$lambda-z-e^{-z}=0$ has one solution in the right half plane
$begingroup$
Let $lambda > 1$ , want to show that the equation $$lambda-z-e^{-z}=0$$ has exactly one solution in the right half plane ${z:Re(z)>0}$. Moreover, the solution must be real.
I tried to use Rouche's theorem on $g(z)=lambda - z$ and $f(z)=e^{-z}$ to get that the number of zeros of $f+g$ and the number of zeros of $g(z)$ is the same, and since $g(z)$ has only one solution then the equation about must also have one solution the problem is I don't know how to choose the correct curve $gamma$ such that this will work.
for the second part I used the IVT to show that $lambda -x-e^{-x}$ has a zero in $(0,lambda)$ to conclude that the solution is real. is this acceptable? Thank you for your help.
complex-analysis roots
edited Jul 3 '13 at 14:16
Antonio Vargas
20.7k245111
asked Mar 4 '13 at 2:04
i.a.mi.a.m
1,595725
$endgroup$
add a comment |
$begingroup$
Let $lambda > 1$ , want to show that the equation $$lambda-z-e^{-z}=0$$ has exactly one solution in the right half plane ${z:Re(z)>0}$. Moreover, the solution must be real.
I tried to use Rouche's theorem on $g(z)=lambda - z$ and $f(z)=e^{-z}$ to get that the number of zeros of $f+g$ and the number of zeros of $g(z)$ is the same, and since $g(z)$ has only one solution then the equation about must also have one solution the problem is I don't know how to choose the correct curve $gamma$ such that this will work.
for the second part I used the IVT to show that $lambda -x-e^{-x}$ has a zero in $(0,lambda)$ to conclude that the solution is real. is this acceptable? Thank you for your help.
complex-analysis roots
edited Jul 3 '13 at 14:16
Antonio Vargas
20.7k245111
asked Mar 4 '13 at 2:04
i.a.mi.a.m
1,595725
$endgroup$
$begingroup$
How did you use Roche's theorem? Don't you need $|f(z)| < |g(z)|$ for say $z = lambda$.
$endgroup$
– muzzlator
Mar 4 '13 at 2:15
$begingroup$
@muzzlator I took a small circle with center $lambda$ and the inequality you wrote should be true for all values on (not in) this circle, and this is that case.
$endgroup$
– i.a.m
Mar 4 '13 at 2:19
$begingroup$
Ah, I see. Yeah that looks like it works
$endgroup$
– muzzlator
Mar 4 '13 at 2:23
$begingroup$
Rouche is a good idea. But if you apply it to a small circle, how does this apply to the whole half-plane?
$endgroup$
– Julien
Mar 4 '13 at 2:56
$begingroup$
@julien I was thinking using part two i.e the real case we established that the equation has a zero in the right half plane, now lets assume we have another zero in the right half plane and take a circle containing both zeros to get a contradictoin, and conclude that there is exactly one solution. the problem I'm facing now is that I can't get the inequality of Rouche's theorem.
$endgroup$
– i.a.m
Mar 4 '13 at 3:02
add a comment |
$begingroup$
Let $lambda > 1$ , want to show that the equation $$lambda-z-e^{-z}=0$$ has exactly one solution in the right half plane ${z:Re(z)>0}$. Moreover, the solution must be real.
I tried to use Rouche's theorem on $g(z)=lambda - z$ and $f(z)=e^{-z}$ to get that the number of zeros of $f+g$ and the number of zeros of $g(z)$ is the same, and since $g(z)$ has only one solution then the equation about must also have one solution the problem is I don't know how to choose the correct curve $gamma$ such that this will work.
for the second part I used the IVT to show that $lambda -x-e^{-x}$ has a zero in $(0,lambda)$ to conclude that the solution is real. is this acceptable? Thank you for your help.
complex-analysis roots
edited Jul 3 '13 at 14:16
Antonio Vargas
20.7k245111
asked Mar 4 '13 at 2:04
i.a.mi.a.m
1,595725
$endgroup$
Let $lambda > 1$ , want to show that the equation $$lambda-z-e^{-z}=0$$ has exactly one solution in the right half plane ${z:Re(z)>0}$. Moreover, the solution must be real.
I tried to use Rouche's theorem on $g(z)=lambda - z$ and $f(z)=e^{-z}$ to get that the number of zeros of $f+g$ and the number of zeros of $g(z)$ is the same, and since $g(z)$ has only one solution then the equation about must also have one solution the problem is I don't know how to choose the correct curve $gamma$ such that this will work.
for the second part I used the IVT to show that $lambda -x-e^{-x}$ has a zero in $(0,lambda)$ to conclude that the solution is real. is this acceptable? Thank you for your help.
complex-analysis roots
complex-analysis roots
edited Jul 3 '13 at 14:16
Antonio Vargas
20.7k245111
asked Mar 4 '13 at 2:04
i.a.mi.a.m
1,595725
edited Jul 3 '13 at 14:16
Antonio Vargas
20.7k245111
asked Mar 4 '13 at 2:04
i.a.mi.a.m
1,595725
edited Jul 3 '13 at 14:16
Antonio Vargas
20.7k245111
edited Jul 3 '13 at 14:16
Antonio Vargas
20.7k245111
edited Jul 3 '13 at 14:16
Antonio Vargas
20.7k245111
20.7k245111
asked Mar 4 '13 at 2:04
i.a.mi.a.m
1,595725
asked Mar 4 '13 at 2:04
i.a.mi.a.m
1,595725
asked Mar 4 '13 at 2:04
i.a.mi.a.m
1,595725
1,595725
$begingroup$
How did you use Roche's theorem? Don't you need $|f(z)| < |g(z)|$ for say $z = lambda$.
$endgroup$
– muzzlator
Mar 4 '13 at 2:15
$begingroup$
@muzzlator I took a small circle with center $lambda$ and the inequality you wrote should be true for all values on (not in) this circle, and this is that case.
$endgroup$
– i.a.m
Mar 4 '13 at 2:19
$begingroup$
Ah, I see. Yeah that looks like it works
$endgroup$
– muzzlator
Mar 4 '13 at 2:23
$begingroup$
Rouche is a good idea. But if you apply it to a small circle, how does this apply to the whole half-plane?
$endgroup$
– Julien
Mar 4 '13 at 2:56
$begingroup$
@julien I was thinking using part two i.e the real case we established that the equation has a zero in the right half plane, now lets assume we have another zero in the right half plane and take a circle containing both zeros to get a contradictoin, and conclude that there is exactly one solution. the problem I'm facing now is that I can't get the inequality of Rouche's theorem.
$endgroup$
– i.a.m
Mar 4 '13 at 3:02
add a comment |
$begingroup$
How did you use Roche's theorem? Don't you need $|f(z)| < |g(z)|$ for say $z = lambda$.
$endgroup$
– muzzlator
Mar 4 '13 at 2:15
$begingroup$
@muzzlator I took a small circle with center $lambda$ and the inequality you wrote should be true for all values on (not in) this circle, and this is that case.
$endgroup$
– i.a.m
Mar 4 '13 at 2:19
$begingroup$
Ah, I see. Yeah that looks like it works
$endgroup$
– muzzlator
Mar 4 '13 at 2:23
$begingroup$
Rouche is a good idea. But if you apply it to a small circle, how does this apply to the whole half-plane?
$endgroup$
– Julien
Mar 4 '13 at 2:56
$begingroup$
@julien I was thinking using part two i.e the real case we established that the equation has a zero in the right half plane, now lets assume we have another zero in the right half plane and take a circle containing both zeros to get a contradictoin, and conclude that there is exactly one solution. the problem I'm facing now is that I can't get the inequality of Rouche's theorem.
$endgroup$
– i.a.m
Mar 4 '13 at 3:02
$begingroup$
How did you use Roche's theorem? Don't you need $|f(z)| < |g(z)|$ for say $z = lambda$.
$endgroup$
– muzzlator
Mar 4 '13 at 2:15
$begingroup$
How did you use Roche's theorem? Don't you need $|f(z)| < |g(z)|$ for say $z = lambda$.
$endgroup$
– muzzlator
Mar 4 '13 at 2:15
$begingroup$
@muzzlator I took a small circle with center $lambda$ and the inequality you wrote should be true for all values on (not in) this circle, and this is that case.
$endgroup$
– i.a.m
Mar 4 '13 at 2:19
$begingroup$
@muzzlator I took a small circle with center $lambda$ and the inequality you wrote should be true for all values on (not in) this circle, and this is that case.
$endgroup$
– i.a.m
Mar 4 '13 at 2:19
$begingroup$
Ah, I see. Yeah that looks like it works
$endgroup$
– muzzlator
Mar 4 '13 at 2:23
$begingroup$
Ah, I see. Yeah that looks like it works
$endgroup$
– muzzlator
Mar 4 '13 at 2:23
$begingroup$
Rouche is a good idea. But if you apply it to a small circle, how does this apply to the whole half-plane?
$endgroup$
– Julien
Mar 4 '13 at 2:56
$begingroup$
Rouche is a good idea. But if you apply it to a small circle, how does this apply to the whole half-plane?
$endgroup$
– Julien
Mar 4 '13 at 2:56
$begingroup$
@julien I was thinking using part two i.e the real case we established that the equation has a zero in the right half plane, now lets assume we have another zero in the right half plane and take a circle containing both zeros to get a contradictoin, and conclude that there is exactly one solution. the problem I'm facing now is that I can't get the inequality of Rouche's theorem.
$endgroup$
– i.a.m
Mar 4 '13 at 3:02
$begingroup$
@julien I was thinking using part two i.e the real case we established that the equation has a zero in the right half plane, now lets assume we have another zero in the right half plane and take a circle containing both zeros to get a contradictoin, and conclude that there is exactly one solution. the problem I'm facing now is that I can't get the inequality of Rouche's theorem.
$endgroup$
– i.a.m
Mar 4 '13 at 3:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A hint.
If $operatorname{Re} z > 0$ and $lambda - z - e^{-z} = 0$ then
$$
|lambda - z| = e^{-operatorname{Re} z} < 1.
$$
In other words, if the equation has any solutions in the right half-plane then they lie in the open disc $|z-lambda|<1$.
answered Mar 4 '13 at 7:33
Antonio VargasAntonio Vargas
20.7k245111
$endgroup$
$begingroup$
so if I take $gamma$ to be the circle $|lambda-z|=1$ then we have $|f(z)|=e^{-Re(z)}<1=|lambda-z|=|g(z)|$, Thus, by Rouche's theorem $g$ and $f+g$ have the same number of zeros in this circle which is 1.
$endgroup$
– i.a.m
Mar 4 '13 at 15:07
1
$begingroup$
@i.a.m You're welcome.
$endgroup$
– Antonio Vargas
Mar 4 '13 at 15:57
$begingroup$
wait, is the question asking us to prove there is only one solution which happens to be in right half plane OR that there exists only 1 solution in right half plane (and possible some solution in left half plane)?
$endgroup$
– Red Floyd
Sep 9 '18 at 7:54
$begingroup$
@AlphaRomeo, the equation has infinitely-many solutions, and exactly one of those is in the right half-plane.
$endgroup$
– Antonio Vargas
Sep 9 '18 at 20:48
$begingroup$
Can you please explain why you say infinitely-many solutions in left half plane? If z is purely real then I plotted and checked it can have only 1 solution. How to analyze when z is not purely real?
$endgroup$
– Red Floyd
Sep 10 '18 at 16:02
|
show 2 more comments
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1 Answer
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active
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1 Answer
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active
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votes
$begingroup$
A hint.
If $operatorname{Re} z > 0$ and $lambda - z - e^{-z} = 0$ then
$$
|lambda - z| = e^{-operatorname{Re} z} < 1.
$$
In other words, if the equation has any solutions in the right half-plane then they lie in the open disc $|z-lambda|<1$.
answered Mar 4 '13 at 7:33
Antonio VargasAntonio Vargas
20.7k245111
$endgroup$
$begingroup$
so if I take $gamma$ to be the circle $|lambda-z|=1$ then we have $|f(z)|=e^{-Re(z)}<1=|lambda-z|=|g(z)|$, Thus, by Rouche's theorem $g$ and $f+g$ have the same number of zeros in this circle which is 1.
$endgroup$
– i.a.m
Mar 4 '13 at 15:07
1
$begingroup$
@i.a.m You're welcome.
$endgroup$
– Antonio Vargas
Mar 4 '13 at 15:57
$begingroup$
wait, is the question asking us to prove there is only one solution which happens to be in right half plane OR that there exists only 1 solution in right half plane (and possible some solution in left half plane)?
$endgroup$
– Red Floyd
Sep 9 '18 at 7:54
$begingroup$
@AlphaRomeo, the equation has infinitely-many solutions, and exactly one of those is in the right half-plane.
$endgroup$
– Antonio Vargas
Sep 9 '18 at 20:48
$begingroup$
Can you please explain why you say infinitely-many solutions in left half plane? If z is purely real then I plotted and checked it can have only 1 solution. How to analyze when z is not purely real?
$endgroup$
– Red Floyd
Sep 10 '18 at 16:02
|
show 2 more comments
$begingroup$
A hint.
If $operatorname{Re} z > 0$ and $lambda - z - e^{-z} = 0$ then
$$
|lambda - z| = e^{-operatorname{Re} z} < 1.
$$
In other words, if the equation has any solutions in the right half-plane then they lie in the open disc $|z-lambda|<1$.
answered Mar 4 '13 at 7:33
Antonio VargasAntonio Vargas
20.7k245111
$endgroup$
$begingroup$
so if I take $gamma$ to be the circle $|lambda-z|=1$ then we have $|f(z)|=e^{-Re(z)}<1=|lambda-z|=|g(z)|$, Thus, by Rouche's theorem $g$ and $f+g$ have the same number of zeros in this circle which is 1.
$endgroup$
– i.a.m
Mar 4 '13 at 15:07
1
$begingroup$
@i.a.m You're welcome.
$endgroup$
– Antonio Vargas
Mar 4 '13 at 15:57
$begingroup$
wait, is the question asking us to prove there is only one solution which happens to be in right half plane OR that there exists only 1 solution in right half plane (and possible some solution in left half plane)?
$endgroup$
– Red Floyd
Sep 9 '18 at 7:54
$begingroup$
@AlphaRomeo, the equation has infinitely-many solutions, and exactly one of those is in the right half-plane.
$endgroup$
– Antonio Vargas
Sep 9 '18 at 20:48
$begingroup$
Can you please explain why you say infinitely-many solutions in left half plane? If z is purely real then I plotted and checked it can have only 1 solution. How to analyze when z is not purely real?
$endgroup$
– Red Floyd
Sep 10 '18 at 16:02
|
show 2 more comments
$begingroup$
A hint.
If $operatorname{Re} z > 0$ and $lambda - z - e^{-z} = 0$ then
$$
|lambda - z| = e^{-operatorname{Re} z} < 1.
$$
In other words, if the equation has any solutions in the right half-plane then they lie in the open disc $|z-lambda|<1$.
answered Mar 4 '13 at 7:33
Antonio VargasAntonio Vargas
20.7k245111
$endgroup$
A hint.
If $operatorname{Re} z > 0$ and $lambda - z - e^{-z} = 0$ then
$$
|lambda - z| = e^{-operatorname{Re} z} < 1.
$$
In other words, if the equation has any solutions in the right half-plane then they lie in the open disc $|z-lambda|<1$.
answered Mar 4 '13 at 7:33
Antonio VargasAntonio Vargas
20.7k245111
edited Mar 4 '13 at 7:51
answered Mar 4 '13 at 7:33
Antonio VargasAntonio Vargas
20.7k245111
answered Mar 4 '13 at 7:33
Antonio VargasAntonio Vargas
20.7k245111
answered Mar 4 '13 at 7:33
Antonio VargasAntonio Vargas
20.7k245111
20.7k245111
$begingroup$
so if I take $gamma$ to be the circle $|lambda-z|=1$ then we have $|f(z)|=e^{-Re(z)}<1=|lambda-z|=|g(z)|$, Thus, by Rouche's theorem $g$ and $f+g$ have the same number of zeros in this circle which is 1.
$endgroup$
– i.a.m
Mar 4 '13 at 15:07
1
$begingroup$
@i.a.m You're welcome.
$endgroup$
– Antonio Vargas
Mar 4 '13 at 15:57
$begingroup$
wait, is the question asking us to prove there is only one solution which happens to be in right half plane OR that there exists only 1 solution in right half plane (and possible some solution in left half plane)?
$endgroup$
– Red Floyd
Sep 9 '18 at 7:54
$begingroup$
@AlphaRomeo, the equation has infinitely-many solutions, and exactly one of those is in the right half-plane.
$endgroup$
– Antonio Vargas
Sep 9 '18 at 20:48
$begingroup$
Can you please explain why you say infinitely-many solutions in left half plane? If z is purely real then I plotted and checked it can have only 1 solution. How to analyze when z is not purely real?
$endgroup$
– Red Floyd
Sep 10 '18 at 16:02
|
show 2 more comments
$begingroup$
so if I take $gamma$ to be the circle $|lambda-z|=1$ then we have $|f(z)|=e^{-Re(z)}<1=|lambda-z|=|g(z)|$, Thus, by Rouche's theorem $g$ and $f+g$ have the same number of zeros in this circle which is 1.
$endgroup$
– i.a.m
Mar 4 '13 at 15:07
1
$begingroup$
@i.a.m You're welcome.
$endgroup$
– Antonio Vargas
Mar 4 '13 at 15:57
$begingroup$
wait, is the question asking us to prove there is only one solution which happens to be in right half plane OR that there exists only 1 solution in right half plane (and possible some solution in left half plane)?
$endgroup$
– Red Floyd
Sep 9 '18 at 7:54
$begingroup$
@AlphaRomeo, the equation has infinitely-many solutions, and exactly one of those is in the right half-plane.
$endgroup$
– Antonio Vargas
Sep 9 '18 at 20:48
$begingroup$
Can you please explain why you say infinitely-many solutions in left half plane? If z is purely real then I plotted and checked it can have only 1 solution. How to analyze when z is not purely real?
$endgroup$
– Red Floyd
Sep 10 '18 at 16:02
$begingroup$
so if I take $gamma$ to be the circle $|lambda-z|=1$ then we have $|f(z)|=e^{-Re(z)}<1=|lambda-z|=|g(z)|$, Thus, by Rouche's theorem $g$ and $f+g$ have the same number of zeros in this circle which is 1.
$endgroup$
– i.a.m
Mar 4 '13 at 15:07
$begingroup$
so if I take $gamma$ to be the circle $|lambda-z|=1$ then we have $|f(z)|=e^{-Re(z)}<1=|lambda-z|=|g(z)|$, Thus, by Rouche's theorem $g$ and $f+g$ have the same number of zeros in this circle which is 1.
$endgroup$
– i.a.m
Mar 4 '13 at 15:07
1
1
$begingroup$
@i.a.m You're welcome.
$endgroup$
– Antonio Vargas
Mar 4 '13 at 15:57
$begingroup$
@i.a.m You're welcome.
$endgroup$
– Antonio Vargas
Mar 4 '13 at 15:57
$begingroup$
wait, is the question asking us to prove there is only one solution which happens to be in right half plane OR that there exists only 1 solution in right half plane (and possible some solution in left half plane)?
$endgroup$
– Red Floyd
Sep 9 '18 at 7:54
$begingroup$
wait, is the question asking us to prove there is only one solution which happens to be in right half plane OR that there exists only 1 solution in right half plane (and possible some solution in left half plane)?
$endgroup$
– Red Floyd
Sep 9 '18 at 7:54
$begingroup$
@AlphaRomeo, the equation has infinitely-many solutions, and exactly one of those is in the right half-plane.
$endgroup$
– Antonio Vargas
Sep 9 '18 at 20:48
$begingroup$
@AlphaRomeo, the equation has infinitely-many solutions, and exactly one of those is in the right half-plane.
$endgroup$
– Antonio Vargas
Sep 9 '18 at 20:48
$begingroup$
Can you please explain why you say infinitely-many solutions in left half plane? If z is purely real then I plotted and checked it can have only 1 solution. How to analyze when z is not purely real?
$endgroup$
– Red Floyd
Sep 10 '18 at 16:02
$begingroup$
Can you please explain why you say infinitely-many solutions in left half plane? If z is purely real then I plotted and checked it can have only 1 solution. How to analyze when z is not purely real?
$endgroup$
– Red Floyd
Sep 10 '18 at 16:02
|
show 2 more comments
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$begingroup$
How did you use Roche's theorem? Don't you need $|f(z)| < |g(z)|$ for say $z = lambda$.
$endgroup$
– muzzlator
Mar 4 '13 at 2:15
$begingroup$
@muzzlator I took a small circle with center $lambda$ and the inequality you wrote should be true for all values on (not in) this circle, and this is that case.
$endgroup$
– i.a.m
Mar 4 '13 at 2:19
$begingroup$
Ah, I see. Yeah that looks like it works
$endgroup$
– muzzlator
Mar 4 '13 at 2:23
$begingroup$
Rouche is a good idea. But if you apply it to a small circle, how does this apply to the whole half-plane?
$endgroup$
– Julien
Mar 4 '13 at 2:56
$begingroup$
@julien I was thinking using part two i.e the real case we established that the equation has a zero in the right half plane, now lets assume we have another zero in the right half plane and take a circle containing both zeros to get a contradictoin, and conclude that there is exactly one solution. the problem I'm facing now is that I can't get the inequality of Rouche's theorem.
$endgroup$
– i.a.m
Mar 4 '13 at 3:02