Find the $limlimits_{k to infty} frac{f(k)}{k^2} $












1












$begingroup$


Let $newcommand{N}{mathbb{N}}k$ be positive integer then we define the set
$$
A(k) := Big{(m,n)in N^2 : bigvert kcosbig(frac{npi}{k}big)bigvert leq m leq ksinbig(frac{npi}{k}big)Big}.
$$
Furthermore, we define the function
begin{align}
f: left{
begin{array}{rcl}
N & to &N, \
k & mapsto & vert A(k) vert,
end{array}
right.
end{align}
where $vert Avert$ is the number of elements of a set $A$.




Question: Does $limlimits_{k to infty} frac{f(k)}{k^2}$ exist and if so how can it be calculated?




My attempts:



enter image description here






real-analysis limits







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$endgroup$












  • $begingroup$
    why are you constantly uploading the same image and edit the question with no difference at all.
    $endgroup$
    – Nathanael Skrepek
    Jan 29 at 16:49
















1












$begingroup$


Let $newcommand{N}{mathbb{N}}k$ be positive integer then we define the set
$$
A(k) := Big{(m,n)in N^2 : bigvert kcosbig(frac{npi}{k}big)bigvert leq m leq ksinbig(frac{npi}{k}big)Big}.
$$
Furthermore, we define the function
begin{align}
f: left{
begin{array}{rcl}
N & to &N, \
k & mapsto & vert A(k) vert,
end{array}
right.
end{align}
where $vert Avert$ is the number of elements of a set $A$.




Question: Does $limlimits_{k to infty} frac{f(k)}{k^2}$ exist and if so how can it be calculated?




My attempts:



enter image description here






real-analysis limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    why are you constantly uploading the same image and edit the question with no difference at all.
    $endgroup$
    – Nathanael Skrepek
    Jan 29 at 16:49














1












1








1





$begingroup$


Let $newcommand{N}{mathbb{N}}k$ be positive integer then we define the set
$$
A(k) := Big{(m,n)in N^2 : bigvert kcosbig(frac{npi}{k}big)bigvert leq m leq ksinbig(frac{npi}{k}big)Big}.
$$
Furthermore, we define the function
begin{align}
f: left{
begin{array}{rcl}
N & to &N, \
k & mapsto & vert A(k) vert,
end{array}
right.
end{align}
where $vert Avert$ is the number of elements of a set $A$.




Question: Does $limlimits_{k to infty} frac{f(k)}{k^2}$ exist and if so how can it be calculated?




My attempts:



enter image description here






real-analysis limits







share|cite|improve this question











$endgroup$




Let $newcommand{N}{mathbb{N}}k$ be positive integer then we define the set
$$
A(k) := Big{(m,n)in N^2 : bigvert kcosbig(frac{npi}{k}big)bigvert leq m leq ksinbig(frac{npi}{k}big)Big}.
$$
Furthermore, we define the function
begin{align}
f: left{
begin{array}{rcl}
N & to &N, \
k & mapsto & vert A(k) vert,
end{array}
right.
end{align}
where $vert Avert$ is the number of elements of a set $A$.




Question: Does $limlimits_{k to infty} frac{f(k)}{k^2}$ exist and if so how can it be calculated?




My attempts:



enter image description here






real-analysis limits




real-analysis limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 29 at 17:53







mina_world

















asked Jan 29 at 16:27









mina_worldmina_world

1799




1799












  • $begingroup$
    why are you constantly uploading the same image and edit the question with no difference at all.
    $endgroup$
    – Nathanael Skrepek
    Jan 29 at 16:49


















  • $begingroup$
    why are you constantly uploading the same image and edit the question with no difference at all.
    $endgroup$
    – Nathanael Skrepek
    Jan 29 at 16:49
















$begingroup$
why are you constantly uploading the same image and edit the question with no difference at all.
$endgroup$
– Nathanael Skrepek
Jan 29 at 16:49




$begingroup$
why are you constantly uploading the same image and edit the question with no difference at all.
$endgroup$
– Nathanael Skrepek
Jan 29 at 16:49










1 Answer
1






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oldest

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1












$begingroup$

Note that for all $a, bin Bbb R,$ it holds that $$
(b-a)^+-1 le|{x:ale xle b}cap Bbb Z|le (b-a)^++1
$$ where $c^+=max{c,0}$.
Hence the number $N(k,n)$ of $min Bbb Z$ satisfying $$
kleft|cos left(frac{npi}{k}right)right|le m le ksinleft(frac{npi}{k}right)
$$ satisfies $$kleft(sinleft(frac{npi}{k}right)-left|cos left(frac{npi}{k}right)right|right)^+-1 le N(k,n)le kleft(sinleft(frac{npi}{k}right)-kleft|cos left(frac{npi}{k}right)right|right)^++1.$$ Since $f(k) =sum_{n=1}^k N(k,n)$, this gives
$$
left|sum_{n=1}^kkleft(sinleft(frac{npi}{k}right)-left|cos left(frac{npi}{k}right)right|right)^+ -f(k)right|le k.
$$ It follows that
$$begin{eqnarray}
lim_{ktoinfty}frac{f(k)}{k^2}&=&lim_{ktoinfty}sum_{n=1}^kleft(sinleft(frac{npi}{k}right)-left|cos left(frac{npi}{k}right)right|right)^+frac{1}{k}\&=&int_0^1 (sin pi x-|cos pi x|)^+mathrm{d}x\
&=&2int_{frac{1}{4}}^{frac{1}{2}}(sinpi x-cospi x) mathrm{d}x\
&=&frac{2}{pi}left[-cos pi x-sin pi xright]^{frac{1}{2}}_{frac{1}{4}}=frac{2}{pi}(sqrt{2}-1).
end{eqnarray}$$






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    1 Answer
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    1 Answer
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    active

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    1












    $begingroup$

    Note that for all $a, bin Bbb R,$ it holds that $$
    (b-a)^+-1 le|{x:ale xle b}cap Bbb Z|le (b-a)^++1
    $$     where $c^+=max{c,0}$.
    Hence the number $N(k,n)$ of $min Bbb Z$ satisfying $$
    kleft|cos left(frac{npi}{k}right)right|le m le ksinleft(frac{npi}{k}right)
    $$     satisfies $$kleft(sinleft(frac{npi}{k}right)-left|cos left(frac{npi}{k}right)right|right)^+-1 le N(k,n)le kleft(sinleft(frac{npi}{k}right)-kleft|cos left(frac{npi}{k}right)right|right)^++1.$$ Since $f(k) =sum_{n=1}^k N(k,n)$, this gives
    $$
    left|sum_{n=1}^kkleft(sinleft(frac{npi}{k}right)-left|cos left(frac{npi}{k}right)right|right)^+ -f(k)right|le k.
    $$     It follows that
    $$begin{eqnarray}
    lim_{ktoinfty}frac{f(k)}{k^2}&=&lim_{ktoinfty}sum_{n=1}^kleft(sinleft(frac{npi}{k}right)-left|cos left(frac{npi}{k}right)right|right)^+frac{1}{k}\&=&int_0^1 (sin pi x-|cos pi x|)^+mathrm{d}x\
    &=&2int_{frac{1}{4}}^{frac{1}{2}}(sinpi x-cospi x) mathrm{d}x\
    &=&frac{2}{pi}left[-cos pi x-sin pi xright]^{frac{1}{2}}_{frac{1}{4}}=frac{2}{pi}(sqrt{2}-1).
    end{eqnarray}$$






    share|cite|improve this answer









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      1












      $begingroup$

      Note that for all $a, bin Bbb R,$ it holds that $$
      (b-a)^+-1 le|{x:ale xle b}cap Bbb Z|le (b-a)^++1
      $$       where $c^+=max{c,0}$.
      Hence the number $N(k,n)$ of $min Bbb Z$ satisfying $$
      kleft|cos left(frac{npi}{k}right)right|le m le ksinleft(frac{npi}{k}right)
      $$       satisfies $$kleft(sinleft(frac{npi}{k}right)-left|cos left(frac{npi}{k}right)right|right)^+-1 le N(k,n)le kleft(sinleft(frac{npi}{k}right)-kleft|cos left(frac{npi}{k}right)right|right)^++1.$$ Since $f(k) =sum_{n=1}^k N(k,n)$, this gives
      $$
      left|sum_{n=1}^kkleft(sinleft(frac{npi}{k}right)-left|cos left(frac{npi}{k}right)right|right)^+ -f(k)right|le k.
      $$       It follows that
      $$begin{eqnarray}
      lim_{ktoinfty}frac{f(k)}{k^2}&=&lim_{ktoinfty}sum_{n=1}^kleft(sinleft(frac{npi}{k}right)-left|cos left(frac{npi}{k}right)right|right)^+frac{1}{k}\&=&int_0^1 (sin pi x-|cos pi x|)^+mathrm{d}x\
      &=&2int_{frac{1}{4}}^{frac{1}{2}}(sinpi x-cospi x) mathrm{d}x\
      &=&frac{2}{pi}left[-cos pi x-sin pi xright]^{frac{1}{2}}_{frac{1}{4}}=frac{2}{pi}(sqrt{2}-1).
      end{eqnarray}$$






      share|cite|improve this answer









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        1





        $begingroup$

        Note that for all $a, bin Bbb R,$ it holds that $$
        (b-a)^+-1 le|{x:ale xle b}cap Bbb Z|le (b-a)^++1
        $$         where $c^+=max{c,0}$.
        Hence the number $N(k,n)$ of $min Bbb Z$ satisfying $$
        kleft|cos left(frac{npi}{k}right)right|le m le ksinleft(frac{npi}{k}right)
        $$         satisfies $$kleft(sinleft(frac{npi}{k}right)-left|cos left(frac{npi}{k}right)right|right)^+-1 le N(k,n)le kleft(sinleft(frac{npi}{k}right)-kleft|cos left(frac{npi}{k}right)right|right)^++1.$$ Since $f(k) =sum_{n=1}^k N(k,n)$, this gives
        $$
        left|sum_{n=1}^kkleft(sinleft(frac{npi}{k}right)-left|cos left(frac{npi}{k}right)right|right)^+ -f(k)right|le k.
        $$         It follows that
        $$begin{eqnarray}
        lim_{ktoinfty}frac{f(k)}{k^2}&=&lim_{ktoinfty}sum_{n=1}^kleft(sinleft(frac{npi}{k}right)-left|cos left(frac{npi}{k}right)right|right)^+frac{1}{k}\&=&int_0^1 (sin pi x-|cos pi x|)^+mathrm{d}x\
        &=&2int_{frac{1}{4}}^{frac{1}{2}}(sinpi x-cospi x) mathrm{d}x\
        &=&frac{2}{pi}left[-cos pi x-sin pi xright]^{frac{1}{2}}_{frac{1}{4}}=frac{2}{pi}(sqrt{2}-1).
        end{eqnarray}$$






        share|cite|improve this answer









        $endgroup$



        Note that for all $a, bin Bbb R,$ it holds that $$
        (b-a)^+-1 le|{x:ale xle b}cap Bbb Z|le (b-a)^++1
        $$         where $c^+=max{c,0}$.
        Hence the number $N(k,n)$ of $min Bbb Z$ satisfying $$
        kleft|cos left(frac{npi}{k}right)right|le m le ksinleft(frac{npi}{k}right)
        $$         satisfies $$kleft(sinleft(frac{npi}{k}right)-left|cos left(frac{npi}{k}right)right|right)^+-1 le N(k,n)le kleft(sinleft(frac{npi}{k}right)-kleft|cos left(frac{npi}{k}right)right|right)^++1.$$ Since $f(k) =sum_{n=1}^k N(k,n)$, this gives
        $$
        left|sum_{n=1}^kkleft(sinleft(frac{npi}{k}right)-left|cos left(frac{npi}{k}right)right|right)^+ -f(k)right|le k.
        $$         It follows that
        $$begin{eqnarray}
        lim_{ktoinfty}frac{f(k)}{k^2}&=&lim_{ktoinfty}sum_{n=1}^kleft(sinleft(frac{npi}{k}right)-left|cos left(frac{npi}{k}right)right|right)^+frac{1}{k}\&=&int_0^1 (sin pi x-|cos pi x|)^+mathrm{d}x\
        &=&2int_{frac{1}{4}}^{frac{1}{2}}(sinpi x-cospi x) mathrm{d}x\
        &=&frac{2}{pi}left[-cos pi x-sin pi xright]^{frac{1}{2}}_{frac{1}{4}}=frac{2}{pi}(sqrt{2}-1).
        end{eqnarray}$$







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        answered Jan 29 at 16:51









        SongSong

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