Lower bound for complex polynomial beyond circle or radius R [duplicate]
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For a polynomial $p(z)$, prove there exist $R>0$, such that if $|z|=R$, then $|p(z)|geq |a_n|R^n/2$
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If we have a polynomial with $c_i$ a complex number
$$c_nz^n + c_n-1z^n-1 + cdots + c_1 z + c_0$$
then $$|P(z)| > fracR^n2$$
When |z| > R for some R
I have tried using the triangle inequality where I obtain,
$|P(Z)| leq |c_n||z|^n + cdots + |c_0|$. But I seem to keep getting stuck. Any hints on how to move forward? Thank you!
complex-analysis polynomials upper-lower-bounds
asked Sep 10 at 17:52
rannoudanames
441614
marked as duplicate by Martin R, Community♦ Sep 10 at 18:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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up vote
3
down vote
favorite
This question already has an answer here:
For a polynomial $p(z)$, prove there exist $R>0$, such that if $|z|=R$, then $|p(z)|geq |a_n|R^n/2$
2 answers
If we have a polynomial with $c_i$ a complex number
$$c_nz^n + c_n-1z^n-1 + cdots + c_1 z + c_0$$
then $$|P(z)| > fracR^n2$$
When |z| > R for some R
I have tried using the triangle inequality where I obtain,
$|P(Z)| leq |c_n||z|^n + cdots + |c_0|$. But I seem to keep getting stuck. Any hints on how to move forward? Thank you!
complex-analysis polynomials upper-lower-bounds
asked Sep 10 at 17:52
rannoudanames
441614
marked as duplicate by Martin R, Community♦ Sep 10 at 18:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This question already has an answer here:
For a polynomial $p(z)$, prove there exist $R>0$, such that if $|z|=R$, then $|p(z)|geq |a_n|R^n/2$
2 answers
If we have a polynomial with $c_i$ a complex number
$$c_nz^n + c_n-1z^n-1 + cdots + c_1 z + c_0$$
then $$|P(z)| > fracR^n2$$
When |z| > R for some R
I have tried using the triangle inequality where I obtain,
$|P(Z)| leq |c_n||z|^n + cdots + |c_0|$. But I seem to keep getting stuck. Any hints on how to move forward? Thank you!
complex-analysis polynomials upper-lower-bounds
asked Sep 10 at 17:52
rannoudanames
441614
This question already has an answer here:
For a polynomial $p(z)$, prove there exist $R>0$, such that if $|z|=R$, then $|p(z)|geq |a_n|R^n/2$
2 answers
If we have a polynomial with $c_i$ a complex number
$$c_nz^n + c_n-1z^n-1 + cdots + c_1 z + c_0$$
then $$|P(z)| > fracR^n2$$
When |z| > R for some R
I have tried using the triangle inequality where I obtain,
$|P(Z)| leq |c_n||z|^n + cdots + |c_0|$. But I seem to keep getting stuck. Any hints on how to move forward? Thank you!
This question already has an answer here:
For a polynomial $p(z)$, prove there exist $R>0$, such that if $|z|=R$, then $|p(z)|geq |a_n|R^n/2$
2 answers
complex-analysis polynomials upper-lower-bounds
complex-analysis polynomials upper-lower-bounds
asked Sep 10 at 17:52
rannoudanames
441614
asked Sep 10 at 17:52
rannoudanames
441614
asked Sep 10 at 17:52
rannoudanames
441614
asked Sep 10 at 17:52
rannoudanames
441614
asked Sep 10 at 17:52
rannoudanames
441614
441614
marked as duplicate by Martin R, Community♦ Sep 10 at 18:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, Community♦ Sep 10 at 18:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
add a comment |Â
2 Answers
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You have
$$fracz
=left|c_n+fracc_n-1z+fracc_n-2z^2+cdots+fracc_0z^nright|.$$
Show that if $|z|$ is large enough, then
$$left|fracc_n-1z+fracc_n-2z^2+cdots+fracc_0z^nright|<frac2$$
etc.
answered Sep 10 at 17:57
Lord Shark the Unknown
90.3k955117
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This is part of Growth lemma, we have
$$|p(z)|=|z^n(c_n +dfrac c_n-1z + cdots + dfracc_1z^n-1 + dfracc_0z^n)|$$
for $|z|=r$
$$|p(z)|geq r^nleft(|c_n|-|dfrac c_n-1z + cdots + dfracc_1z^n-1 + dfracc_0z^n)|right)$$
but
$$|dfrac c_n-1z + cdots + dfracc_1z^n-1 + dfracc_0z^n)|leqdfrac1rleft(|c_n-1+cdots+|c_0|right)=dfracalphar$$
therefore
$$|p(z)|geq r^nleft(|c_n|-dfracalpharright)$$
now let $r>dfrac2alpha$.
answered Sep 10 at 18:07
Nosrati
22.8k61951
(+1) nice. Would you be interested in supporting the proposal of a "Math Challenges" site? Please feel free to drop in this chatroom for discussion :)
– TheSimpliFire
Sep 11 at 6:59
@TheSimpliFire thank you very much.
– Nosrati
Sep 11 at 8:22
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You have
$$fracz
=left|c_n+fracc_n-1z+fracc_n-2z^2+cdots+fracc_0z^nright|.$$
Show that if $|z|$ is large enough, then
$$left|fracc_n-1z+fracc_n-2z^2+cdots+fracc_0z^nright|<frac2$$
etc.
answered Sep 10 at 17:57
Lord Shark the Unknown
90.3k955117
add a comment |Â
up vote
1
down vote
accepted
You have
$$fracz
=left|c_n+fracc_n-1z+fracc_n-2z^2+cdots+fracc_0z^nright|.$$
Show that if $|z|$ is large enough, then
$$left|fracc_n-1z+fracc_n-2z^2+cdots+fracc_0z^nright|<frac2$$
etc.
answered Sep 10 at 17:57
Lord Shark the Unknown
90.3k955117
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You have
$$fracz
=left|c_n+fracc_n-1z+fracc_n-2z^2+cdots+fracc_0z^nright|.$$
Show that if $|z|$ is large enough, then
$$left|fracc_n-1z+fracc_n-2z^2+cdots+fracc_0z^nright|<frac2$$
etc.
answered Sep 10 at 17:57
Lord Shark the Unknown
90.3k955117
You have
$$fracz
=left|c_n+fracc_n-1z+fracc_n-2z^2+cdots+fracc_0z^nright|.$$
Show that if $|z|$ is large enough, then
$$left|fracc_n-1z+fracc_n-2z^2+cdots+fracc_0z^nright|<frac2$$
etc.
answered Sep 10 at 17:57
Lord Shark the Unknown
90.3k955117
answered Sep 10 at 17:57
Lord Shark the Unknown
90.3k955117
answered Sep 10 at 17:57
Lord Shark the Unknown
90.3k955117
answered Sep 10 at 17:57
Lord Shark the Unknown
90.3k955117
90.3k955117
add a comment |Â
add a comment |Â
up vote
3
down vote
This is part of Growth lemma, we have
$$|p(z)|=|z^n(c_n +dfrac c_n-1z + cdots + dfracc_1z^n-1 + dfracc_0z^n)|$$
for $|z|=r$
$$|p(z)|geq r^nleft(|c_n|-|dfrac c_n-1z + cdots + dfracc_1z^n-1 + dfracc_0z^n)|right)$$
but
$$|dfrac c_n-1z + cdots + dfracc_1z^n-1 + dfracc_0z^n)|leqdfrac1rleft(|c_n-1+cdots+|c_0|right)=dfracalphar$$
therefore
$$|p(z)|geq r^nleft(|c_n|-dfracalpharright)$$
now let $r>dfrac2alpha$.
answered Sep 10 at 18:07
Nosrati
22.8k61951
(+1) nice. Would you be interested in supporting the proposal of a "Math Challenges" site? Please feel free to drop in this chatroom for discussion :)
– TheSimpliFire
Sep 11 at 6:59
@TheSimpliFire thank you very much.
– Nosrati
Sep 11 at 8:22
add a comment |Â
up vote
3
down vote
This is part of Growth lemma, we have
$$|p(z)|=|z^n(c_n +dfrac c_n-1z + cdots + dfracc_1z^n-1 + dfracc_0z^n)|$$
for $|z|=r$
$$|p(z)|geq r^nleft(|c_n|-|dfrac c_n-1z + cdots + dfracc_1z^n-1 + dfracc_0z^n)|right)$$
but
$$|dfrac c_n-1z + cdots + dfracc_1z^n-1 + dfracc_0z^n)|leqdfrac1rleft(|c_n-1+cdots+|c_0|right)=dfracalphar$$
therefore
$$|p(z)|geq r^nleft(|c_n|-dfracalpharright)$$
now let $r>dfrac2alpha$.
answered Sep 10 at 18:07
Nosrati
22.8k61951
(+1) nice. Would you be interested in supporting the proposal of a "Math Challenges" site? Please feel free to drop in this chatroom for discussion :)
– TheSimpliFire
Sep 11 at 6:59
@TheSimpliFire thank you very much.
– Nosrati
Sep 11 at 8:22
add a comment |Â
up vote
3
down vote
up vote
3
down vote
This is part of Growth lemma, we have
$$|p(z)|=|z^n(c_n +dfrac c_n-1z + cdots + dfracc_1z^n-1 + dfracc_0z^n)|$$
for $|z|=r$
$$|p(z)|geq r^nleft(|c_n|-|dfrac c_n-1z + cdots + dfracc_1z^n-1 + dfracc_0z^n)|right)$$
but
$$|dfrac c_n-1z + cdots + dfracc_1z^n-1 + dfracc_0z^n)|leqdfrac1rleft(|c_n-1+cdots+|c_0|right)=dfracalphar$$
therefore
$$|p(z)|geq r^nleft(|c_n|-dfracalpharright)$$
now let $r>dfrac2alpha$.
answered Sep 10 at 18:07
Nosrati
22.8k61951
This is part of Growth lemma, we have
$$|p(z)|=|z^n(c_n +dfrac c_n-1z + cdots + dfracc_1z^n-1 + dfracc_0z^n)|$$
for $|z|=r$
$$|p(z)|geq r^nleft(|c_n|-|dfrac c_n-1z + cdots + dfracc_1z^n-1 + dfracc_0z^n)|right)$$
but
$$|dfrac c_n-1z + cdots + dfracc_1z^n-1 + dfracc_0z^n)|leqdfrac1rleft(|c_n-1+cdots+|c_0|right)=dfracalphar$$
therefore
$$|p(z)|geq r^nleft(|c_n|-dfracalpharright)$$
now let $r>dfrac2alpha$.
answered Sep 10 at 18:07
Nosrati
22.8k61951
answered Sep 10 at 18:07
Nosrati
22.8k61951
answered Sep 10 at 18:07
Nosrati
22.8k61951
answered Sep 10 at 18:07
Nosrati
22.8k61951
22.8k61951
(+1) nice. Would you be interested in supporting the proposal of a "Math Challenges" site? Please feel free to drop in this chatroom for discussion :)
– TheSimpliFire
Sep 11 at 6:59
@TheSimpliFire thank you very much.
– Nosrati
Sep 11 at 8:22
add a comment |Â
(+1) nice. Would you be interested in supporting the proposal of a "Math Challenges" site? Please feel free to drop in this chatroom for discussion :)
– TheSimpliFire
Sep 11 at 6:59
@TheSimpliFire thank you very much.
– Nosrati
Sep 11 at 8:22
(+1) nice. Would you be interested in supporting the proposal of a "Math Challenges" site? Please feel free to drop in this chatroom for discussion :)
– TheSimpliFire
Sep 11 at 6:59
(+1) nice. Would you be interested in supporting the proposal of a "Math Challenges" site? Please feel free to drop in this chatroom for discussion :)
– TheSimpliFire
Sep 11 at 6:59
@TheSimpliFire thank you very much.
– Nosrati
Sep 11 at 8:22
@TheSimpliFire thank you very much.
– Nosrati
Sep 11 at 8:22
add a comment |Â
