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
$e^{theta_{11}}=a_{11}a_{1bar{1}}a_{1bar{2}}a_{2bar{1}}a_{bar{1}bar{2}}$.
calculus limits
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up vote
0
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favorite
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
add a comment |
up vote
0
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up vote
0
down vote
favorite
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
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
calculus limits
edited Nov 20 at 6:44
asked Nov 20 at 5:57
A. Riaz
254
254
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2 Answers
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0
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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 $.
@ Fabio, if the limit do not exist, then how the answer can be obtained as described above
– A. Riaz
Nov 20 at 6:46
add a comment |
up vote
0
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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.
@ Fred, if the limit do not exist, then how the answer can be obtained as described above
– A. Riaz
Nov 20 at 6:45
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $.
@ Fabio, if the limit do not exist, then how the answer can be obtained as described above
– A. Riaz
Nov 20 at 6:46
add a comment |
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 $.
@ Fabio, if the limit do not exist, then how the answer can be obtained as described above
– A. Riaz
Nov 20 at 6:46
add a comment |
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 $.
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 $.
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
add a comment |
@ 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
add a comment |
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.
@ Fred, if the limit do not exist, then how the answer can be obtained as described above
– A. Riaz
Nov 20 at 6:45
add a comment |
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.
@ Fred, if the limit do not exist, then how the answer can be obtained as described above
– A. Riaz
Nov 20 at 6:45
add a comment |
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.
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.
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
add a comment |
@ 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
add a comment |
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