Prove that $ze^lambda - z - 1$ has a real root in the unit diskShow $z e^lambda-z-1$ has only one real root in the unit disk.Proving $f$ has at least one zero inside unit diskAll the zeroes of $p(z)$ lie inside the unit diskIf $|a_0|<|a_n|$, Prove that $P(z)=sum^n_k=0a_kz^k$ has at least one zero inside the unit disk.Show $z e^lambda-z-1$ has only one real root in the unit disk.Zeros of $f(z) = z^5+3z^4+9z^3+10$ in the unit diskMultiplicity of zeros of $4z^10-e^z$ in unit diskShow that the equation $lambda - z - e^-z = 0$ has exactly one solution in the half plane.Complex roots of $z^10 - 6z^9 + 6^9$ inside the disk of radius sixQuestion on Rouche's Theorem - Boundary Included?How many roots does $g(z)=z^7-2z^5+6z^3-z+1$ have inside the unit disk - Rouche's Theorem Application Verification
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Prove that $ze^lambda - z - 1$ has a real root in the unit disk
Show $z e^lambda-z-1$ has only one real root in the unit disk.Proving $f$ has at least one zero inside unit diskAll the zeroes of $p(z)$ lie inside the unit diskIf $|a_0|<|a_n|$, Prove that $P(z)=sum^n_k=0a_kz^k$ has at least one zero inside the unit disk.Show $z e^lambda-z-1$ has only one real root in the unit disk.Zeros of $f(z) = z^5+3z^4+9z^3+10$ in the unit diskMultiplicity of zeros of $4z^10-e^z$ in unit diskShow that the equation $lambda - z - e^-z = 0$ has exactly one solution in the half plane.Complex roots of $z^10 - 6z^9 + 6^9$ inside the disk of radius sixQuestion on Rouche's Theorem - Boundary Included?How many roots does $g(z)=z^7-2z^5+6z^3-z+1$ have inside the unit disk - Rouche's Theorem Application Verification
$begingroup$
I am trying to show that $ f(z) = ze^lambda - z - 1$, $lambda > 1$ has a real root inside the disk. I have already showed, using Rouche's Theorem, that there is exactly one root inside the disk. But I am not sure how to show that this root is real.
I have considered the real-valued analog: $xe^lambda - x - 1$ and tried appealing to calculus. I was able to show that this function is increasing on the interval $(-infty, 1)$. So If I could find two values in this interval such that $f(x) < 0$ and $f(x) > 0$, I would be done by appealing to the intermediate value theorem.
Any tips?
Note that I have already looked at Show $z e^lambda-z-1$ has only one real root in the unit disk. but that post does not discuss how to show the root is real.
calculus complex-analysis rouches-theorem
asked Mar 29 at 0:59
Nicholas RobertsNicholas Roberts
144112
$endgroup$
|
show 3 more comments
$begingroup$
I am trying to show that $ f(z) = ze^lambda - z - 1$, $lambda > 1$ has a real root inside the disk. I have already showed, using Rouche's Theorem, that there is exactly one root inside the disk. But I am not sure how to show that this root is real.
I have considered the real-valued analog: $xe^lambda - x - 1$ and tried appealing to calculus. I was able to show that this function is increasing on the interval $(-infty, 1)$. So If I could find two values in this interval such that $f(x) < 0$ and $f(x) > 0$, I would be done by appealing to the intermediate value theorem.
Any tips?
Note that I have already looked at Show $z e^lambda-z-1$ has only one real root in the unit disk. but that post does not discuss how to show the root is real.
calculus complex-analysis rouches-theorem
asked Mar 29 at 0:59
Nicholas RobertsNicholas Roberts
144112
$endgroup$
3
$begingroup$
Other suggestion, which builds upon your own comment. $0$ and $1$ satisfy the conditions, right?
$endgroup$
– Stan Tendijck
Mar 29 at 1:06
1
$begingroup$
@StanTendijck Yes, I see that $f(0) < 0$ and $f(1) > 0$. So by IVT, there must be a root between $0$ and $1$. Thanks for the tip!
$endgroup$
– Nicholas Roberts
Mar 29 at 1:10
1
$begingroup$
Sure. Say that you use the IVT to find a root $tildex$ of $xe^lambda - x - 1$ with $lvert tildex rvert < 1$. Then, letting $z=tildex + i0$ will give you a (real) root in $mathbbC$. Since there is only one root in the unit disk, then the $z$ above must be it.
$endgroup$
– Gary Moon
Mar 29 at 1:49
1
$begingroup$
Since $lambda$ is real, the Taylor coefficients of $f(z) = ze^lambda - z - 1=zSigmafrac(lambda-z)^nn!-1$ are real, so $f(barz)=barf(z)$, so if $f(z)$ is real, $f(z)=f(barz)$, so in particular if $f(z)=0, f(barz)=0$, so by unicity you are done
$endgroup$
– Conrad
Mar 29 at 2:53
1
$begingroup$
@Conrad Why not an official answer? By the way, you do not need the Taylor series of $f$, it is well known that $overlinee^w = e^overlinew$.
$endgroup$
– Paul Frost
Mar 29 at 16:53
|
show 3 more comments
$begingroup$
I am trying to show that $ f(z) = ze^lambda - z - 1$, $lambda > 1$ has a real root inside the disk. I have already showed, using Rouche's Theorem, that there is exactly one root inside the disk. But I am not sure how to show that this root is real.
I have considered the real-valued analog: $xe^lambda - x - 1$ and tried appealing to calculus. I was able to show that this function is increasing on the interval $(-infty, 1)$. So If I could find two values in this interval such that $f(x) < 0$ and $f(x) > 0$, I would be done by appealing to the intermediate value theorem.
Any tips?
Note that I have already looked at Show $z e^lambda-z-1$ has only one real root in the unit disk. but that post does not discuss how to show the root is real.
calculus complex-analysis rouches-theorem
asked Mar 29 at 0:59
Nicholas RobertsNicholas Roberts
144112
$endgroup$
I am trying to show that $ f(z) = ze^lambda - z - 1$, $lambda > 1$ has a real root inside the disk. I have already showed, using Rouche's Theorem, that there is exactly one root inside the disk. But I am not sure how to show that this root is real.
I have considered the real-valued analog: $xe^lambda - x - 1$ and tried appealing to calculus. I was able to show that this function is increasing on the interval $(-infty, 1)$. So If I could find two values in this interval such that $f(x) < 0$ and $f(x) > 0$, I would be done by appealing to the intermediate value theorem.
Any tips?
Note that I have already looked at Show $z e^lambda-z-1$ has only one real root in the unit disk. but that post does not discuss how to show the root is real.
calculus complex-analysis rouches-theorem
calculus complex-analysis rouches-theorem
asked Mar 29 at 0:59
Nicholas RobertsNicholas Roberts
144112
asked Mar 29 at 0:59
Nicholas RobertsNicholas Roberts
144112
asked Mar 29 at 0:59
Nicholas RobertsNicholas Roberts
144112
asked Mar 29 at 0:59
Nicholas RobertsNicholas Roberts
144112
asked Mar 29 at 0:59
Nicholas RobertsNicholas Roberts
144112
144112
3
$begingroup$
Other suggestion, which builds upon your own comment. $0$ and $1$ satisfy the conditions, right?
$endgroup$
– Stan Tendijck
Mar 29 at 1:06
1
$begingroup$
@StanTendijck Yes, I see that $f(0) < 0$ and $f(1) > 0$. So by IVT, there must be a root between $0$ and $1$. Thanks for the tip!
$endgroup$
– Nicholas Roberts
Mar 29 at 1:10
1
$begingroup$
Sure. Say that you use the IVT to find a root $tildex$ of $xe^lambda - x - 1$ with $lvert tildex rvert < 1$. Then, letting $z=tildex + i0$ will give you a (real) root in $mathbbC$. Since there is only one root in the unit disk, then the $z$ above must be it.
$endgroup$
– Gary Moon
Mar 29 at 1:49
1
$begingroup$
Since $lambda$ is real, the Taylor coefficients of $f(z) = ze^lambda - z - 1=zSigmafrac(lambda-z)^nn!-1$ are real, so $f(barz)=barf(z)$, so if $f(z)$ is real, $f(z)=f(barz)$, so in particular if $f(z)=0, f(barz)=0$, so by unicity you are done
$endgroup$
– Conrad
Mar 29 at 2:53
1
$begingroup$
@Conrad Why not an official answer? By the way, you do not need the Taylor series of $f$, it is well known that $overlinee^w = e^overlinew$.
$endgroup$
– Paul Frost
Mar 29 at 16:53
|
show 3 more comments
3
$begingroup$
Other suggestion, which builds upon your own comment. $0$ and $1$ satisfy the conditions, right?
$endgroup$
– Stan Tendijck
Mar 29 at 1:06
1
$begingroup$
@StanTendijck Yes, I see that $f(0) < 0$ and $f(1) > 0$. So by IVT, there must be a root between $0$ and $1$. Thanks for the tip!
$endgroup$
– Nicholas Roberts
Mar 29 at 1:10
1
$begingroup$
Sure. Say that you use the IVT to find a root $tildex$ of $xe^lambda - x - 1$ with $lvert tildex rvert < 1$. Then, letting $z=tildex + i0$ will give you a (real) root in $mathbbC$. Since there is only one root in the unit disk, then the $z$ above must be it.
$endgroup$
– Gary Moon
Mar 29 at 1:49
1
$begingroup$
Since $lambda$ is real, the Taylor coefficients of $f(z) = ze^lambda - z - 1=zSigmafrac(lambda-z)^nn!-1$ are real, so $f(barz)=barf(z)$, so if $f(z)$ is real, $f(z)=f(barz)$, so in particular if $f(z)=0, f(barz)=0$, so by unicity you are done
$endgroup$
– Conrad
Mar 29 at 2:53
1
$begingroup$
@Conrad Why not an official answer? By the way, you do not need the Taylor series of $f$, it is well known that $overlinee^w = e^overlinew$.
$endgroup$
– Paul Frost
Mar 29 at 16:53
3
3
$begingroup$
Other suggestion, which builds upon your own comment. $0$ and $1$ satisfy the conditions, right?
$endgroup$
– Stan Tendijck
Mar 29 at 1:06
$begingroup$
Other suggestion, which builds upon your own comment. $0$ and $1$ satisfy the conditions, right?
$endgroup$
– Stan Tendijck
Mar 29 at 1:06
1
1
$begingroup$
@StanTendijck Yes, I see that $f(0) < 0$ and $f(1) > 0$. So by IVT, there must be a root between $0$ and $1$. Thanks for the tip!
$endgroup$
– Nicholas Roberts
Mar 29 at 1:10
$begingroup$
@StanTendijck Yes, I see that $f(0) < 0$ and $f(1) > 0$. So by IVT, there must be a root between $0$ and $1$. Thanks for the tip!
$endgroup$
– Nicholas Roberts
Mar 29 at 1:10
1
1
$begingroup$
Sure. Say that you use the IVT to find a root $tildex$ of $xe^lambda - x - 1$ with $lvert tildex rvert < 1$. Then, letting $z=tildex + i0$ will give you a (real) root in $mathbbC$. Since there is only one root in the unit disk, then the $z$ above must be it.
$endgroup$
– Gary Moon
Mar 29 at 1:49
$begingroup$
Sure. Say that you use the IVT to find a root $tildex$ of $xe^lambda - x - 1$ with $lvert tildex rvert < 1$. Then, letting $z=tildex + i0$ will give you a (real) root in $mathbbC$. Since there is only one root in the unit disk, then the $z$ above must be it.
$endgroup$
– Gary Moon
Mar 29 at 1:49
1
1
$begingroup$
Since $lambda$ is real, the Taylor coefficients of $f(z) = ze^lambda - z - 1=zSigmafrac(lambda-z)^nn!-1$ are real, so $f(barz)=barf(z)$, so if $f(z)$ is real, $f(z)=f(barz)$, so in particular if $f(z)=0, f(barz)=0$, so by unicity you are done
$endgroup$
– Conrad
Mar 29 at 2:53
$begingroup$
Since $lambda$ is real, the Taylor coefficients of $f(z) = ze^lambda - z - 1=zSigmafrac(lambda-z)^nn!-1$ are real, so $f(barz)=barf(z)$, so if $f(z)$ is real, $f(z)=f(barz)$, so in particular if $f(z)=0, f(barz)=0$, so by unicity you are done
$endgroup$
– Conrad
Mar 29 at 2:53
1
1
$begingroup$
@Conrad Why not an official answer? By the way, you do not need the Taylor series of $f$, it is well known that $overlinee^w = e^overlinew$.
$endgroup$
– Paul Frost
Mar 29 at 16:53
$begingroup$
@Conrad Why not an official answer? By the way, you do not need the Taylor series of $f$, it is well known that $overlinee^w = e^overlinew$.
$endgroup$
– Paul Frost
Mar 29 at 16:53
|
show 3 more comments
0
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3
$begingroup$
Other suggestion, which builds upon your own comment. $0$ and $1$ satisfy the conditions, right?
$endgroup$
– Stan Tendijck
Mar 29 at 1:06
1
$begingroup$
@StanTendijck Yes, I see that $f(0) < 0$ and $f(1) > 0$. So by IVT, there must be a root between $0$ and $1$. Thanks for the tip!
$endgroup$
– Nicholas Roberts
Mar 29 at 1:10
1
$begingroup$
Sure. Say that you use the IVT to find a root $tildex$ of $xe^lambda - x - 1$ with $lvert tildex rvert < 1$. Then, letting $z=tildex + i0$ will give you a (real) root in $mathbbC$. Since there is only one root in the unit disk, then the $z$ above must be it.
$endgroup$
– Gary Moon
Mar 29 at 1:49
1
$begingroup$
Since $lambda$ is real, the Taylor coefficients of $f(z) = ze^lambda - z - 1=zSigmafrac(lambda-z)^nn!-1$ are real, so $f(barz)=barf(z)$, so if $f(z)$ is real, $f(z)=f(barz)$, so in particular if $f(z)=0, f(barz)=0$, so by unicity you are done
$endgroup$
– Conrad
Mar 29 at 2:53
1
$begingroup$
@Conrad Why not an official answer? By the way, you do not need the Taylor series of $f$, it is well known that $overlinee^w = e^overlinew$.
$endgroup$
– Paul Frost
Mar 29 at 16:53