Mixing
Roots and relation between polynomials and their derivatives
Clash Royale CLAN TAG#URR8PPP
up vote
9
down vote
favorite
This is probably easy but it might be interesting. Here goes $dots$
Let $PinmathbbR[x]$ be a polynomial of degree $n>2$ and $P'=fracdPdx$. If $x_1, x_2, dots, x_n$ are the roots of $P(x)$, including multiplicities, consider the multi-variable expression
$$V_n(P)=sum_1leq i<jleq n(x_i-x_j)^2.$$
Question: Is there a constant $c(n)$, independent of $P$, such that
$V_n-1(P')=c(n)V_n(P)$? If so, what is $c(n)$?
polynomials elementary-proofs
asked Aug 29 at 0:11
T. Amdeberhan
15.7k225119
add a comment |Â
up vote
9
down vote
favorite
This is probably easy but it might be interesting. Here goes $dots$
Let $PinmathbbR[x]$ be a polynomial of degree $n>2$ and $P'=fracdPdx$. If $x_1, x_2, dots, x_n$ are the roots of $P(x)$, including multiplicities, consider the multi-variable expression
$$V_n(P)=sum_1leq i<jleq n(x_i-x_j)^2.$$
Question: Is there a constant $c(n)$, independent of $P$, such that
$V_n-1(P')=c(n)V_n(P)$? If so, what is $c(n)$?
polynomials elementary-proofs
asked Aug 29 at 0:11
T. Amdeberhan
15.7k225119
2
Note that the condition $n > 2$ can be improved to $n > 0$, as Gjergji's answer shows. The degenerate cases are a bit boring with all the empty sums, but the formula still works :)
– darij grinberg
Aug 29 at 1:03
Being curious: what is the motivation, how did you arrive to this question? Is there a heuristic argument that shows why it would be natural to have such ac(n)?
– Basj
Aug 29 at 12:52
add a comment |Â
up vote
9
down vote
favorite
up vote
9
down vote
favorite
This is probably easy but it might be interesting. Here goes $dots$
Let $PinmathbbR[x]$ be a polynomial of degree $n>2$ and $P'=fracdPdx$. If $x_1, x_2, dots, x_n$ are the roots of $P(x)$, including multiplicities, consider the multi-variable expression
$$V_n(P)=sum_1leq i<jleq n(x_i-x_j)^2.$$
Question: Is there a constant $c(n)$, independent of $P$, such that
$V_n-1(P')=c(n)V_n(P)$? If so, what is $c(n)$?
polynomials elementary-proofs
asked Aug 29 at 0:11
T. Amdeberhan
15.7k225119
This is probably easy but it might be interesting. Here goes $dots$
Let $PinmathbbR[x]$ be a polynomial of degree $n>2$ and $P'=fracdPdx$. If $x_1, x_2, dots, x_n$ are the roots of $P(x)$, including multiplicities, consider the multi-variable expression
$$V_n(P)=sum_1leq i<jleq n(x_i-x_j)^2.$$
Question: Is there a constant $c(n)$, independent of $P$, such that
$V_n-1(P')=c(n)V_n(P)$? If so, what is $c(n)$?
polynomials elementary-proofs
asked Aug 29 at 0:11
T. Amdeberhan
15.7k225119
edited Aug 29 at 14:18
asked Aug 29 at 0:11
T. Amdeberhan
15.7k225119
asked Aug 29 at 0:11
T. Amdeberhan
15.7k225119
asked Aug 29 at 0:11
T. Amdeberhan
15.7k225119
15.7k225119
2
Note that the condition $n > 2$ can be improved to $n > 0$, as Gjergji's answer shows. The degenerate cases are a bit boring with all the empty sums, but the formula still works :)
– darij grinberg
Aug 29 at 1:03
Being curious: what is the motivation, how did you arrive to this question? Is there a heuristic argument that shows why it would be natural to have such ac(n)?
– Basj
Aug 29 at 12:52
add a comment |Â
2
Note that the condition $n > 2$ can be improved to $n > 0$, as Gjergji's answer shows. The degenerate cases are a bit boring with all the empty sums, but the formula still works :)
– darij grinberg
Aug 29 at 1:03
Being curious: what is the motivation, how did you arrive to this question? Is there a heuristic argument that shows why it would be natural to have such ac(n)?
– Basj
Aug 29 at 12:52
2
2
Note that the condition $n > 2$ can be improved to $n > 0$, as Gjergji's answer shows. The degenerate cases are a bit boring with all the empty sums, but the formula still works :)
– darij grinberg
Aug 29 at 1:03
Note that the condition $n > 2$ can be improved to $n > 0$, as Gjergji's answer shows. The degenerate cases are a bit boring with all the empty sums, but the formula still works :)
– darij grinberg
Aug 29 at 1:03
Being curious: what is the motivation, how did you arrive to this question? Is there a heuristic argument that shows why it would be natural to have such a
c(n)?– Basj
Aug 29 at 12:52
Being curious: what is the motivation, how did you arrive to this question? Is there a heuristic argument that shows why it would be natural to have such a
c(n)?– Basj
Aug 29 at 12:52
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
20
down vote
accepted
Suppose that we have
$$P(x)=x^n-ax^n-1+bx^n-2+cdots$$
where we can take $P$ to be monic since it doesn't affect $V_n(P)$. From Vieta's formula we have
$$a=sum_i=1^n x_i quad , quad b=sum_1le i<jle n x_ix_j$$
so we can find that $V_n(P)=(n-1)a^2-2nb$. Similarly we have
$$V_n-1(P')=(n-2)frac(n-1)^2a^2n^2-2(n-1)frac(n-2)bn$$
$$=frac(n-1)(n-2)n^2left((n-1)a^2-2nbright)=frac(n-1)(n-2)n^2V_n(P)$$
so we get $c(n)=1-frac3n+frac2n^2$.
edited Aug 29 at 0:58
darij grinberg
17.5k368175
answered Aug 29 at 0:52
Gjergji Zaimi
58.4k3152290
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
20
down vote
accepted
Suppose that we have
$$P(x)=x^n-ax^n-1+bx^n-2+cdots$$
where we can take $P$ to be monic since it doesn't affect $V_n(P)$. From Vieta's formula we have
$$a=sum_i=1^n x_i quad , quad b=sum_1le i<jle n x_ix_j$$
so we can find that $V_n(P)=(n-1)a^2-2nb$. Similarly we have
$$V_n-1(P')=(n-2)frac(n-1)^2a^2n^2-2(n-1)frac(n-2)bn$$
$$=frac(n-1)(n-2)n^2left((n-1)a^2-2nbright)=frac(n-1)(n-2)n^2V_n(P)$$
so we get $c(n)=1-frac3n+frac2n^2$.
edited Aug 29 at 0:58
darij grinberg
17.5k368175
answered Aug 29 at 0:52
Gjergji Zaimi
58.4k3152290
add a comment |Â
up vote
20
down vote
accepted
Suppose that we have
$$P(x)=x^n-ax^n-1+bx^n-2+cdots$$
where we can take $P$ to be monic since it doesn't affect $V_n(P)$. From Vieta's formula we have
$$a=sum_i=1^n x_i quad , quad b=sum_1le i<jle n x_ix_j$$
so we can find that $V_n(P)=(n-1)a^2-2nb$. Similarly we have
$$V_n-1(P')=(n-2)frac(n-1)^2a^2n^2-2(n-1)frac(n-2)bn$$
$$=frac(n-1)(n-2)n^2left((n-1)a^2-2nbright)=frac(n-1)(n-2)n^2V_n(P)$$
so we get $c(n)=1-frac3n+frac2n^2$.
edited Aug 29 at 0:58
darij grinberg
17.5k368175
answered Aug 29 at 0:52
Gjergji Zaimi
58.4k3152290
add a comment |Â
up vote
20
down vote
accepted
up vote
20
down vote
accepted
Suppose that we have
$$P(x)=x^n-ax^n-1+bx^n-2+cdots$$
where we can take $P$ to be monic since it doesn't affect $V_n(P)$. From Vieta's formula we have
$$a=sum_i=1^n x_i quad , quad b=sum_1le i<jle n x_ix_j$$
so we can find that $V_n(P)=(n-1)a^2-2nb$. Similarly we have
$$V_n-1(P')=(n-2)frac(n-1)^2a^2n^2-2(n-1)frac(n-2)bn$$
$$=frac(n-1)(n-2)n^2left((n-1)a^2-2nbright)=frac(n-1)(n-2)n^2V_n(P)$$
so we get $c(n)=1-frac3n+frac2n^2$.
edited Aug 29 at 0:58
darij grinberg
17.5k368175
answered Aug 29 at 0:52
Gjergji Zaimi
58.4k3152290
Suppose that we have
$$P(x)=x^n-ax^n-1+bx^n-2+cdots$$
where we can take $P$ to be monic since it doesn't affect $V_n(P)$. From Vieta's formula we have
$$a=sum_i=1^n x_i quad , quad b=sum_1le i<jle n x_ix_j$$
so we can find that $V_n(P)=(n-1)a^2-2nb$. Similarly we have
$$V_n-1(P')=(n-2)frac(n-1)^2a^2n^2-2(n-1)frac(n-2)bn$$
$$=frac(n-1)(n-2)n^2left((n-1)a^2-2nbright)=frac(n-1)(n-2)n^2V_n(P)$$
so we get $c(n)=1-frac3n+frac2n^2$.
edited Aug 29 at 0:58
darij grinberg
17.5k368175
answered Aug 29 at 0:52
Gjergji Zaimi
58.4k3152290
edited Aug 29 at 0:58
darij grinberg
17.5k368175
edited Aug 29 at 0:58
darij grinberg
17.5k368175
edited Aug 29 at 0:58
darij grinberg
17.5k368175
17.5k368175
answered Aug 29 at 0:52
Gjergji Zaimi
58.4k3152290
answered Aug 29 at 0:52
Gjergji Zaimi
58.4k3152290
answered Aug 29 at 0:52
Gjergji Zaimi
58.4k3152290
58.4k3152290
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f309340%2froots-and-relation-between-polynomials-and-their-derivatives%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
Note that the condition $n > 2$ can be improved to $n > 0$, as Gjergji's answer shows. The degenerate cases are a bit boring with all the empty sums, but the formula still works :)
– darij grinberg
Aug 29 at 1:03
Being curious: what is the motivation, how did you arrive to this question? Is there a heuristic argument that shows why it would be natural to have such a
c(n)?– Basj
Aug 29 at 12:52