Complex Polynomial Inequality Proof
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I'm trying to solve this question but I am having trouble connecting the dots. The question reads:
Assume that we have a complex polynomial:
$$P(z) = a_0+a_1z+...+a_nz^n$$
Satisfies $|P(z)|leq 1$ whenever $|z|=1$. Show that $|a_n|leq 1 >>forall n$.
So, I have simplified $|P(z)|geq|a_n|left|1-frac{|a_{n-1}|}{|a_n|} -...-frac{|a_0|}{|a_n|}right|$, using the fact that $|z|=1$.
Now, I'm confused by how to proceed. There's no indication that the sequence is decreasing? Am I on the right track?
sequences-and-series complex-analysis
edited Nov 12 at 5:10
Wesley Strik
1,378322
asked Nov 12 at 4:25
Felicio Grande
499619
add a comment |
up vote
2
down vote
favorite
I'm trying to solve this question but I am having trouble connecting the dots. The question reads:
Assume that we have a complex polynomial:
$$P(z) = a_0+a_1z+...+a_nz^n$$
Satisfies $|P(z)|leq 1$ whenever $|z|=1$. Show that $|a_n|leq 1 >>forall n$.
So, I have simplified $|P(z)|geq|a_n|left|1-frac{|a_{n-1}|}{|a_n|} -...-frac{|a_0|}{|a_n|}right|$, using the fact that $|z|=1$.
Now, I'm confused by how to proceed. There's no indication that the sequence is decreasing? Am I on the right track?
sequences-and-series complex-analysis
edited Nov 12 at 5:10
Wesley Strik
1,378322
asked Nov 12 at 4:25
Felicio Grande
499619
1
Maybe it pays to think about integrating $z^{-k}P(z)$ around the unit circle.
– Gerry Myerson
Nov 12 at 4:55
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm trying to solve this question but I am having trouble connecting the dots. The question reads:
Assume that we have a complex polynomial:
$$P(z) = a_0+a_1z+...+a_nz^n$$
Satisfies $|P(z)|leq 1$ whenever $|z|=1$. Show that $|a_n|leq 1 >>forall n$.
So, I have simplified $|P(z)|geq|a_n|left|1-frac{|a_{n-1}|}{|a_n|} -...-frac{|a_0|}{|a_n|}right|$, using the fact that $|z|=1$.
Now, I'm confused by how to proceed. There's no indication that the sequence is decreasing? Am I on the right track?
sequences-and-series complex-analysis
edited Nov 12 at 5:10
Wesley Strik
1,378322
asked Nov 12 at 4:25
Felicio Grande
499619
I'm trying to solve this question but I am having trouble connecting the dots. The question reads:
Assume that we have a complex polynomial:
$$P(z) = a_0+a_1z+...+a_nz^n$$
Satisfies $|P(z)|leq 1$ whenever $|z|=1$. Show that $|a_n|leq 1 >>forall n$.
So, I have simplified $|P(z)|geq|a_n|left|1-frac{|a_{n-1}|}{|a_n|} -...-frac{|a_0|}{|a_n|}right|$, using the fact that $|z|=1$.
Now, I'm confused by how to proceed. There's no indication that the sequence is decreasing? Am I on the right track?
sequences-and-series complex-analysis
sequences-and-series complex-analysis
edited Nov 12 at 5:10
Wesley Strik
1,378322
asked Nov 12 at 4:25
Felicio Grande
499619
edited Nov 12 at 5:10
Wesley Strik
1,378322
asked Nov 12 at 4:25
Felicio Grande
499619
edited Nov 12 at 5:10
Wesley Strik
1,378322
edited Nov 12 at 5:10
Wesley Strik
1,378322
edited Nov 12 at 5:10
Wesley Strik
1,378322
1,378322
asked Nov 12 at 4:25
Felicio Grande
499619
asked Nov 12 at 4:25
Felicio Grande
499619
asked Nov 12 at 4:25
Felicio Grande
499619
499619
1
Maybe it pays to think about integrating $z^{-k}P(z)$ around the unit circle.
– Gerry Myerson
Nov 12 at 4:55
add a comment |
1
Maybe it pays to think about integrating $z^{-k}P(z)$ around the unit circle.
– Gerry Myerson
Nov 12 at 4:55
1
1
Maybe it pays to think about integrating $z^{-k}P(z)$ around the unit circle.
– Gerry Myerson
Nov 12 at 4:55
Maybe it pays to think about integrating $z^{-k}P(z)$ around the unit circle.
– Gerry Myerson
Nov 12 at 4:55
add a comment |
2 Answers
2
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up vote
4
down vote
This is very easy with a simple trick. Let $q(z)=a_n+a_{n-1}z+cdots+ a_0z^{n}$. Then $q(z)=z^{n}p(frac 1 z)$. $q$ is a polynomial and $|q(z)| leq 1$ if $|z|=1$. By MMP $|q(0)| leq 1$ which is what we need.
answered Nov 12 at 5:31
Kavi Rama Murthy
46.6k31854
Ah, I misunderstood the question – I thought it was asking for $|a_k|le1$ for $0le kle n$.
– Gerry Myerson
Nov 13 at 10:06
add a comment |
up vote
2
down vote
One way to show this is via Cauchy estimates.
From Cauchy integral formula, for the $n$-th derivative of $P$ we have
$$
P^{(n)}(z) = frac{n!}{2pi i } int_{|xi| = 1} frac{P(xi)}{(z - xi)^{n+1}}dxi.
$$
Hence, using the fact that $|P|leq 1$ on $|xi| = 1$, we obtain
$$
n! |a_n| leq frac{n!}{2pi } int_{|xi| = 1} frac{|dxi| }{|z - xi|^{n+1}} =
frac{n!}{2pi} intlimits_0^{2pi} left|frac{ie^{itheta}} {(e^{itheta})^{n+1}}right| dtheta = n!,
$$
which gives $|a_n|leq 1$.
answered Nov 12 at 5:04
Hayk
2,022212
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
This is very easy with a simple trick. Let $q(z)=a_n+a_{n-1}z+cdots+ a_0z^{n}$. Then $q(z)=z^{n}p(frac 1 z)$. $q$ is a polynomial and $|q(z)| leq 1$ if $|z|=1$. By MMP $|q(0)| leq 1$ which is what we need.
answered Nov 12 at 5:31
Kavi Rama Murthy
46.6k31854
Ah, I misunderstood the question – I thought it was asking for $|a_k|le1$ for $0le kle n$.
– Gerry Myerson
Nov 13 at 10:06
add a comment |
up vote
4
down vote
This is very easy with a simple trick. Let $q(z)=a_n+a_{n-1}z+cdots+ a_0z^{n}$. Then $q(z)=z^{n}p(frac 1 z)$. $q$ is a polynomial and $|q(z)| leq 1$ if $|z|=1$. By MMP $|q(0)| leq 1$ which is what we need.
answered Nov 12 at 5:31
Kavi Rama Murthy
46.6k31854
Ah, I misunderstood the question – I thought it was asking for $|a_k|le1$ for $0le kle n$.
– Gerry Myerson
Nov 13 at 10:06
add a comment |
up vote
4
down vote
up vote
4
down vote
This is very easy with a simple trick. Let $q(z)=a_n+a_{n-1}z+cdots+ a_0z^{n}$. Then $q(z)=z^{n}p(frac 1 z)$. $q$ is a polynomial and $|q(z)| leq 1$ if $|z|=1$. By MMP $|q(0)| leq 1$ which is what we need.
answered Nov 12 at 5:31
Kavi Rama Murthy
46.6k31854
This is very easy with a simple trick. Let $q(z)=a_n+a_{n-1}z+cdots+ a_0z^{n}$. Then $q(z)=z^{n}p(frac 1 z)$. $q$ is a polynomial and $|q(z)| leq 1$ if $|z|=1$. By MMP $|q(0)| leq 1$ which is what we need.
answered Nov 12 at 5:31
Kavi Rama Murthy
46.6k31854
answered Nov 12 at 5:31
Kavi Rama Murthy
46.6k31854
answered Nov 12 at 5:31
Kavi Rama Murthy
46.6k31854
answered Nov 12 at 5:31
Kavi Rama Murthy
46.6k31854
46.6k31854
Ah, I misunderstood the question – I thought it was asking for $|a_k|le1$ for $0le kle n$.
– Gerry Myerson
Nov 13 at 10:06
add a comment |
Ah, I misunderstood the question – I thought it was asking for $|a_k|le1$ for $0le kle n$.
– Gerry Myerson
Nov 13 at 10:06
Ah, I misunderstood the question – I thought it was asking for $|a_k|le1$ for $0le kle n$.
– Gerry Myerson
Nov 13 at 10:06
Ah, I misunderstood the question – I thought it was asking for $|a_k|le1$ for $0le kle n$.
– Gerry Myerson
Nov 13 at 10:06
add a comment |
up vote
2
down vote
One way to show this is via Cauchy estimates.
From Cauchy integral formula, for the $n$-th derivative of $P$ we have
$$
P^{(n)}(z) = frac{n!}{2pi i } int_{|xi| = 1} frac{P(xi)}{(z - xi)^{n+1}}dxi.
$$
Hence, using the fact that $|P|leq 1$ on $|xi| = 1$, we obtain
$$
n! |a_n| leq frac{n!}{2pi } int_{|xi| = 1} frac{|dxi| }{|z - xi|^{n+1}} =
frac{n!}{2pi} intlimits_0^{2pi} left|frac{ie^{itheta}} {(e^{itheta})^{n+1}}right| dtheta = n!,
$$
which gives $|a_n|leq 1$.
answered Nov 12 at 5:04
Hayk
2,022212
add a comment |
up vote
2
down vote
One way to show this is via Cauchy estimates.
From Cauchy integral formula, for the $n$-th derivative of $P$ we have
$$
P^{(n)}(z) = frac{n!}{2pi i } int_{|xi| = 1} frac{P(xi)}{(z - xi)^{n+1}}dxi.
$$
Hence, using the fact that $|P|leq 1$ on $|xi| = 1$, we obtain
$$
n! |a_n| leq frac{n!}{2pi } int_{|xi| = 1} frac{|dxi| }{|z - xi|^{n+1}} =
frac{n!}{2pi} intlimits_0^{2pi} left|frac{ie^{itheta}} {(e^{itheta})^{n+1}}right| dtheta = n!,
$$
which gives $|a_n|leq 1$.
answered Nov 12 at 5:04
Hayk
2,022212
add a comment |
up vote
2
down vote
up vote
2
down vote
One way to show this is via Cauchy estimates.
From Cauchy integral formula, for the $n$-th derivative of $P$ we have
$$
P^{(n)}(z) = frac{n!}{2pi i } int_{|xi| = 1} frac{P(xi)}{(z - xi)^{n+1}}dxi.
$$
Hence, using the fact that $|P|leq 1$ on $|xi| = 1$, we obtain
$$
n! |a_n| leq frac{n!}{2pi } int_{|xi| = 1} frac{|dxi| }{|z - xi|^{n+1}} =
frac{n!}{2pi} intlimits_0^{2pi} left|frac{ie^{itheta}} {(e^{itheta})^{n+1}}right| dtheta = n!,
$$
which gives $|a_n|leq 1$.
answered Nov 12 at 5:04
Hayk
2,022212
One way to show this is via Cauchy estimates.
From Cauchy integral formula, for the $n$-th derivative of $P$ we have
$$
P^{(n)}(z) = frac{n!}{2pi i } int_{|xi| = 1} frac{P(xi)}{(z - xi)^{n+1}}dxi.
$$
Hence, using the fact that $|P|leq 1$ on $|xi| = 1$, we obtain
$$
n! |a_n| leq frac{n!}{2pi } int_{|xi| = 1} frac{|dxi| }{|z - xi|^{n+1}} =
frac{n!}{2pi} intlimits_0^{2pi} left|frac{ie^{itheta}} {(e^{itheta})^{n+1}}right| dtheta = n!,
$$
which gives $|a_n|leq 1$.
answered Nov 12 at 5:04
Hayk
2,022212
answered Nov 12 at 5:04
Hayk
2,022212
answered Nov 12 at 5:04
Hayk
2,022212
answered Nov 12 at 5:04
Hayk
2,022212
2,022212
add a comment |
add a comment |
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1
Maybe it pays to think about integrating $z^{-k}P(z)$ around the unit circle.
– Gerry Myerson
Nov 12 at 4:55