If $f(x)=sum_{i=0}^{n}binom{n}{i}a_{i}x^i$ then also $g(x)=sum_{i=0}^{n}binom{n}{i}b_{i}x^i$ [closed]
$begingroup$
Let $$f(x)=sum_{i=0}^{n}binom{n}{i}a_{i}x^i,~~~~~~~~g(x)=sum_{i=0}^{n}binom{n}{i}b_{i}x^i$$where $b_{i}in (a_{i}- varepsilon,a_{i}+ varepsilon),a_{i}in mathbb{R},(i=1,2,cdots,n),varepsilonto 0^{+}$
If there exists $zin mathbb{C}backslash mathbb{R}$, such that $f(z)=0$,
$f$ is not all real roots,show $g$ is not all real roots
I conjecture:
there also exists $z'in mathbb{C}backslash mathbb{R}$ such that
$g(z')=0$.
on the other words: if $f$ is not all real roots,show $g$ is not all real roots?
Can you help me to prove it? Thanks.
real-analysis
asked Dec 5 '18 at 5:59
function sugfunction sug
3631438
$endgroup$
closed as off-topic by Kavi Rama Murthy, user10354138, Rebellos, Vidyanshu Mishra, Brahadeesh Dec 5 '18 at 8:01
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, Rebellos, Vidyanshu Mishra
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $$f(x)=sum_{i=0}^{n}binom{n}{i}a_{i}x^i,~~~~~~~~g(x)=sum_{i=0}^{n}binom{n}{i}b_{i}x^i$$where $b_{i}in (a_{i}- varepsilon,a_{i}+ varepsilon),a_{i}in mathbb{R},(i=1,2,cdots,n),varepsilonto 0^{+}$
If there exists $zin mathbb{C}backslash mathbb{R}$, such that $f(z)=0$,
$f$ is not all real roots,show $g$ is not all real roots
I conjecture:
there also exists $z'in mathbb{C}backslash mathbb{R}$ such that
$g(z')=0$.
on the other words: if $f$ is not all real roots,show $g$ is not all real roots?
Can you help me to prove it? Thanks.
real-analysis
asked Dec 5 '18 at 5:59
function sugfunction sug
3631438
$endgroup$
closed as off-topic by Kavi Rama Murthy, user10354138, Rebellos, Vidyanshu Mishra, Brahadeesh Dec 5 '18 at 8:01
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, Rebellos, Vidyanshu Mishra
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
As I understand it, you want to prove that for any polynomial $f$ with a non-real root, then all polynomials $g$ with the same degree and coefficients which are "sufficiently close" to that of $f$ will also have such a non-real root. Sufficiently closed means their "distance" is smaller than some threshold, possibly depending on $f$, where the distance is $sup_i |a_i-b_i|$. Is my understanding correct?
$endgroup$
– Mike Earnest
Dec 5 '18 at 6:12
add a comment |
$begingroup$
Let $$f(x)=sum_{i=0}^{n}binom{n}{i}a_{i}x^i,~~~~~~~~g(x)=sum_{i=0}^{n}binom{n}{i}b_{i}x^i$$where $b_{i}in (a_{i}- varepsilon,a_{i}+ varepsilon),a_{i}in mathbb{R},(i=1,2,cdots,n),varepsilonto 0^{+}$
If there exists $zin mathbb{C}backslash mathbb{R}$, such that $f(z)=0$,
$f$ is not all real roots,show $g$ is not all real roots
I conjecture:
there also exists $z'in mathbb{C}backslash mathbb{R}$ such that
$g(z')=0$.
on the other words: if $f$ is not all real roots,show $g$ is not all real roots?
Can you help me to prove it? Thanks.
real-analysis
asked Dec 5 '18 at 5:59
function sugfunction sug
3631438
$endgroup$
Let $$f(x)=sum_{i=0}^{n}binom{n}{i}a_{i}x^i,~~~~~~~~g(x)=sum_{i=0}^{n}binom{n}{i}b_{i}x^i$$where $b_{i}in (a_{i}- varepsilon,a_{i}+ varepsilon),a_{i}in mathbb{R},(i=1,2,cdots,n),varepsilonto 0^{+}$
If there exists $zin mathbb{C}backslash mathbb{R}$, such that $f(z)=0$,
$f$ is not all real roots,show $g$ is not all real roots
I conjecture:
there also exists $z'in mathbb{C}backslash mathbb{R}$ such that
$g(z')=0$.
on the other words: if $f$ is not all real roots,show $g$ is not all real roots?
Can you help me to prove it? Thanks.
real-analysis
real-analysis
asked Dec 5 '18 at 5:59
function sugfunction sug
3631438
asked Dec 5 '18 at 5:59
function sugfunction sug
3631438
edited Dec 5 '18 at 6:32
function sug
asked Dec 5 '18 at 5:59
function sugfunction sug
3631438
asked Dec 5 '18 at 5:59
function sugfunction sug
3631438
asked Dec 5 '18 at 5:59
function sugfunction sug
3631438
3631438
closed as off-topic by Kavi Rama Murthy, user10354138, Rebellos, Vidyanshu Mishra, Brahadeesh Dec 5 '18 at 8:01
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, Rebellos, Vidyanshu Mishra
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Kavi Rama Murthy, user10354138, Rebellos, Vidyanshu Mishra, Brahadeesh Dec 5 '18 at 8:01
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, Rebellos, Vidyanshu Mishra
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
As I understand it, you want to prove that for any polynomial $f$ with a non-real root, then all polynomials $g$ with the same degree and coefficients which are "sufficiently close" to that of $f$ will also have such a non-real root. Sufficiently closed means their "distance" is smaller than some threshold, possibly depending on $f$, where the distance is $sup_i |a_i-b_i|$. Is my understanding correct?
$endgroup$
– Mike Earnest
Dec 5 '18 at 6:12
add a comment |
$begingroup$
As I understand it, you want to prove that for any polynomial $f$ with a non-real root, then all polynomials $g$ with the same degree and coefficients which are "sufficiently close" to that of $f$ will also have such a non-real root. Sufficiently closed means their "distance" is smaller than some threshold, possibly depending on $f$, where the distance is $sup_i |a_i-b_i|$. Is my understanding correct?
$endgroup$
– Mike Earnest
Dec 5 '18 at 6:12
$begingroup$
As I understand it, you want to prove that for any polynomial $f$ with a non-real root, then all polynomials $g$ with the same degree and coefficients which are "sufficiently close" to that of $f$ will also have such a non-real root. Sufficiently closed means their "distance" is smaller than some threshold, possibly depending on $f$, where the distance is $sup_i |a_i-b_i|$. Is my understanding correct?
$endgroup$
– Mike Earnest
Dec 5 '18 at 6:12
$begingroup$
As I understand it, you want to prove that for any polynomial $f$ with a non-real root, then all polynomials $g$ with the same degree and coefficients which are "sufficiently close" to that of $f$ will also have such a non-real root. Sufficiently closed means their "distance" is smaller than some threshold, possibly depending on $f$, where the distance is $sup_i |a_i-b_i|$. Is my understanding correct?
$endgroup$
– Mike Earnest
Dec 5 '18 at 6:12
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Let's try the simplest case $n=2$ (as every unary function with real coefficients has just one root which is real).
Let $g(x)=x^2+2x+1$, that is, $b_0=b_1=b_2=1$. Then $-1$ is a double root, so no complex roots.
Let $f(x)=x^2+2x+1+varepsilon$, that is, $a_0=a_1=1$ and $a_2=1+varepsilon$, for $varepsilon>0$.
These have two complex roots.
answered Dec 5 '18 at 6:07
A. PongráczA. Pongrácz
5,9531929
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's try the simplest case $n=2$ (as every unary function with real coefficients has just one root which is real).
Let $g(x)=x^2+2x+1$, that is, $b_0=b_1=b_2=1$. Then $-1$ is a double root, so no complex roots.
Let $f(x)=x^2+2x+1+varepsilon$, that is, $a_0=a_1=1$ and $a_2=1+varepsilon$, for $varepsilon>0$.
These have two complex roots.
answered Dec 5 '18 at 6:07
A. PongráczA. Pongrácz
5,9531929
$endgroup$
add a comment |
$begingroup$
Let's try the simplest case $n=2$ (as every unary function with real coefficients has just one root which is real).
Let $g(x)=x^2+2x+1$, that is, $b_0=b_1=b_2=1$. Then $-1$ is a double root, so no complex roots.
Let $f(x)=x^2+2x+1+varepsilon$, that is, $a_0=a_1=1$ and $a_2=1+varepsilon$, for $varepsilon>0$.
These have two complex roots.
answered Dec 5 '18 at 6:07
A. PongráczA. Pongrácz
5,9531929
$endgroup$
add a comment |
$begingroup$
Let's try the simplest case $n=2$ (as every unary function with real coefficients has just one root which is real).
Let $g(x)=x^2+2x+1$, that is, $b_0=b_1=b_2=1$. Then $-1$ is a double root, so no complex roots.
Let $f(x)=x^2+2x+1+varepsilon$, that is, $a_0=a_1=1$ and $a_2=1+varepsilon$, for $varepsilon>0$.
These have two complex roots.
answered Dec 5 '18 at 6:07
A. PongráczA. Pongrácz
5,9531929
$endgroup$
Let's try the simplest case $n=2$ (as every unary function with real coefficients has just one root which is real).
Let $g(x)=x^2+2x+1$, that is, $b_0=b_1=b_2=1$. Then $-1$ is a double root, so no complex roots.
Let $f(x)=x^2+2x+1+varepsilon$, that is, $a_0=a_1=1$ and $a_2=1+varepsilon$, for $varepsilon>0$.
These have two complex roots.
answered Dec 5 '18 at 6:07
A. PongráczA. Pongrácz
5,9531929
edited Dec 5 '18 at 6:14
answered Dec 5 '18 at 6:07
A. PongráczA. Pongrácz
5,9531929
answered Dec 5 '18 at 6:07
A. PongráczA. Pongrácz
5,9531929
answered Dec 5 '18 at 6:07
A. PongráczA. Pongrácz
5,9531929
5,9531929
add a comment |
add a comment |
$begingroup$
As I understand it, you want to prove that for any polynomial $f$ with a non-real root, then all polynomials $g$ with the same degree and coefficients which are "sufficiently close" to that of $f$ will also have such a non-real root. Sufficiently closed means their "distance" is smaller than some threshold, possibly depending on $f$, where the distance is $sup_i |a_i-b_i|$. Is my understanding correct?
$endgroup$
– Mike Earnest
Dec 5 '18 at 6:12