asymptotic limit of exponential function











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What is the limit of following exponential function



$e^{dotimath k} $
when $k_{R}rightarrow -infty$ and $k_{R}rightarrow infty$?



where $k_{R}$ means real part of $k$.





In the book,
begin{equation}
r=frac{g_{1}+g_{2}}{1+f_{1}+f_{2}}, quad (1)
end{equation}

where
begin{eqnarray}
&& g_{1}=e^{eta_{1}+eta_{2}}, \
&& g_{2} = a_{12bar{1}}e^{eta_{1}+bar{eta}_{1}+eta_{2}}+
a_{12bar{2}}e^{eta_{1}+bar{eta}_{2}+eta_{2}}, \
&& f_{1} = a_{1bar{1}}e^{eta_{1}+bar{eta}_{1}} +
a_{1bar{2}}e^{eta_{1}+bar{eta}_{2}} +
a_{2bar{1}}e^{eta_{2}+bar{eta}_{1}} +
a_{2bar{2}}e^{eta_{2}+bar{eta}_{2}}, \
&& f_{2} =
a_{12bar{1}bar{2}}e^{eta_{1}+bar{eta}_{1}+eta_{2}+bar{eta}_{2}},
\
&& a_{ijbar{k}} = a_{ij} a_{ibar{k}} a_{jbar{k}}, quad
a_{ijbar{k}bar{l}} = a_{ij} a_{ibar{k}} a_{ibar{l}} a_{jbar{k}}
a_{jbar{l}} a_{bar{k}bar{l}}, \
&&a_{ij} = (p_{i}-p_{j})^{2}, ; a_{ibar{j}} =
frac{(p_{i}bar{p}_{j})^{2}}{(p_{i}+bar{p}_{j})^{2}}, ;
a_{bar{i}bar{j}} = (p_{i}-bar{p}_{j})^{2}
end{eqnarray}

where $eta_{i} = p_{i}^{-1}x + p_{i}t,; i=1,;2$. Asymptotic form
of equation (1):
begin{equation}
frac{p_{1R}}{|p_{1}|^{2}}textrm{sech}left(frac{eta_{1}+bar{eta}_{1}+theta_{11}}{2}right)
; eta_{2R}rightarrow infty ; eta_{1} sim O(1)
end{equation}

with
$e^{theta_{11}}=a_{11}a_{1bar{1}}a_{1bar{2}}a_{2bar{1}}a_{bar{1}bar{2}}$.










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    What is the limit of following exponential function



    $e^{dotimath k} $
    when $k_{R}rightarrow -infty$ and $k_{R}rightarrow infty$?



    where $k_{R}$ means real part of $k$.





    In the book,
    begin{equation}
    r=frac{g_{1}+g_{2}}{1+f_{1}+f_{2}}, quad (1)
    end{equation}

    where
    begin{eqnarray}
    && g_{1}=e^{eta_{1}+eta_{2}}, \
    && g_{2} = a_{12bar{1}}e^{eta_{1}+bar{eta}_{1}+eta_{2}}+
    a_{12bar{2}}e^{eta_{1}+bar{eta}_{2}+eta_{2}}, \
    && f_{1} = a_{1bar{1}}e^{eta_{1}+bar{eta}_{1}} +
    a_{1bar{2}}e^{eta_{1}+bar{eta}_{2}} +
    a_{2bar{1}}e^{eta_{2}+bar{eta}_{1}} +
    a_{2bar{2}}e^{eta_{2}+bar{eta}_{2}}, \
    && f_{2} =
    a_{12bar{1}bar{2}}e^{eta_{1}+bar{eta}_{1}+eta_{2}+bar{eta}_{2}},
    \
    && a_{ijbar{k}} = a_{ij} a_{ibar{k}} a_{jbar{k}}, quad
    a_{ijbar{k}bar{l}} = a_{ij} a_{ibar{k}} a_{ibar{l}} a_{jbar{k}}
    a_{jbar{l}} a_{bar{k}bar{l}}, \
    &&a_{ij} = (p_{i}-p_{j})^{2}, ; a_{ibar{j}} =
    frac{(p_{i}bar{p}_{j})^{2}}{(p_{i}+bar{p}_{j})^{2}}, ;
    a_{bar{i}bar{j}} = (p_{i}-bar{p}_{j})^{2}
    end{eqnarray}

    where $eta_{i} = p_{i}^{-1}x + p_{i}t,; i=1,;2$. Asymptotic form
    of equation (1):
    begin{equation}
    frac{p_{1R}}{|p_{1}|^{2}}textrm{sech}left(frac{eta_{1}+bar{eta}_{1}+theta_{11}}{2}right)
    ; eta_{2R}rightarrow infty ; eta_{1} sim O(1)
    end{equation}

    with
    $e^{theta_{11}}=a_{11}a_{1bar{1}}a_{1bar{2}}a_{2bar{1}}a_{bar{1}bar{2}}$.










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      up vote
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      What is the limit of following exponential function



      $e^{dotimath k} $
      when $k_{R}rightarrow -infty$ and $k_{R}rightarrow infty$?



      where $k_{R}$ means real part of $k$.





      In the book,
      begin{equation}
      r=frac{g_{1}+g_{2}}{1+f_{1}+f_{2}}, quad (1)
      end{equation}

      where
      begin{eqnarray}
      && g_{1}=e^{eta_{1}+eta_{2}}, \
      && g_{2} = a_{12bar{1}}e^{eta_{1}+bar{eta}_{1}+eta_{2}}+
      a_{12bar{2}}e^{eta_{1}+bar{eta}_{2}+eta_{2}}, \
      && f_{1} = a_{1bar{1}}e^{eta_{1}+bar{eta}_{1}} +
      a_{1bar{2}}e^{eta_{1}+bar{eta}_{2}} +
      a_{2bar{1}}e^{eta_{2}+bar{eta}_{1}} +
      a_{2bar{2}}e^{eta_{2}+bar{eta}_{2}}, \
      && f_{2} =
      a_{12bar{1}bar{2}}e^{eta_{1}+bar{eta}_{1}+eta_{2}+bar{eta}_{2}},
      \
      && a_{ijbar{k}} = a_{ij} a_{ibar{k}} a_{jbar{k}}, quad
      a_{ijbar{k}bar{l}} = a_{ij} a_{ibar{k}} a_{ibar{l}} a_{jbar{k}}
      a_{jbar{l}} a_{bar{k}bar{l}}, \
      &&a_{ij} = (p_{i}-p_{j})^{2}, ; a_{ibar{j}} =
      frac{(p_{i}bar{p}_{j})^{2}}{(p_{i}+bar{p}_{j})^{2}}, ;
      a_{bar{i}bar{j}} = (p_{i}-bar{p}_{j})^{2}
      end{eqnarray}

      where $eta_{i} = p_{i}^{-1}x + p_{i}t,; i=1,;2$. Asymptotic form
      of equation (1):
      begin{equation}
      frac{p_{1R}}{|p_{1}|^{2}}textrm{sech}left(frac{eta_{1}+bar{eta}_{1}+theta_{11}}{2}right)
      ; eta_{2R}rightarrow infty ; eta_{1} sim O(1)
      end{equation}

      with
      $e^{theta_{11}}=a_{11}a_{1bar{1}}a_{1bar{2}}a_{2bar{1}}a_{bar{1}bar{2}}$.










      share|cite|improve this question

















      What is the limit of following exponential function



      $e^{dotimath k} $
      when $k_{R}rightarrow -infty$ and $k_{R}rightarrow infty$?



      where $k_{R}$ means real part of $k$.





      In the book,
      begin{equation}
      r=frac{g_{1}+g_{2}}{1+f_{1}+f_{2}}, quad (1)
      end{equation}

      where
      begin{eqnarray}
      && g_{1}=e^{eta_{1}+eta_{2}}, \
      && g_{2} = a_{12bar{1}}e^{eta_{1}+bar{eta}_{1}+eta_{2}}+
      a_{12bar{2}}e^{eta_{1}+bar{eta}_{2}+eta_{2}}, \
      && f_{1} = a_{1bar{1}}e^{eta_{1}+bar{eta}_{1}} +
      a_{1bar{2}}e^{eta_{1}+bar{eta}_{2}} +
      a_{2bar{1}}e^{eta_{2}+bar{eta}_{1}} +
      a_{2bar{2}}e^{eta_{2}+bar{eta}_{2}}, \
      && f_{2} =
      a_{12bar{1}bar{2}}e^{eta_{1}+bar{eta}_{1}+eta_{2}+bar{eta}_{2}},
      \
      && a_{ijbar{k}} = a_{ij} a_{ibar{k}} a_{jbar{k}}, quad
      a_{ijbar{k}bar{l}} = a_{ij} a_{ibar{k}} a_{ibar{l}} a_{jbar{k}}
      a_{jbar{l}} a_{bar{k}bar{l}}, \
      &&a_{ij} = (p_{i}-p_{j})^{2}, ; a_{ibar{j}} =
      frac{(p_{i}bar{p}_{j})^{2}}{(p_{i}+bar{p}_{j})^{2}}, ;
      a_{bar{i}bar{j}} = (p_{i}-bar{p}_{j})^{2}
      end{eqnarray}

      where $eta_{i} = p_{i}^{-1}x + p_{i}t,; i=1,;2$. Asymptotic form
      of equation (1):
      begin{equation}
      frac{p_{1R}}{|p_{1}|^{2}}textrm{sech}left(frac{eta_{1}+bar{eta}_{1}+theta_{11}}{2}right)
      ; eta_{2R}rightarrow infty ; eta_{1} sim O(1)
      end{equation}

      with
      $e^{theta_{11}}=a_{11}a_{1bar{1}}a_{1bar{2}}a_{2bar{1}}a_{bar{1}bar{2}}$.







      calculus limits






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      edited Nov 20 at 6:44

























      asked Nov 20 at 5:57









      A. Riaz

      254




      254






















          2 Answers
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          If $k=a+ib $, then
          begin{align}
          e^{ik}
          &=e^{ia-b}\
          &=e^{-b}(cos (a)+isin (a))
          end{align}

          hence the limit doesn't exists for $atopminfty $.






          share|cite|improve this answer





















          • @ Fabio, if the limit do not exist, then how the answer can be obtained as described above
            – A. Riaz
            Nov 20 at 6:46


















          up vote
          0
          down vote













          Let $k=x+iy$ with $x,y in mathbb R$. Then $e^{ik}=e^{ix}e^{-y}$



          Let $x_n:=n pi$ for $n in mathbb N$. Then $e^{i x_n}=(-1)^n=e^{- ix_n}$.



          This shows that the limits in question do not exist.






          share|cite|improve this answer





















          • @ Fred, if the limit do not exist, then how the answer can be obtained as described above
            – A. Riaz
            Nov 20 at 6:45











          Your Answer





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          2 Answers
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          2 Answers
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          up vote
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          down vote













          If $k=a+ib $, then
          begin{align}
          e^{ik}
          &=e^{ia-b}\
          &=e^{-b}(cos (a)+isin (a))
          end{align}

          hence the limit doesn't exists for $atopminfty $.






          share|cite|improve this answer





















          • @ Fabio, if the limit do not exist, then how the answer can be obtained as described above
            – A. Riaz
            Nov 20 at 6:46















          up vote
          0
          down vote













          If $k=a+ib $, then
          begin{align}
          e^{ik}
          &=e^{ia-b}\
          &=e^{-b}(cos (a)+isin (a))
          end{align}

          hence the limit doesn't exists for $atopminfty $.






          share|cite|improve this answer





















          • @ Fabio, if the limit do not exist, then how the answer can be obtained as described above
            – A. Riaz
            Nov 20 at 6:46













          up vote
          0
          down vote










          up vote
          0
          down vote









          If $k=a+ib $, then
          begin{align}
          e^{ik}
          &=e^{ia-b}\
          &=e^{-b}(cos (a)+isin (a))
          end{align}

          hence the limit doesn't exists for $atopminfty $.






          share|cite|improve this answer












          If $k=a+ib $, then
          begin{align}
          e^{ik}
          &=e^{ia-b}\
          &=e^{-b}(cos (a)+isin (a))
          end{align}

          hence the limit doesn't exists for $atopminfty $.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 20 at 6:07









          Fabio Lucchini

          7,81311326




          7,81311326












          • @ Fabio, if the limit do not exist, then how the answer can be obtained as described above
            – A. Riaz
            Nov 20 at 6:46


















          • @ Fabio, if the limit do not exist, then how the answer can be obtained as described above
            – A. Riaz
            Nov 20 at 6:46
















          @ Fabio, if the limit do not exist, then how the answer can be obtained as described above
          – A. Riaz
          Nov 20 at 6:46




          @ Fabio, if the limit do not exist, then how the answer can be obtained as described above
          – A. Riaz
          Nov 20 at 6:46










          up vote
          0
          down vote













          Let $k=x+iy$ with $x,y in mathbb R$. Then $e^{ik}=e^{ix}e^{-y}$



          Let $x_n:=n pi$ for $n in mathbb N$. Then $e^{i x_n}=(-1)^n=e^{- ix_n}$.



          This shows that the limits in question do not exist.






          share|cite|improve this answer





















          • @ Fred, if the limit do not exist, then how the answer can be obtained as described above
            – A. Riaz
            Nov 20 at 6:45















          up vote
          0
          down vote













          Let $k=x+iy$ with $x,y in mathbb R$. Then $e^{ik}=e^{ix}e^{-y}$



          Let $x_n:=n pi$ for $n in mathbb N$. Then $e^{i x_n}=(-1)^n=e^{- ix_n}$.



          This shows that the limits in question do not exist.






          share|cite|improve this answer





















          • @ Fred, if the limit do not exist, then how the answer can be obtained as described above
            – A. Riaz
            Nov 20 at 6:45













          up vote
          0
          down vote










          up vote
          0
          down vote









          Let $k=x+iy$ with $x,y in mathbb R$. Then $e^{ik}=e^{ix}e^{-y}$



          Let $x_n:=n pi$ for $n in mathbb N$. Then $e^{i x_n}=(-1)^n=e^{- ix_n}$.



          This shows that the limits in question do not exist.






          share|cite|improve this answer












          Let $k=x+iy$ with $x,y in mathbb R$. Then $e^{ik}=e^{ix}e^{-y}$



          Let $x_n:=n pi$ for $n in mathbb N$. Then $e^{i x_n}=(-1)^n=e^{- ix_n}$.



          This shows that the limits in question do not exist.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 20 at 6:08









          Fred

          43.3k1644




          43.3k1644












          • @ Fred, if the limit do not exist, then how the answer can be obtained as described above
            – A. Riaz
            Nov 20 at 6:45


















          • @ Fred, if the limit do not exist, then how the answer can be obtained as described above
            – A. Riaz
            Nov 20 at 6:45
















          @ Fred, if the limit do not exist, then how the answer can be obtained as described above
          – A. Riaz
          Nov 20 at 6:45




          @ Fred, if the limit do not exist, then how the answer can be obtained as described above
          – A. Riaz
          Nov 20 at 6:45


















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