Horrible limit envolving floor function












10












$begingroup$


Let $xin [0,1]$, $ellinmathbb{Z}$ and $tau>0$. I want to calculate
$$lim_{Ltoinfty}sum_{k=0}^{lfloor tau L^2rfloor}frac{1}{2^{lfloor tau L^2rfloor}}binom{lfloor tau L^2rfloor}{k}cosleft(2pi ellfrac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L}right).$$



I think the result is $exp(-2pi^2ell^2tau)cos(2piell x)$ but I have no idea in how to prove it.



I tried estimating the binomial with Stirling's aproximation but without success.






real-analysis limits







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    Wow, this is really horrible.
    $endgroup$
    – thesagniksaha
    Dec 15 '18 at 21:25






  • 1




    $begingroup$
    Have you tried replacing the cosine by an complex exponential, using the Newton formula and studying the different factors?
    $endgroup$
    – Mindlack
    Dec 15 '18 at 21:36












  • $begingroup$
    @Mindlack Which formula by Newton?
    $endgroup$
    – Gabriel Ribeiro
    Dec 15 '18 at 22:12










  • $begingroup$
    I meant $(1+z)^n=ldots$.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 22:22






  • 1




    $begingroup$
    Factor out everything that does not depend on $k$.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 22:29
















10












$begingroup$


Let $xin [0,1]$, $ellinmathbb{Z}$ and $tau>0$. I want to calculate
$$lim_{Ltoinfty}sum_{k=0}^{lfloor tau L^2rfloor}frac{1}{2^{lfloor tau L^2rfloor}}binom{lfloor tau L^2rfloor}{k}cosleft(2pi ellfrac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L}right).$$



I think the result is $exp(-2pi^2ell^2tau)cos(2piell x)$ but I have no idea in how to prove it.



I tried estimating the binomial with Stirling's aproximation but without success.






real-analysis limits







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    Wow, this is really horrible.
    $endgroup$
    – thesagniksaha
    Dec 15 '18 at 21:25






  • 1




    $begingroup$
    Have you tried replacing the cosine by an complex exponential, using the Newton formula and studying the different factors?
    $endgroup$
    – Mindlack
    Dec 15 '18 at 21:36












  • $begingroup$
    @Mindlack Which formula by Newton?
    $endgroup$
    – Gabriel Ribeiro
    Dec 15 '18 at 22:12










  • $begingroup$
    I meant $(1+z)^n=ldots$.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 22:22






  • 1




    $begingroup$
    Factor out everything that does not depend on $k$.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 22:29














10












10








10


1



$begingroup$


Let $xin [0,1]$, $ellinmathbb{Z}$ and $tau>0$. I want to calculate
$$lim_{Ltoinfty}sum_{k=0}^{lfloor tau L^2rfloor}frac{1}{2^{lfloor tau L^2rfloor}}binom{lfloor tau L^2rfloor}{k}cosleft(2pi ellfrac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L}right).$$



I think the result is $exp(-2pi^2ell^2tau)cos(2piell x)$ but I have no idea in how to prove it.



I tried estimating the binomial with Stirling's aproximation but without success.






real-analysis limits







share|cite|improve this question









$endgroup$




Let $xin [0,1]$, $ellinmathbb{Z}$ and $tau>0$. I want to calculate
$$lim_{Ltoinfty}sum_{k=0}^{lfloor tau L^2rfloor}frac{1}{2^{lfloor tau L^2rfloor}}binom{lfloor tau L^2rfloor}{k}cosleft(2pi ellfrac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L}right).$$



I think the result is $exp(-2pi^2ell^2tau)cos(2piell x)$ but I have no idea in how to prove it.



I tried estimating the binomial with Stirling's aproximation but without success.






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










asked Dec 15 '18 at 21:24









Gabriel RibeiroGabriel Ribeiro

1,454523




1,454523








  • 4




    $begingroup$
    Wow, this is really horrible.
    $endgroup$
    – thesagniksaha
    Dec 15 '18 at 21:25






  • 1




    $begingroup$
    Have you tried replacing the cosine by an complex exponential, using the Newton formula and studying the different factors?
    $endgroup$
    – Mindlack
    Dec 15 '18 at 21:36












  • $begingroup$
    @Mindlack Which formula by Newton?
    $endgroup$
    – Gabriel Ribeiro
    Dec 15 '18 at 22:12










  • $begingroup$
    I meant $(1+z)^n=ldots$.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 22:22






  • 1




    $begingroup$
    Factor out everything that does not depend on $k$.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 22:29














  • 4




    $begingroup$
    Wow, this is really horrible.
    $endgroup$
    – thesagniksaha
    Dec 15 '18 at 21:25






  • 1




    $begingroup$
    Have you tried replacing the cosine by an complex exponential, using the Newton formula and studying the different factors?
    $endgroup$
    – Mindlack
    Dec 15 '18 at 21:36












  • $begingroup$
    @Mindlack Which formula by Newton?
    $endgroup$
    – Gabriel Ribeiro
    Dec 15 '18 at 22:12










  • $begingroup$
    I meant $(1+z)^n=ldots$.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 22:22






  • 1




    $begingroup$
    Factor out everything that does not depend on $k$.
    $endgroup$
    – Mindlack
    Dec 15 '18 at 22:29








4




4




$begingroup$
Wow, this is really horrible.
$endgroup$
– thesagniksaha
Dec 15 '18 at 21:25




$begingroup$
Wow, this is really horrible.
$endgroup$
– thesagniksaha
Dec 15 '18 at 21:25




1




1




$begingroup$
Have you tried replacing the cosine by an complex exponential, using the Newton formula and studying the different factors?
$endgroup$
– Mindlack
Dec 15 '18 at 21:36






$begingroup$
Have you tried replacing the cosine by an complex exponential, using the Newton formula and studying the different factors?
$endgroup$
– Mindlack
Dec 15 '18 at 21:36














$begingroup$
@Mindlack Which formula by Newton?
$endgroup$
– Gabriel Ribeiro
Dec 15 '18 at 22:12




$begingroup$
@Mindlack Which formula by Newton?
$endgroup$
– Gabriel Ribeiro
Dec 15 '18 at 22:12












$begingroup$
I meant $(1+z)^n=ldots$.
$endgroup$
– Mindlack
Dec 15 '18 at 22:22




$begingroup$
I meant $(1+z)^n=ldots$.
$endgroup$
– Mindlack
Dec 15 '18 at 22:22




1




1




$begingroup$
Factor out everything that does not depend on $k$.
$endgroup$
– Mindlack
Dec 15 '18 at 22:29




$begingroup$
Factor out everything that does not depend on $k$.
$endgroup$
– Mindlack
Dec 15 '18 at 22:29










1 Answer
1






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10












$begingroup$

You are correct. We have that
$$begin{aligned}
& sum_{k=0}^{lfloor tau L^2rfloor}frac{1}{2^{lfloor tau L^2rfloor}}binom{lfloor tau L^2rfloor}{k}cosleft(2pi ellfrac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L}right) \
= &sum_{k=0}^{lfloor tau L^2rfloor}frac{1}{2^{lfloor tau L^2rfloor}}binom{lfloor tau L^2rfloor}{k}Releft(exp(i2piell frac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L})right) \
=& frac{1}{2^{lfloor tau L^2rfloor}}Releft(sum_{k=0}^{lfloor tau L^2rfloor}binom{lfloor tau L^2rfloor}{k}exp(i2piell frac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L})right) \
=& frac{1}{2^{lfloor tau L^2rfloor}}Releft(expleft(i2piellfrac{lfloor xLrfloor -lfloortau L^2rfloor}{L}right)left(1+expleft(ifrac{4piell}{L}right)right)^{lfloor tau L^2rfloor}right).
end{aligned}$$
Since we are taking the limit as $Ltoinfty$ we have that
$$frac{lfloor xLrfloor -lfloortau L^2rfloor}{L}to x-tau L$$
so we can consider the expression
$$frac{1}{2^{lfloor tau L^2rfloor}}Releft(expleft(i2piell(x-tau L)right)left(1+expleft(ifrac{4piell}{L}right)right)^{lfloor tau L^2rfloor}right).$$
Again since we only care about $Ltoinfty$ we can let $Lmapsto L/sqrt{tau}$ so that we are considering the expression
$$begin{aligned}
&frac{1}{2^{L^2}}Releft(expleft(i2piell(x-sqrt{tau} L)right)left(1+expleft(ifrac{4piellsqrt{tau}}{L}right)right)^{L^2}right) \
=&Releft(exp(i2piell x)left(frac{expleft(-ifrac{2piellsqrt{tau}}{L}right)+expleft(ifrac{2piellsqrt{tau}}{L}right)}{2}right)^{L^2}right) \
=&cosleft(frac{2piellsqrt{tau}}{L}right)^{L^2}Releft(exp(i2piell x)right) \
=& cosleft(frac{2piellsqrt{tau}}{L}right)^{L^2}cos(2piell x).
end{aligned}$$
Taking the limit we indeed get
$$lim_{Ltoinfty} cosleft(frac{2piellsqrt{tau}}{L}right)^{L^2}cos(2piell x)=exp(-2pi^2ell^2tau)cos(2piell x).$$






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    10












    $begingroup$

    You are correct. We have that
    $$begin{aligned}
    & sum_{k=0}^{lfloor tau L^2rfloor}frac{1}{2^{lfloor tau L^2rfloor}}binom{lfloor tau L^2rfloor}{k}cosleft(2pi ellfrac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L}right) \
    = &sum_{k=0}^{lfloor tau L^2rfloor}frac{1}{2^{lfloor tau L^2rfloor}}binom{lfloor tau L^2rfloor}{k}Releft(exp(i2piell frac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L})right) \
    =& frac{1}{2^{lfloor tau L^2rfloor}}Releft(sum_{k=0}^{lfloor tau L^2rfloor}binom{lfloor tau L^2rfloor}{k}exp(i2piell frac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L})right) \
    =& frac{1}{2^{lfloor tau L^2rfloor}}Releft(expleft(i2piellfrac{lfloor xLrfloor -lfloortau L^2rfloor}{L}right)left(1+expleft(ifrac{4piell}{L}right)right)^{lfloor tau L^2rfloor}right).
    end{aligned}$$
    Since we are taking the limit as $Ltoinfty$ we have that
    $$frac{lfloor xLrfloor -lfloortau L^2rfloor}{L}to x-tau L$$
    so we can consider the expression
    $$frac{1}{2^{lfloor tau L^2rfloor}}Releft(expleft(i2piell(x-tau L)right)left(1+expleft(ifrac{4piell}{L}right)right)^{lfloor tau L^2rfloor}right).$$
    Again since we only care about $Ltoinfty$ we can let $Lmapsto L/sqrt{tau}$ so that we are considering the expression
    $$begin{aligned}
    &frac{1}{2^{L^2}}Releft(expleft(i2piell(x-sqrt{tau} L)right)left(1+expleft(ifrac{4piellsqrt{tau}}{L}right)right)^{L^2}right) \
    =&Releft(exp(i2piell x)left(frac{expleft(-ifrac{2piellsqrt{tau}}{L}right)+expleft(ifrac{2piellsqrt{tau}}{L}right)}{2}right)^{L^2}right) \
    =&cosleft(frac{2piellsqrt{tau}}{L}right)^{L^2}Releft(exp(i2piell x)right) \
    =& cosleft(frac{2piellsqrt{tau}}{L}right)^{L^2}cos(2piell x).
    end{aligned}$$
    Taking the limit we indeed get
    $$lim_{Ltoinfty} cosleft(frac{2piellsqrt{tau}}{L}right)^{L^2}cos(2piell x)=exp(-2pi^2ell^2tau)cos(2piell x).$$






    share|cite|improve this answer









    $endgroup$


















      10












      $begingroup$

      You are correct. We have that
      $$begin{aligned}
      & sum_{k=0}^{lfloor tau L^2rfloor}frac{1}{2^{lfloor tau L^2rfloor}}binom{lfloor tau L^2rfloor}{k}cosleft(2pi ellfrac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L}right) \
      = &sum_{k=0}^{lfloor tau L^2rfloor}frac{1}{2^{lfloor tau L^2rfloor}}binom{lfloor tau L^2rfloor}{k}Releft(exp(i2piell frac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L})right) \
      =& frac{1}{2^{lfloor tau L^2rfloor}}Releft(sum_{k=0}^{lfloor tau L^2rfloor}binom{lfloor tau L^2rfloor}{k}exp(i2piell frac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L})right) \
      =& frac{1}{2^{lfloor tau L^2rfloor}}Releft(expleft(i2piellfrac{lfloor xLrfloor -lfloortau L^2rfloor}{L}right)left(1+expleft(ifrac{4piell}{L}right)right)^{lfloor tau L^2rfloor}right).
      end{aligned}$$
      Since we are taking the limit as $Ltoinfty$ we have that
      $$frac{lfloor xLrfloor -lfloortau L^2rfloor}{L}to x-tau L$$
      so we can consider the expression
      $$frac{1}{2^{lfloor tau L^2rfloor}}Releft(expleft(i2piell(x-tau L)right)left(1+expleft(ifrac{4piell}{L}right)right)^{lfloor tau L^2rfloor}right).$$
      Again since we only care about $Ltoinfty$ we can let $Lmapsto L/sqrt{tau}$ so that we are considering the expression
      $$begin{aligned}
      &frac{1}{2^{L^2}}Releft(expleft(i2piell(x-sqrt{tau} L)right)left(1+expleft(ifrac{4piellsqrt{tau}}{L}right)right)^{L^2}right) \
      =&Releft(exp(i2piell x)left(frac{expleft(-ifrac{2piellsqrt{tau}}{L}right)+expleft(ifrac{2piellsqrt{tau}}{L}right)}{2}right)^{L^2}right) \
      =&cosleft(frac{2piellsqrt{tau}}{L}right)^{L^2}Releft(exp(i2piell x)right) \
      =& cosleft(frac{2piellsqrt{tau}}{L}right)^{L^2}cos(2piell x).
      end{aligned}$$
      Taking the limit we indeed get
      $$lim_{Ltoinfty} cosleft(frac{2piellsqrt{tau}}{L}right)^{L^2}cos(2piell x)=exp(-2pi^2ell^2tau)cos(2piell x).$$






      share|cite|improve this answer









      $endgroup$
















        10












        10








        10





        $begingroup$

        You are correct. We have that
        $$begin{aligned}
        & sum_{k=0}^{lfloor tau L^2rfloor}frac{1}{2^{lfloor tau L^2rfloor}}binom{lfloor tau L^2rfloor}{k}cosleft(2pi ellfrac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L}right) \
        = &sum_{k=0}^{lfloor tau L^2rfloor}frac{1}{2^{lfloor tau L^2rfloor}}binom{lfloor tau L^2rfloor}{k}Releft(exp(i2piell frac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L})right) \
        =& frac{1}{2^{lfloor tau L^2rfloor}}Releft(sum_{k=0}^{lfloor tau L^2rfloor}binom{lfloor tau L^2rfloor}{k}exp(i2piell frac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L})right) \
        =& frac{1}{2^{lfloor tau L^2rfloor}}Releft(expleft(i2piellfrac{lfloor xLrfloor -lfloortau L^2rfloor}{L}right)left(1+expleft(ifrac{4piell}{L}right)right)^{lfloor tau L^2rfloor}right).
        end{aligned}$$
        Since we are taking the limit as $Ltoinfty$ we have that
        $$frac{lfloor xLrfloor -lfloortau L^2rfloor}{L}to x-tau L$$
        so we can consider the expression
        $$frac{1}{2^{lfloor tau L^2rfloor}}Releft(expleft(i2piell(x-tau L)right)left(1+expleft(ifrac{4piell}{L}right)right)^{lfloor tau L^2rfloor}right).$$
        Again since we only care about $Ltoinfty$ we can let $Lmapsto L/sqrt{tau}$ so that we are considering the expression
        $$begin{aligned}
        &frac{1}{2^{L^2}}Releft(expleft(i2piell(x-sqrt{tau} L)right)left(1+expleft(ifrac{4piellsqrt{tau}}{L}right)right)^{L^2}right) \
        =&Releft(exp(i2piell x)left(frac{expleft(-ifrac{2piellsqrt{tau}}{L}right)+expleft(ifrac{2piellsqrt{tau}}{L}right)}{2}right)^{L^2}right) \
        =&cosleft(frac{2piellsqrt{tau}}{L}right)^{L^2}Releft(exp(i2piell x)right) \
        =& cosleft(frac{2piellsqrt{tau}}{L}right)^{L^2}cos(2piell x).
        end{aligned}$$
        Taking the limit we indeed get
        $$lim_{Ltoinfty} cosleft(frac{2piellsqrt{tau}}{L}right)^{L^2}cos(2piell x)=exp(-2pi^2ell^2tau)cos(2piell x).$$






        share|cite|improve this answer









        $endgroup$



        You are correct. We have that
        $$begin{aligned}
        & sum_{k=0}^{lfloor tau L^2rfloor}frac{1}{2^{lfloor tau L^2rfloor}}binom{lfloor tau L^2rfloor}{k}cosleft(2pi ellfrac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L}right) \
        = &sum_{k=0}^{lfloor tau L^2rfloor}frac{1}{2^{lfloor tau L^2rfloor}}binom{lfloor tau L^2rfloor}{k}Releft(exp(i2piell frac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L})right) \
        =& frac{1}{2^{lfloor tau L^2rfloor}}Releft(sum_{k=0}^{lfloor tau L^2rfloor}binom{lfloor tau L^2rfloor}{k}exp(i2piell frac{lfloor xLrfloor-lfloor tau L^2rfloor+2k}{L})right) \
        =& frac{1}{2^{lfloor tau L^2rfloor}}Releft(expleft(i2piellfrac{lfloor xLrfloor -lfloortau L^2rfloor}{L}right)left(1+expleft(ifrac{4piell}{L}right)right)^{lfloor tau L^2rfloor}right).
        end{aligned}$$
        Since we are taking the limit as $Ltoinfty$ we have that
        $$frac{lfloor xLrfloor -lfloortau L^2rfloor}{L}to x-tau L$$
        so we can consider the expression
        $$frac{1}{2^{lfloor tau L^2rfloor}}Releft(expleft(i2piell(x-tau L)right)left(1+expleft(ifrac{4piell}{L}right)right)^{lfloor tau L^2rfloor}right).$$
        Again since we only care about $Ltoinfty$ we can let $Lmapsto L/sqrt{tau}$ so that we are considering the expression
        $$begin{aligned}
        &frac{1}{2^{L^2}}Releft(expleft(i2piell(x-sqrt{tau} L)right)left(1+expleft(ifrac{4piellsqrt{tau}}{L}right)right)^{L^2}right) \
        =&Releft(exp(i2piell x)left(frac{expleft(-ifrac{2piellsqrt{tau}}{L}right)+expleft(ifrac{2piellsqrt{tau}}{L}right)}{2}right)^{L^2}right) \
        =&cosleft(frac{2piellsqrt{tau}}{L}right)^{L^2}Releft(exp(i2piell x)right) \
        =& cosleft(frac{2piellsqrt{tau}}{L}right)^{L^2}cos(2piell x).
        end{aligned}$$
        Taking the limit we indeed get
        $$lim_{Ltoinfty} cosleft(frac{2piellsqrt{tau}}{L}right)^{L^2}cos(2piell x)=exp(-2pi^2ell^2tau)cos(2piell x).$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '18 at 0:27









        Will FisherWill Fisher

        4,04811032




        4,04811032






























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            Ida-Boy-Ed-Garten

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            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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