Algebraic vs. analytic definition of the multiplicity of a polynomial's root The Next CEO of Stack OverflowFour polynomials with single-rooted sumsLogic behind cubic resolution of $x^4+px^2+qx+r=0$Show that a complex polynomial of degree $n$ doesn't have zeros in a unit ballRoots of polynomials: Vieta's FormulaCan the roots of the derivative of the polynomial in complex variable be as close as we want them to be from the roots of the polynomial itself?Factor ring of polynomials by an ideal that corresponds to multiple rootsWhen do non-real roots of polynomials P(x) and R(x) have a real component in common?Do roots of a monic polynomial contain an open cube?All integer polynomials of the form $x^d_t-x^d_1-1-x^d_2-1…x^d_t-d_t$ with the same maximal (real) root.Proof: For every $alpha in mathbbC$, there exists a unique $beta in mathbbC$ such that $alpha + beta = 0$
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Algebraic vs. analytic definition of the multiplicity of a polynomial's root
The Next CEO of Stack OverflowFour polynomials with single-rooted sumsLogic behind cubic resolution of $x^4+px^2+qx+r=0$Show that a complex polynomial of degree $n$ doesn't have zeros in a unit ballRoots of polynomials: Vieta's FormulaCan the roots of the derivative of the polynomial in complex variable be as close as we want them to be from the roots of the polynomial itself?Factor ring of polynomials by an ideal that corresponds to multiple rootsWhen do non-real roots of polynomials P(x) and R(x) have a real component in common?Do roots of a monic polynomial contain an open cube?All integer polynomials of the form $x^d_t-x^d_1-1-x^d_2-1…x^d_t-d_t$ with the same maximal (real) root.Proof: For every $alpha in mathbbC$, there exists a unique $beta in mathbbC$ such that $alpha + beta = 0$
$begingroup$
Let $f(x) = a(x - c_1)^d_1(x - c_2)^d_2 dots (x - c_n)^d_n$ be a polynomial over the complex numbers ($n, d_i in 1, 2, dots$, $a in mathbbCsetminus 0$), where the roots $c_1, c_2, dots, c_n$ are pairwise distinct. It is known that for every $i in 1, 2, dots, n$ $c_i$ is a root of $f^(m)$ for $m in 0, 1, dots, d_i - 1$. Is it possible that $c_i$ is a root of $f^(d_i)$?
complex-analysis derivatives polynomials roots factoring
asked Apr 10 '15 at 12:28
Evan AadEvan Aad
5,62911854
$endgroup$
add a comment |
$begingroup$
Let $f(x) = a(x - c_1)^d_1(x - c_2)^d_2 dots (x - c_n)^d_n$ be a polynomial over the complex numbers ($n, d_i in 1, 2, dots$, $a in mathbbCsetminus 0$), where the roots $c_1, c_2, dots, c_n$ are pairwise distinct. It is known that for every $i in 1, 2, dots, n$ $c_i$ is a root of $f^(m)$ for $m in 0, 1, dots, d_i - 1$. Is it possible that $c_i$ is a root of $f^(d_i)$?
complex-analysis derivatives polynomials roots factoring
asked Apr 10 '15 at 12:28
Evan AadEvan Aad
5,62911854
$endgroup$
1
$begingroup$
You mean a root of $f^(d_i)$?
$endgroup$
– Jean-Claude Arbaut
Apr 10 '15 at 12:35
$begingroup$
@Jean-ClaudeArbaut: Yes, I do. I've corrected the question. Thanks.
$endgroup$
– Evan Aad
Apr 10 '15 at 12:36
add a comment |
$begingroup$
Let $f(x) = a(x - c_1)^d_1(x - c_2)^d_2 dots (x - c_n)^d_n$ be a polynomial over the complex numbers ($n, d_i in 1, 2, dots$, $a in mathbbCsetminus 0$), where the roots $c_1, c_2, dots, c_n$ are pairwise distinct. It is known that for every $i in 1, 2, dots, n$ $c_i$ is a root of $f^(m)$ for $m in 0, 1, dots, d_i - 1$. Is it possible that $c_i$ is a root of $f^(d_i)$?
complex-analysis derivatives polynomials roots factoring
asked Apr 10 '15 at 12:28
Evan AadEvan Aad
5,62911854
$endgroup$
Let $f(x) = a(x - c_1)^d_1(x - c_2)^d_2 dots (x - c_n)^d_n$ be a polynomial over the complex numbers ($n, d_i in 1, 2, dots$, $a in mathbbCsetminus 0$), where the roots $c_1, c_2, dots, c_n$ are pairwise distinct. It is known that for every $i in 1, 2, dots, n$ $c_i$ is a root of $f^(m)$ for $m in 0, 1, dots, d_i - 1$. Is it possible that $c_i$ is a root of $f^(d_i)$?
complex-analysis derivatives polynomials roots factoring
complex-analysis derivatives polynomials roots factoring
asked Apr 10 '15 at 12:28
Evan AadEvan Aad
5,62911854
asked Apr 10 '15 at 12:28
Evan AadEvan Aad
5,62911854
edited Apr 10 '15 at 12:35
Evan Aad
asked Apr 10 '15 at 12:28
Evan AadEvan Aad
5,62911854
asked Apr 10 '15 at 12:28
Evan AadEvan Aad
5,62911854
asked Apr 10 '15 at 12:28
Evan AadEvan Aad
5,62911854
5,62911854
1
$begingroup$
You mean a root of $f^(d_i)$?
$endgroup$
– Jean-Claude Arbaut
Apr 10 '15 at 12:35
$begingroup$
@Jean-ClaudeArbaut: Yes, I do. I've corrected the question. Thanks.
$endgroup$
– Evan Aad
Apr 10 '15 at 12:36
add a comment |
1
$begingroup$
You mean a root of $f^(d_i)$?
$endgroup$
– Jean-Claude Arbaut
Apr 10 '15 at 12:35
$begingroup$
@Jean-ClaudeArbaut: Yes, I do. I've corrected the question. Thanks.
$endgroup$
– Evan Aad
Apr 10 '15 at 12:36
1
1
$begingroup$
You mean a root of $f^(d_i)$?
$endgroup$
– Jean-Claude Arbaut
Apr 10 '15 at 12:35
$begingroup$
You mean a root of $f^(d_i)$?
$endgroup$
– Jean-Claude Arbaut
Apr 10 '15 at 12:35
$begingroup$
@Jean-ClaudeArbaut: Yes, I do. I've corrected the question. Thanks.
$endgroup$
– Evan Aad
Apr 10 '15 at 12:36
$begingroup$
@Jean-ClaudeArbaut: Yes, I do. I've corrected the question. Thanks.
$endgroup$
– Evan Aad
Apr 10 '15 at 12:36
add a comment |
1 Answer
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$begingroup$
If I understand the question correctly:
You can use the Taylor formula for the polynomial $f$ of degree $n$, at $x=c$:
$$f(x)=f(c)+(x-c)f'(c)+dots+frac(x-c)^nf^(n)(c)n!$$
Thus, if $c$ is a root of $f^(k)$ for $k in 0, dots d$, then
$$f(x)=frac(x-c)^d+1f^d+1(c)(d+1)!+dots+frac(x-c)^nf^(n)(c)n!$$
$$f(x)=(x-c)^d+1left(fracf^d+1(c)(d+1)!+dots+frac(x-c)^n-d-1f^(n)(c)n!right)$$
And there is a factor $(x-c)^d+1$ in $f$, hence the multiplicity of the root $c$ is at least $d+1$.
edited Mar 19 at 10:21
Cardioid_Ass_22
47815
answered Apr 10 '15 at 12:41
Jean-Claude ArbautJean-Claude Arbaut
14.9k63464
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If I understand the question correctly:
You can use the Taylor formula for the polynomial $f$ of degree $n$, at $x=c$:
$$f(x)=f(c)+(x-c)f'(c)+dots+frac(x-c)^nf^(n)(c)n!$$
Thus, if $c$ is a root of $f^(k)$ for $k in 0, dots d$, then
$$f(x)=frac(x-c)^d+1f^d+1(c)(d+1)!+dots+frac(x-c)^nf^(n)(c)n!$$
$$f(x)=(x-c)^d+1left(fracf^d+1(c)(d+1)!+dots+frac(x-c)^n-d-1f^(n)(c)n!right)$$
And there is a factor $(x-c)^d+1$ in $f$, hence the multiplicity of the root $c$ is at least $d+1$.
edited Mar 19 at 10:21
Cardioid_Ass_22
47815
answered Apr 10 '15 at 12:41
Jean-Claude ArbautJean-Claude Arbaut
14.9k63464
$endgroup$
add a comment |
$begingroup$
If I understand the question correctly:
You can use the Taylor formula for the polynomial $f$ of degree $n$, at $x=c$:
$$f(x)=f(c)+(x-c)f'(c)+dots+frac(x-c)^nf^(n)(c)n!$$
Thus, if $c$ is a root of $f^(k)$ for $k in 0, dots d$, then
$$f(x)=frac(x-c)^d+1f^d+1(c)(d+1)!+dots+frac(x-c)^nf^(n)(c)n!$$
$$f(x)=(x-c)^d+1left(fracf^d+1(c)(d+1)!+dots+frac(x-c)^n-d-1f^(n)(c)n!right)$$
And there is a factor $(x-c)^d+1$ in $f$, hence the multiplicity of the root $c$ is at least $d+1$.
edited Mar 19 at 10:21
Cardioid_Ass_22
47815
answered Apr 10 '15 at 12:41
Jean-Claude ArbautJean-Claude Arbaut
14.9k63464
$endgroup$
add a comment |
$begingroup$
If I understand the question correctly:
You can use the Taylor formula for the polynomial $f$ of degree $n$, at $x=c$:
$$f(x)=f(c)+(x-c)f'(c)+dots+frac(x-c)^nf^(n)(c)n!$$
Thus, if $c$ is a root of $f^(k)$ for $k in 0, dots d$, then
$$f(x)=frac(x-c)^d+1f^d+1(c)(d+1)!+dots+frac(x-c)^nf^(n)(c)n!$$
$$f(x)=(x-c)^d+1left(fracf^d+1(c)(d+1)!+dots+frac(x-c)^n-d-1f^(n)(c)n!right)$$
And there is a factor $(x-c)^d+1$ in $f$, hence the multiplicity of the root $c$ is at least $d+1$.
edited Mar 19 at 10:21
Cardioid_Ass_22
47815
answered Apr 10 '15 at 12:41
Jean-Claude ArbautJean-Claude Arbaut
14.9k63464
$endgroup$
If I understand the question correctly:
You can use the Taylor formula for the polynomial $f$ of degree $n$, at $x=c$:
$$f(x)=f(c)+(x-c)f'(c)+dots+frac(x-c)^nf^(n)(c)n!$$
Thus, if $c$ is a root of $f^(k)$ for $k in 0, dots d$, then
$$f(x)=frac(x-c)^d+1f^d+1(c)(d+1)!+dots+frac(x-c)^nf^(n)(c)n!$$
$$f(x)=(x-c)^d+1left(fracf^d+1(c)(d+1)!+dots+frac(x-c)^n-d-1f^(n)(c)n!right)$$
And there is a factor $(x-c)^d+1$ in $f$, hence the multiplicity of the root $c$ is at least $d+1$.
edited Mar 19 at 10:21
Cardioid_Ass_22
47815
answered Apr 10 '15 at 12:41
Jean-Claude ArbautJean-Claude Arbaut
14.9k63464
edited Mar 19 at 10:21
Cardioid_Ass_22
47815
edited Mar 19 at 10:21
Cardioid_Ass_22
47815
edited Mar 19 at 10:21
Cardioid_Ass_22
47815
47815
answered Apr 10 '15 at 12:41
Jean-Claude ArbautJean-Claude Arbaut
14.9k63464
answered Apr 10 '15 at 12:41
Jean-Claude ArbautJean-Claude Arbaut
14.9k63464
answered Apr 10 '15 at 12:41
Jean-Claude ArbautJean-Claude Arbaut
14.9k63464
14.9k63464
add a comment |
add a comment |
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$begingroup$
You mean a root of $f^(d_i)$?
$endgroup$
– Jean-Claude Arbaut
Apr 10 '15 at 12:35
$begingroup$
@Jean-ClaudeArbaut: Yes, I do. I've corrected the question. Thanks.
$endgroup$
– Evan Aad
Apr 10 '15 at 12:36