Is the sequence , 0,1/2,0,1/3,2/3,0,1/4,2/4,3/4,0… equidistributed?
$begingroup$
Is the sequence ,
$$0,1/2,0,1/3,2/3,0,1/4,2/4,3/4,0..... $$
equidistributed ?
A sequence ${xi_n}$ is equidistributed in $[0,1),$ that is if
$$lim_{Nrightarrowinfty}frac{Card{1leq nleq N|xi_nin(a,b)}}{N}=b-a$$
Well ,the sequence write in a formula is :
$$xi_n=0,;; whenever; n=k(k-1)/2+1$$
$$xi_n=frac{k}{m},;;whenever;n=frac{(m-1)(m–2)}{2}+k$$
$forall N,(a,b)$,firt of all 0 is in the $(a,b)$, so
$$Card{k|frac{k(k-1)}{2}+1leq N}=O(sqrt{N})$$
Next we let
$$frac{(m-1)(m–2)}{2}+kleq N$$
and
$$amleq kleq bm$$
Just like a nonlinear arrangement problem, suppose the $m$ is the $x$-axis, the $k$ is the $y$-axis, the question now , is to estimate the integer points of this area .
I don't know how to do next
number-theory
asked Dec 23 '18 at 6:57
Alexander LauAlexander Lau
1318
$endgroup$
add a comment |
$begingroup$
Is the sequence ,
$$0,1/2,0,1/3,2/3,0,1/4,2/4,3/4,0..... $$
equidistributed ?
A sequence ${xi_n}$ is equidistributed in $[0,1),$ that is if
$$lim_{Nrightarrowinfty}frac{Card{1leq nleq N|xi_nin(a,b)}}{N}=b-a$$
Well ,the sequence write in a formula is :
$$xi_n=0,;; whenever; n=k(k-1)/2+1$$
$$xi_n=frac{k}{m},;;whenever;n=frac{(m-1)(m–2)}{2}+k$$
$forall N,(a,b)$,firt of all 0 is in the $(a,b)$, so
$$Card{k|frac{k(k-1)}{2}+1leq N}=O(sqrt{N})$$
Next we let
$$frac{(m-1)(m–2)}{2}+kleq N$$
and
$$amleq kleq bm$$
Just like a nonlinear arrangement problem, suppose the $m$ is the $x$-axis, the $k$ is the $y$-axis, the question now , is to estimate the integer points of this area .
I don't know how to do next
number-theory
asked Dec 23 '18 at 6:57
Alexander LauAlexander Lau
1318
$endgroup$
2
$begingroup$
Can you show us what progress you've made?
$endgroup$
– zoidberg
Dec 23 '18 at 7:27
$begingroup$
@norfair of course, I show it in the answer ..
$endgroup$
– Alexander Lau
Dec 23 '18 at 7:47
add a comment |
$begingroup$
Is the sequence ,
$$0,1/2,0,1/3,2/3,0,1/4,2/4,3/4,0..... $$
equidistributed ?
A sequence ${xi_n}$ is equidistributed in $[0,1),$ that is if
$$lim_{Nrightarrowinfty}frac{Card{1leq nleq N|xi_nin(a,b)}}{N}=b-a$$
Well ,the sequence write in a formula is :
$$xi_n=0,;; whenever; n=k(k-1)/2+1$$
$$xi_n=frac{k}{m},;;whenever;n=frac{(m-1)(m–2)}{2}+k$$
$forall N,(a,b)$,firt of all 0 is in the $(a,b)$, so
$$Card{k|frac{k(k-1)}{2}+1leq N}=O(sqrt{N})$$
Next we let
$$frac{(m-1)(m–2)}{2}+kleq N$$
and
$$amleq kleq bm$$
Just like a nonlinear arrangement problem, suppose the $m$ is the $x$-axis, the $k$ is the $y$-axis, the question now , is to estimate the integer points of this area .
I don't know how to do next
number-theory
asked Dec 23 '18 at 6:57
Alexander LauAlexander Lau
1318
$endgroup$
Is the sequence ,
$$0,1/2,0,1/3,2/3,0,1/4,2/4,3/4,0..... $$
equidistributed ?
A sequence ${xi_n}$ is equidistributed in $[0,1),$ that is if
$$lim_{Nrightarrowinfty}frac{Card{1leq nleq N|xi_nin(a,b)}}{N}=b-a$$
Well ,the sequence write in a formula is :
$$xi_n=0,;; whenever; n=k(k-1)/2+1$$
$$xi_n=frac{k}{m},;;whenever;n=frac{(m-1)(m–2)}{2}+k$$
$forall N,(a,b)$,firt of all 0 is in the $(a,b)$, so
$$Card{k|frac{k(k-1)}{2}+1leq N}=O(sqrt{N})$$
Next we let
$$frac{(m-1)(m–2)}{2}+kleq N$$
and
$$amleq kleq bm$$
Just like a nonlinear arrangement problem, suppose the $m$ is the $x$-axis, the $k$ is the $y$-axis, the question now , is to estimate the integer points of this area .
I don't know how to do next
number-theory
number-theory
asked Dec 23 '18 at 6:57
Alexander LauAlexander Lau
1318
asked Dec 23 '18 at 6:57
Alexander LauAlexander Lau
1318
edited Dec 23 '18 at 8:38
Alexander Lau
asked Dec 23 '18 at 6:57
Alexander LauAlexander Lau
1318
asked Dec 23 '18 at 6:57
Alexander LauAlexander Lau
1318
asked Dec 23 '18 at 6:57
Alexander LauAlexander Lau
1318
1318
2
$begingroup$
Can you show us what progress you've made?
$endgroup$
– zoidberg
Dec 23 '18 at 7:27
$begingroup$
@norfair of course, I show it in the answer ..
$endgroup$
– Alexander Lau
Dec 23 '18 at 7:47
add a comment |
2
$begingroup$
Can you show us what progress you've made?
$endgroup$
– zoidberg
Dec 23 '18 at 7:27
$begingroup$
@norfair of course, I show it in the answer ..
$endgroup$
– Alexander Lau
Dec 23 '18 at 7:47
2
2
$begingroup$
Can you show us what progress you've made?
$endgroup$
– zoidberg
Dec 23 '18 at 7:27
$begingroup$
Can you show us what progress you've made?
$endgroup$
– zoidberg
Dec 23 '18 at 7:27
$begingroup$
@norfair of course, I show it in the answer ..
$endgroup$
– Alexander Lau
Dec 23 '18 at 7:47
$begingroup$
@norfair of course, I show it in the answer ..
$endgroup$
– Alexander Lau
Dec 23 '18 at 7:47
add a comment |
1 Answer
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$begingroup$
Well ,the sequence write in a formula is :
$$xi_n=0,;; whenever; n=k(k-1)/2+1$$
$$xi_n=frac{k}{m},;;whenever;n=frac{(m-1)(m–2)}{2}+k$$
$forall N,(a,b)$,firt of all 0 is in the $(a,b)$, so
$$Card{k|frac{k(k-1)}{2}+1leq N}=O(sqrt{N})$$
Next we let
$$frac{(m-1)(m–2)}{2}+kleq N$$
and
$$amleq kleq bm$$
Just like a nonlinear arrangement problem, suppose the $m$ is the $x$-axis, the $k$ is the $y$-axis, the question now , is to estimate the integer points of this area .
answered Dec 23 '18 at 7:46
Alexander LauAlexander Lau
1318
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Well ,the sequence write in a formula is :
$$xi_n=0,;; whenever; n=k(k-1)/2+1$$
$$xi_n=frac{k}{m},;;whenever;n=frac{(m-1)(m–2)}{2}+k$$
$forall N,(a,b)$,firt of all 0 is in the $(a,b)$, so
$$Card{k|frac{k(k-1)}{2}+1leq N}=O(sqrt{N})$$
Next we let
$$frac{(m-1)(m–2)}{2}+kleq N$$
and
$$amleq kleq bm$$
Just like a nonlinear arrangement problem, suppose the $m$ is the $x$-axis, the $k$ is the $y$-axis, the question now , is to estimate the integer points of this area .
answered Dec 23 '18 at 7:46
Alexander LauAlexander Lau
1318
$endgroup$
add a comment |
$begingroup$
Well ,the sequence write in a formula is :
$$xi_n=0,;; whenever; n=k(k-1)/2+1$$
$$xi_n=frac{k}{m},;;whenever;n=frac{(m-1)(m–2)}{2}+k$$
$forall N,(a,b)$,firt of all 0 is in the $(a,b)$, so
$$Card{k|frac{k(k-1)}{2}+1leq N}=O(sqrt{N})$$
Next we let
$$frac{(m-1)(m–2)}{2}+kleq N$$
and
$$amleq kleq bm$$
Just like a nonlinear arrangement problem, suppose the $m$ is the $x$-axis, the $k$ is the $y$-axis, the question now , is to estimate the integer points of this area .
answered Dec 23 '18 at 7:46
Alexander LauAlexander Lau
1318
$endgroup$
add a comment |
$begingroup$
Well ,the sequence write in a formula is :
$$xi_n=0,;; whenever; n=k(k-1)/2+1$$
$$xi_n=frac{k}{m},;;whenever;n=frac{(m-1)(m–2)}{2}+k$$
$forall N,(a,b)$,firt of all 0 is in the $(a,b)$, so
$$Card{k|frac{k(k-1)}{2}+1leq N}=O(sqrt{N})$$
Next we let
$$frac{(m-1)(m–2)}{2}+kleq N$$
and
$$amleq kleq bm$$
Just like a nonlinear arrangement problem, suppose the $m$ is the $x$-axis, the $k$ is the $y$-axis, the question now , is to estimate the integer points of this area .
answered Dec 23 '18 at 7:46
Alexander LauAlexander Lau
1318
$endgroup$
Well ,the sequence write in a formula is :
$$xi_n=0,;; whenever; n=k(k-1)/2+1$$
$$xi_n=frac{k}{m},;;whenever;n=frac{(m-1)(m–2)}{2}+k$$
$forall N,(a,b)$,firt of all 0 is in the $(a,b)$, so
$$Card{k|frac{k(k-1)}{2}+1leq N}=O(sqrt{N})$$
Next we let
$$frac{(m-1)(m–2)}{2}+kleq N$$
and
$$amleq kleq bm$$
Just like a nonlinear arrangement problem, suppose the $m$ is the $x$-axis, the $k$ is the $y$-axis, the question now , is to estimate the integer points of this area .
answered Dec 23 '18 at 7:46
Alexander LauAlexander Lau
1318
answered Dec 23 '18 at 7:46
Alexander LauAlexander Lau
1318
answered Dec 23 '18 at 7:46
Alexander LauAlexander Lau
1318
answered Dec 23 '18 at 7:46
Alexander LauAlexander Lau
1318
1318
add a comment |
add a comment |
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2
$begingroup$
Can you show us what progress you've made?
$endgroup$
– zoidberg
Dec 23 '18 at 7:27
$begingroup$
@norfair of course, I show it in the answer ..
$endgroup$
– Alexander Lau
Dec 23 '18 at 7:47