Mixing
asymptotic limit of exponential function
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
asked Nov 20 '18 at 5:57
A. Riaz
254
add a comment |
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
asked Nov 20 '18 at 5:57
A. Riaz
254
Hi Riaz: It's generally considered pretty rude at MSE to get an answer to your question, and then change the question so that the answers are no longer valid. If you're still interested in the modified version, please ask a new question. In the mean time, I've reverted your edit. You can view the old version in the edit history (just above this comment), so you won't have to type it all out again :P
– aleph_two
yesterday
add a comment |
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
asked Nov 20 '18 at 5:57
A. Riaz
254
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
asked Nov 20 '18 at 5:57
A. Riaz
254
asked Nov 20 '18 at 5:57
A. Riaz
254
edited Nov 20 '18 at 6:44
asked Nov 20 '18 at 5:57
A. Riaz
254
asked Nov 20 '18 at 5:57
A. Riaz
254
asked Nov 20 '18 at 5:57
A. Riaz
254
254
Hi Riaz: It's generally considered pretty rude at MSE to get an answer to your question, and then change the question so that the answers are no longer valid. If you're still interested in the modified version, please ask a new question. In the mean time, I've reverted your edit. You can view the old version in the edit history (just above this comment), so you won't have to type it all out again :P
– aleph_two
yesterday
add a comment |
Hi Riaz: It's generally considered pretty rude at MSE to get an answer to your question, and then change the question so that the answers are no longer valid. If you're still interested in the modified version, please ask a new question. In the mean time, I've reverted your edit. You can view the old version in the edit history (just above this comment), so you won't have to type it all out again :P
– aleph_two
yesterday
Hi Riaz: It's generally considered pretty rude at MSE to get an answer to your question, and then change the question so that the answers are no longer valid. If you're still interested in the modified version, please ask a new question. In the mean time, I've reverted your edit. You can view the old version in the edit history (just above this comment), so you won't have to type it all out again :P
– aleph_two
yesterday
Hi Riaz: It's generally considered pretty rude at MSE to get an answer to your question, and then change the question so that the answers are no longer valid. If you're still interested in the modified version, please ask a new question. In the mean time, I've reverted your edit. You can view the old version in the edit history (just above this comment), so you won't have to type it all out again :P
– aleph_two
yesterday
add a comment |
2 Answers
2
active
oldest
votes
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 '18 at 6:07
Fabio Lucchini
7,83811426
@ Fabio, if the limit do not exist, then how the answer can be obtained as described above
– A. Riaz
Nov 20 '18 at 6:46
add a comment |
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 '18 at 6:08
Fred
44.2k1845
@ Fred, if the limit do not exist, then how the answer can be obtained as described above
– A. Riaz
Nov 20 '18 at 6:45
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 '18 at 6:07
Fabio Lucchini
7,83811426
@ Fabio, if the limit do not exist, then how the answer can be obtained as described above
– A. Riaz
Nov 20 '18 at 6:46
add a comment |
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 '18 at 6:07
Fabio Lucchini
7,83811426
@ Fabio, if the limit do not exist, then how the answer can be obtained as described above
– A. Riaz
Nov 20 '18 at 6:46
add a comment |
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 '18 at 6:07
Fabio Lucchini
7,83811426
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 '18 at 6:07
Fabio Lucchini
7,83811426
answered Nov 20 '18 at 6:07
Fabio Lucchini
7,83811426
answered Nov 20 '18 at 6:07
Fabio Lucchini
7,83811426
answered Nov 20 '18 at 6:07
Fabio Lucchini
7,83811426
7,83811426
@ Fabio, if the limit do not exist, then how the answer can be obtained as described above
– A. Riaz
Nov 20 '18 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 '18 at 6:46
@ Fabio, if the limit do not exist, then how the answer can be obtained as described above
– A. Riaz
Nov 20 '18 at 6:46
@ Fabio, if the limit do not exist, then how the answer can be obtained as described above
– A. Riaz
Nov 20 '18 at 6:46
add a comment |
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 '18 at 6:08
Fred
44.2k1845
@ Fred, if the limit do not exist, then how the answer can be obtained as described above
– A. Riaz
Nov 20 '18 at 6:45
add a comment |
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 '18 at 6:08
Fred
44.2k1845
@ Fred, if the limit do not exist, then how the answer can be obtained as described above
– A. Riaz
Nov 20 '18 at 6:45
add a comment |
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 '18 at 6:08
Fred
44.2k1845
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 '18 at 6:08
Fred
44.2k1845
answered Nov 20 '18 at 6:08
Fred
44.2k1845
answered Nov 20 '18 at 6:08
Fred
44.2k1845
answered Nov 20 '18 at 6:08
Fred
44.2k1845
44.2k1845
@ Fred, if the limit do not exist, then how the answer can be obtained as described above
– A. Riaz
Nov 20 '18 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 '18 at 6:45
@ Fred, if the limit do not exist, then how the answer can be obtained as described above
– A. Riaz
Nov 20 '18 at 6:45
@ Fred, if the limit do not exist, then how the answer can be obtained as described above
– A. Riaz
Nov 20 '18 at 6:45
add a comment |
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Hi Riaz: It's generally considered pretty rude at MSE to get an answer to your question, and then change the question so that the answers are no longer valid. If you're still interested in the modified version, please ask a new question. In the mean time, I've reverted your edit. You can view the old version in the edit history (just above this comment), so you won't have to type it all out again :P
– aleph_two
yesterday