How to prove that a polynomial is bounded below [closed]
Show that if
$$P(z) = {a_n}z^{n} + {a_{n-1}}z^{n-1} + ... + {a_1}z + {a_0},$$
is a polynomial of degree $n$ then there is ${R_0} > 0$ such that for $|z| > {R_0}$ the polynomial can be bounded below as
$$|P(z)| > frac{1}{2}|{a_n}||z|^{n}.$$
How do I have to start this proof?
complex-analysis polynomials
edited Nov 29 at 14:15
Tianlalu
3,01021038
asked Nov 29 at 12:59
Peter van de Berg
198
closed as off-topic by Did, Brahadeesh, Alexander Gruber♦ Nov 30 at 3:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Brahadeesh, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Show that if
$$P(z) = {a_n}z^{n} + {a_{n-1}}z^{n-1} + ... + {a_1}z + {a_0},$$
is a polynomial of degree $n$ then there is ${R_0} > 0$ such that for $|z| > {R_0}$ the polynomial can be bounded below as
$$|P(z)| > frac{1}{2}|{a_n}||z|^{n}.$$
How do I have to start this proof?
complex-analysis polynomials
edited Nov 29 at 14:15
Tianlalu
3,01021038
asked Nov 29 at 12:59
Peter van de Berg
198
closed as off-topic by Did, Brahadeesh, Alexander Gruber♦ Nov 30 at 3:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Brahadeesh, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
$P(z)=a_nz^nleft(1+frac{a_{n-1}}{a_nz}+...+frac{a_0}{a_nz^n}right)$ and $frac{a_i}{a_n z^k}to 0$ when $|z|to infty $.
– Surb
Nov 29 at 13:05
add a comment |
Show that if
$$P(z) = {a_n}z^{n} + {a_{n-1}}z^{n-1} + ... + {a_1}z + {a_0},$$
is a polynomial of degree $n$ then there is ${R_0} > 0$ such that for $|z| > {R_0}$ the polynomial can be bounded below as
$$|P(z)| > frac{1}{2}|{a_n}||z|^{n}.$$
How do I have to start this proof?
complex-analysis polynomials
edited Nov 29 at 14:15
Tianlalu
3,01021038
asked Nov 29 at 12:59
Peter van de Berg
198
Show that if
$$P(z) = {a_n}z^{n} + {a_{n-1}}z^{n-1} + ... + {a_1}z + {a_0},$$
is a polynomial of degree $n$ then there is ${R_0} > 0$ such that for $|z| > {R_0}$ the polynomial can be bounded below as
$$|P(z)| > frac{1}{2}|{a_n}||z|^{n}.$$
How do I have to start this proof?
complex-analysis polynomials
complex-analysis polynomials
edited Nov 29 at 14:15
Tianlalu
3,01021038
asked Nov 29 at 12:59
Peter van de Berg
198
edited Nov 29 at 14:15
Tianlalu
3,01021038
asked Nov 29 at 12:59
Peter van de Berg
198
edited Nov 29 at 14:15
Tianlalu
3,01021038
edited Nov 29 at 14:15
Tianlalu
3,01021038
edited Nov 29 at 14:15
Tianlalu
3,01021038
3,01021038
asked Nov 29 at 12:59
Peter van de Berg
198
asked Nov 29 at 12:59
Peter van de Berg
198
asked Nov 29 at 12:59
Peter van de Berg
198
198
closed as off-topic by Did, Brahadeesh, Alexander Gruber♦ Nov 30 at 3:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Brahadeesh, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, Brahadeesh, Alexander Gruber♦ Nov 30 at 3:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Brahadeesh, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
$P(z)=a_nz^nleft(1+frac{a_{n-1}}{a_nz}+...+frac{a_0}{a_nz^n}right)$ and $frac{a_i}{a_n z^k}to 0$ when $|z|to infty $.
– Surb
Nov 29 at 13:05
add a comment |
$P(z)=a_nz^nleft(1+frac{a_{n-1}}{a_nz}+...+frac{a_0}{a_nz^n}right)$ and $frac{a_i}{a_n z^k}to 0$ when $|z|to infty $.
– Surb
Nov 29 at 13:05
$P(z)=a_nz^nleft(1+frac{a_{n-1}}{a_nz}+...+frac{a_0}{a_nz^n}right)$ and $frac{a_i}{a_n z^k}to 0$ when $|z|to infty $.
– Surb
Nov 29 at 13:05
$P(z)=a_nz^nleft(1+frac{a_{n-1}}{a_nz}+...+frac{a_0}{a_nz^n}right)$ and $frac{a_i}{a_n z^k}to 0$ when $|z|to infty $.
– Surb
Nov 29 at 13:05
add a comment |
1 Answer
1
active
oldest
votes
You can start with the triangle inequality:
$$|P(z)|geq |a_n||z|^n-|a_{n-1}||z|^{n-1}-cdots-|a_0|$$
so that
$$|P(z)|geq |a_n||z|^nleft(1-dfrac{|a_{n-1}|}{|z|}-cdots-dfrac{|a_0|}{|z|^n}right)$$
so we just need to show that
$$dfrac{|a_{n-1}|}{|z|}+cdots+dfrac{|a_0|}{|z|^n}leqdfrac{1}{2}$$
if $|z|$ is sufficiently large. But this holds since
$$lim_{|z|rightarrowinfty}dfrac{|a_{n-1}|}{|z|}+cdots+dfrac{|a_0|}{|z|^n}=0$$
You might want to formalize that last bit with $varepsilon$ and $R_0$ depending how rigorous your proof needs to be.
answered Nov 29 at 13:06
Olivier Moschetta
2,7761411
I shall try to start, the proof needs to be very precisely by the way, but thanks!
– Peter van de Berg
Nov 29 at 13:12
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can start with the triangle inequality:
$$|P(z)|geq |a_n||z|^n-|a_{n-1}||z|^{n-1}-cdots-|a_0|$$
so that
$$|P(z)|geq |a_n||z|^nleft(1-dfrac{|a_{n-1}|}{|z|}-cdots-dfrac{|a_0|}{|z|^n}right)$$
so we just need to show that
$$dfrac{|a_{n-1}|}{|z|}+cdots+dfrac{|a_0|}{|z|^n}leqdfrac{1}{2}$$
if $|z|$ is sufficiently large. But this holds since
$$lim_{|z|rightarrowinfty}dfrac{|a_{n-1}|}{|z|}+cdots+dfrac{|a_0|}{|z|^n}=0$$
You might want to formalize that last bit with $varepsilon$ and $R_0$ depending how rigorous your proof needs to be.
answered Nov 29 at 13:06
Olivier Moschetta
2,7761411
I shall try to start, the proof needs to be very precisely by the way, but thanks!
– Peter van de Berg
Nov 29 at 13:12
add a comment |
You can start with the triangle inequality:
$$|P(z)|geq |a_n||z|^n-|a_{n-1}||z|^{n-1}-cdots-|a_0|$$
so that
$$|P(z)|geq |a_n||z|^nleft(1-dfrac{|a_{n-1}|}{|z|}-cdots-dfrac{|a_0|}{|z|^n}right)$$
so we just need to show that
$$dfrac{|a_{n-1}|}{|z|}+cdots+dfrac{|a_0|}{|z|^n}leqdfrac{1}{2}$$
if $|z|$ is sufficiently large. But this holds since
$$lim_{|z|rightarrowinfty}dfrac{|a_{n-1}|}{|z|}+cdots+dfrac{|a_0|}{|z|^n}=0$$
You might want to formalize that last bit with $varepsilon$ and $R_0$ depending how rigorous your proof needs to be.
answered Nov 29 at 13:06
Olivier Moschetta
2,7761411
I shall try to start, the proof needs to be very precisely by the way, but thanks!
– Peter van de Berg
Nov 29 at 13:12
add a comment |
You can start with the triangle inequality:
$$|P(z)|geq |a_n||z|^n-|a_{n-1}||z|^{n-1}-cdots-|a_0|$$
so that
$$|P(z)|geq |a_n||z|^nleft(1-dfrac{|a_{n-1}|}{|z|}-cdots-dfrac{|a_0|}{|z|^n}right)$$
so we just need to show that
$$dfrac{|a_{n-1}|}{|z|}+cdots+dfrac{|a_0|}{|z|^n}leqdfrac{1}{2}$$
if $|z|$ is sufficiently large. But this holds since
$$lim_{|z|rightarrowinfty}dfrac{|a_{n-1}|}{|z|}+cdots+dfrac{|a_0|}{|z|^n}=0$$
You might want to formalize that last bit with $varepsilon$ and $R_0$ depending how rigorous your proof needs to be.
answered Nov 29 at 13:06
Olivier Moschetta
2,7761411
You can start with the triangle inequality:
$$|P(z)|geq |a_n||z|^n-|a_{n-1}||z|^{n-1}-cdots-|a_0|$$
so that
$$|P(z)|geq |a_n||z|^nleft(1-dfrac{|a_{n-1}|}{|z|}-cdots-dfrac{|a_0|}{|z|^n}right)$$
so we just need to show that
$$dfrac{|a_{n-1}|}{|z|}+cdots+dfrac{|a_0|}{|z|^n}leqdfrac{1}{2}$$
if $|z|$ is sufficiently large. But this holds since
$$lim_{|z|rightarrowinfty}dfrac{|a_{n-1}|}{|z|}+cdots+dfrac{|a_0|}{|z|^n}=0$$
You might want to formalize that last bit with $varepsilon$ and $R_0$ depending how rigorous your proof needs to be.
answered Nov 29 at 13:06
Olivier Moschetta
2,7761411
answered Nov 29 at 13:06
Olivier Moschetta
2,7761411
answered Nov 29 at 13:06
Olivier Moschetta
2,7761411
answered Nov 29 at 13:06
Olivier Moschetta
2,7761411
2,7761411
I shall try to start, the proof needs to be very precisely by the way, but thanks!
– Peter van de Berg
Nov 29 at 13:12
add a comment |
I shall try to start, the proof needs to be very precisely by the way, but thanks!
– Peter van de Berg
Nov 29 at 13:12
I shall try to start, the proof needs to be very precisely by the way, but thanks!
– Peter van de Berg
Nov 29 at 13:12
I shall try to start, the proof needs to be very precisely by the way, but thanks!
– Peter van de Berg
Nov 29 at 13:12
add a comment |
$P(z)=a_nz^nleft(1+frac{a_{n-1}}{a_nz}+...+frac{a_0}{a_nz^n}right)$ and $frac{a_i}{a_n z^k}to 0$ when $|z|to infty $.
– Surb
Nov 29 at 13:05