ vert lambda^n -1 vert=vert e^{2pi i n theta}-1vert < 2 pi vert theta n -pvert,quad forall p,n in mathbb{Z}$$.
We have
$$
begin{align}
vert e^{2 pi i n theta}-1 vert& = vert e^{pi i ntheta}vert vert e^{2 pi i n theta}-1 vert\
& = vert e^{ pi i n theta}- e^{-pi i n theta} vert \
& =2 vert frac{e^{ pi i n theta}- e^{-pi i n theta}}{2i} vert \
& = 2 vertsin{pi ntheta}vert \
& = {vert sin{x}vert text{ is } pitext{-periodic}}\
& = 2 vert sin{pi n theta - pi p}vert\
& = { text{ Use that } vert sin{x}vert leq vert x vert quad forall x in mathbb{R} }\
& = 2vert pi (ntheta -p)vert\
& = 2pivert n theta -pvert
end{align}
$$
So given this inequality we can first prove the first direction. Assume $theta$ is not bad approximal. Then for all $c>0$ and $ mu < infty$ we have some $pin mathbb{Z},qin mathbb{Z}^*$ such that
$$vert theta - frac{p}{q}vert < frac{c}{q^{mu}}$$
But then we have that
$$
begin{align}
vert lambda^n-1 vert & < 2 pi vert n theta -pvert\
& = 2pi nvert theta -frac{p}{n}vert\
& < 2 pi n frac{c}{n^{mu }}\
& = 2 pi cn^{1-mu}
end{align}
$$
For the converse i can not see if one can apply the same metod.
answered Jan 13 at 9:09
tomasfotomasfo
63
$endgroup$
I have thought a bit and think i have solved one direction myself. The proof is based on proving following inequality.
$$ vert lambda^n -1 vert=vert e^{2pi i n theta}-1vert < 2 pi vert theta n -pvert,quad forall p,n in mathbb{Z}$$.
We have
$$
begin{align}
vert e^{2 pi i n theta}-1 vert& = vert e^{pi i ntheta}vert vert e^{2 pi i n theta}-1 vert\
& = vert e^{ pi i n theta}- e^{-pi i n theta} vert \
& =2 vert frac{e^{ pi i n theta}- e^{-pi i n theta}}{2i} vert \
& = 2 vertsin{pi ntheta}vert \
& = {vert sin{x}vert text{ is } pitext{-periodic}}\
& = 2 vert sin{pi n theta - pi p}vert\
& = { text{ Use that } vert sin{x}vert leq vert x vert quad forall x in mathbb{R} }\
& = 2vert pi (ntheta -p)vert\
& = 2pivert n theta -pvert
end{align}
$$
So given this inequality we can first prove the first direction. Assume $theta$ is not bad approximal. Then for all $c>0$ and $ mu < infty$ we have some $pin mathbb{Z},qin mathbb{Z}^*$ such that
$$vert theta - frac{p}{q}vert < frac{c}{q^{mu}}$$
But then we have that
$$
begin{align}
vert lambda^n-1 vert & < 2 pi vert n theta -pvert\
& = 2pi nvert theta -frac{p}{n}vert\
& < 2 pi n frac{c}{n^{mu }}\
& = 2 pi cn^{1-mu}
end{align}
$$
For the converse i can not see if one can apply the same metod.
answered Jan 13 at 9:09
tomasfotomasfo
63
answered Jan 13 at 9:09
tomasfotomasfo
63
answered Jan 13 at 9:09
tomasfotomasfo
63
answered Jan 13 at 9:09
tomasfotomasfo
63
63
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068907%2fbad-approximation-alternative-definiton%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Did you mean $forall n geq 1$? Also, is the thing you want to show equivalence to the statement that (for some given bad approximale $theta$) for any n ($geq 1 ?$) there exist $c$ and $mu$ such that $|lambda^n-1|>cn^{1-mu}$ ?
$endgroup$
– Cardioid_Ass_22
Jan 10 at 19:45