Two polynomials with complex coefficients whose level sets at $0$ and $1$ are the same
$begingroup$
For a polynomial $f(X)in mathbb C[X]$, and $ain mathbb C$, let $f^{-1} (a):={mu in mathbb C : f(mu)=a}$.
Now let $f(X), g(X) in mathbb C[X]$ be non-constant polynomials such that $f^{-1}(0)=g^{-1}(0)$ and $f^{-1}(1)=g^{-1}(1)$, then is it true that $f=g$ ?
Let $f^{-1}(0)={mu_1,...,mu_k}; f^{-1}(1)={gamma_1,...,gamma_l}$.
Then $f(X)=c_1prod_{i=1}^k(X-mu_i)^{n_i}, f(X)-1=c_1'prod_{i=1}^l(X-gamma_i)^{n_i'}$
$g(X)=c_2prod_{i=1}^k(X-mu_i)^{m_i}; g(X)-1=c_2'prod_{i=1}^l(X-gamma_i)^{m_i'}$
But I don't know how to proceed further.
Please help
polynomials
asked Dec 9 '18 at 1:41
user521337user521337
1,1981417
$endgroup$
add a comment |
$begingroup$
For a polynomial $f(X)in mathbb C[X]$, and $ain mathbb C$, let $f^{-1} (a):={mu in mathbb C : f(mu)=a}$.
Now let $f(X), g(X) in mathbb C[X]$ be non-constant polynomials such that $f^{-1}(0)=g^{-1}(0)$ and $f^{-1}(1)=g^{-1}(1)$, then is it true that $f=g$ ?
Let $f^{-1}(0)={mu_1,...,mu_k}; f^{-1}(1)={gamma_1,...,gamma_l}$.
Then $f(X)=c_1prod_{i=1}^k(X-mu_i)^{n_i}, f(X)-1=c_1'prod_{i=1}^l(X-gamma_i)^{n_i'}$
$g(X)=c_2prod_{i=1}^k(X-mu_i)^{m_i}; g(X)-1=c_2'prod_{i=1}^l(X-gamma_i)^{m_i'}$
But I don't know how to proceed further.
Please help
polynomials
asked Dec 9 '18 at 1:41
user521337user521337
1,1981417
$endgroup$
$begingroup$
A polynomial is determined by its zeroes, counted by multiplicity, along with one more constant factor. Knowing $f^{-1}(0)$ determines the zeros, but not their multiplicity.
$endgroup$
– Jair Taylor
Dec 9 '18 at 1:46
1
$begingroup$
@JairTaylor but different multiplicities will tend to put the solutions of $f(x)=1$ at different points.
$endgroup$
– Empy2
Dec 9 '18 at 2:05
$begingroup$
@Empy2 Ah, you are right. I see now based on the answer below.
$endgroup$
– Jair Taylor
Dec 9 '18 at 17:48
add a comment |
$begingroup$
For a polynomial $f(X)in mathbb C[X]$, and $ain mathbb C$, let $f^{-1} (a):={mu in mathbb C : f(mu)=a}$.
Now let $f(X), g(X) in mathbb C[X]$ be non-constant polynomials such that $f^{-1}(0)=g^{-1}(0)$ and $f^{-1}(1)=g^{-1}(1)$, then is it true that $f=g$ ?
Let $f^{-1}(0)={mu_1,...,mu_k}; f^{-1}(1)={gamma_1,...,gamma_l}$.
Then $f(X)=c_1prod_{i=1}^k(X-mu_i)^{n_i}, f(X)-1=c_1'prod_{i=1}^l(X-gamma_i)^{n_i'}$
$g(X)=c_2prod_{i=1}^k(X-mu_i)^{m_i}; g(X)-1=c_2'prod_{i=1}^l(X-gamma_i)^{m_i'}$
But I don't know how to proceed further.
Please help
polynomials
asked Dec 9 '18 at 1:41
user521337user521337
1,1981417
$endgroup$
For a polynomial $f(X)in mathbb C[X]$, and $ain mathbb C$, let $f^{-1} (a):={mu in mathbb C : f(mu)=a}$.
Now let $f(X), g(X) in mathbb C[X]$ be non-constant polynomials such that $f^{-1}(0)=g^{-1}(0)$ and $f^{-1}(1)=g^{-1}(1)$, then is it true that $f=g$ ?
Let $f^{-1}(0)={mu_1,...,mu_k}; f^{-1}(1)={gamma_1,...,gamma_l}$.
Then $f(X)=c_1prod_{i=1}^k(X-mu_i)^{n_i}, f(X)-1=c_1'prod_{i=1}^l(X-gamma_i)^{n_i'}$
$g(X)=c_2prod_{i=1}^k(X-mu_i)^{m_i}; g(X)-1=c_2'prod_{i=1}^l(X-gamma_i)^{m_i'}$
But I don't know how to proceed further.
Please help
polynomials
polynomials
asked Dec 9 '18 at 1:41
user521337user521337
1,1981417
asked Dec 9 '18 at 1:41
user521337user521337
1,1981417
edited Dec 9 '18 at 4:06
user521337
asked Dec 9 '18 at 1:41
user521337user521337
1,1981417
asked Dec 9 '18 at 1:41
user521337user521337
1,1981417
asked Dec 9 '18 at 1:41
user521337user521337
1,1981417
1,1981417
$begingroup$
A polynomial is determined by its zeroes, counted by multiplicity, along with one more constant factor. Knowing $f^{-1}(0)$ determines the zeros, but not their multiplicity.
$endgroup$
– Jair Taylor
Dec 9 '18 at 1:46
1
$begingroup$
@JairTaylor but different multiplicities will tend to put the solutions of $f(x)=1$ at different points.
$endgroup$
– Empy2
Dec 9 '18 at 2:05
$begingroup$
@Empy2 Ah, you are right. I see now based on the answer below.
$endgroup$
– Jair Taylor
Dec 9 '18 at 17:48
add a comment |
$begingroup$
A polynomial is determined by its zeroes, counted by multiplicity, along with one more constant factor. Knowing $f^{-1}(0)$ determines the zeros, but not their multiplicity.
$endgroup$
– Jair Taylor
Dec 9 '18 at 1:46
1
$begingroup$
@JairTaylor but different multiplicities will tend to put the solutions of $f(x)=1$ at different points.
$endgroup$
– Empy2
Dec 9 '18 at 2:05
$begingroup$
@Empy2 Ah, you are right. I see now based on the answer below.
$endgroup$
– Jair Taylor
Dec 9 '18 at 17:48
$begingroup$
A polynomial is determined by its zeroes, counted by multiplicity, along with one more constant factor. Knowing $f^{-1}(0)$ determines the zeros, but not their multiplicity.
$endgroup$
– Jair Taylor
Dec 9 '18 at 1:46
$begingroup$
A polynomial is determined by its zeroes, counted by multiplicity, along with one more constant factor. Knowing $f^{-1}(0)$ determines the zeros, but not their multiplicity.
$endgroup$
– Jair Taylor
Dec 9 '18 at 1:46
1
1
$begingroup$
@JairTaylor but different multiplicities will tend to put the solutions of $f(x)=1$ at different points.
$endgroup$
– Empy2
Dec 9 '18 at 2:05
$begingroup$
@JairTaylor but different multiplicities will tend to put the solutions of $f(x)=1$ at different points.
$endgroup$
– Empy2
Dec 9 '18 at 2:05
$begingroup$
@Empy2 Ah, you are right. I see now based on the answer below.
$endgroup$
– Jair Taylor
Dec 9 '18 at 17:48
$begingroup$
@Empy2 Ah, you are right. I see now based on the answer below.
$endgroup$
– Jair Taylor
Dec 9 '18 at 17:48
add a comment |
1 Answer
1
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votes
$begingroup$
We may assume without loss of generality that $deg fgeq deg g$. Let $d=deg f$, let $A=f^{-1}(0)=g^{-1}(0)$, and let $B=f^{-1}(1)=g^{-1}(1)$. Note that $|A|=d-m$ where $m$ is the number of roots that $f'$ has in $A$ (counting multiplicity). Similarly, $|B|=d-n$ where $n$ is the number of roots that $f'$ has in $B$. Thus $$|A|+|B|=2d-(m+n)geq 2d-(d-1)=d+1$$ (since $f'$ has $deg f'=d-1$ roots in total).
But now notice that every point of $Acup B$ is a root of $f-g$. Since $f-g$ has degree at most $d$ and $|Acup B|=|A|+|B|geq d+1$, this implies that $f-g=0$ and so $f=g$.
answered Dec 9 '18 at 3:57
Eric WofseyEric Wofsey
190k14216347
$endgroup$
$begingroup$
Beat me to it by 8 s! +1
$endgroup$
– Macavity
Dec 9 '18 at 3:58
$begingroup$
Generalizing slightly, you've shown for any $c,d in mathbb{C}$ and $p in mathbb{C}[x]$ we have $|p^{-1}(c)| + |p^{-1}(d)| geq 1 + deg p$. Very nice!
$endgroup$
– Jair Taylor
Dec 9 '18 at 18:08
add a comment |
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$begingroup$
We may assume without loss of generality that $deg fgeq deg g$. Let $d=deg f$, let $A=f^{-1}(0)=g^{-1}(0)$, and let $B=f^{-1}(1)=g^{-1}(1)$. Note that $|A|=d-m$ where $m$ is the number of roots that $f'$ has in $A$ (counting multiplicity). Similarly, $|B|=d-n$ where $n$ is the number of roots that $f'$ has in $B$. Thus $$|A|+|B|=2d-(m+n)geq 2d-(d-1)=d+1$$ (since $f'$ has $deg f'=d-1$ roots in total).
But now notice that every point of $Acup B$ is a root of $f-g$. Since $f-g$ has degree at most $d$ and $|Acup B|=|A|+|B|geq d+1$, this implies that $f-g=0$ and so $f=g$.
answered Dec 9 '18 at 3:57
Eric WofseyEric Wofsey
190k14216347
$endgroup$
$begingroup$
Beat me to it by 8 s! +1
$endgroup$
– Macavity
Dec 9 '18 at 3:58
$begingroup$
Generalizing slightly, you've shown for any $c,d in mathbb{C}$ and $p in mathbb{C}[x]$ we have $|p^{-1}(c)| + |p^{-1}(d)| geq 1 + deg p$. Very nice!
$endgroup$
– Jair Taylor
Dec 9 '18 at 18:08
add a comment |
$begingroup$
We may assume without loss of generality that $deg fgeq deg g$. Let $d=deg f$, let $A=f^{-1}(0)=g^{-1}(0)$, and let $B=f^{-1}(1)=g^{-1}(1)$. Note that $|A|=d-m$ where $m$ is the number of roots that $f'$ has in $A$ (counting multiplicity). Similarly, $|B|=d-n$ where $n$ is the number of roots that $f'$ has in $B$. Thus $$|A|+|B|=2d-(m+n)geq 2d-(d-1)=d+1$$ (since $f'$ has $deg f'=d-1$ roots in total).
But now notice that every point of $Acup B$ is a root of $f-g$. Since $f-g$ has degree at most $d$ and $|Acup B|=|A|+|B|geq d+1$, this implies that $f-g=0$ and so $f=g$.
answered Dec 9 '18 at 3:57
Eric WofseyEric Wofsey
190k14216347
$endgroup$
$begingroup$
Beat me to it by 8 s! +1
$endgroup$
– Macavity
Dec 9 '18 at 3:58
$begingroup$
Generalizing slightly, you've shown for any $c,d in mathbb{C}$ and $p in mathbb{C}[x]$ we have $|p^{-1}(c)| + |p^{-1}(d)| geq 1 + deg p$. Very nice!
$endgroup$
– Jair Taylor
Dec 9 '18 at 18:08
add a comment |
$begingroup$
We may assume without loss of generality that $deg fgeq deg g$. Let $d=deg f$, let $A=f^{-1}(0)=g^{-1}(0)$, and let $B=f^{-1}(1)=g^{-1}(1)$. Note that $|A|=d-m$ where $m$ is the number of roots that $f'$ has in $A$ (counting multiplicity). Similarly, $|B|=d-n$ where $n$ is the number of roots that $f'$ has in $B$. Thus $$|A|+|B|=2d-(m+n)geq 2d-(d-1)=d+1$$ (since $f'$ has $deg f'=d-1$ roots in total).
But now notice that every point of $Acup B$ is a root of $f-g$. Since $f-g$ has degree at most $d$ and $|Acup B|=|A|+|B|geq d+1$, this implies that $f-g=0$ and so $f=g$.
answered Dec 9 '18 at 3:57
Eric WofseyEric Wofsey
190k14216347
$endgroup$
We may assume without loss of generality that $deg fgeq deg g$. Let $d=deg f$, let $A=f^{-1}(0)=g^{-1}(0)$, and let $B=f^{-1}(1)=g^{-1}(1)$. Note that $|A|=d-m$ where $m$ is the number of roots that $f'$ has in $A$ (counting multiplicity). Similarly, $|B|=d-n$ where $n$ is the number of roots that $f'$ has in $B$. Thus $$|A|+|B|=2d-(m+n)geq 2d-(d-1)=d+1$$ (since $f'$ has $deg f'=d-1$ roots in total).
But now notice that every point of $Acup B$ is a root of $f-g$. Since $f-g$ has degree at most $d$ and $|Acup B|=|A|+|B|geq d+1$, this implies that $f-g=0$ and so $f=g$.
answered Dec 9 '18 at 3:57
Eric WofseyEric Wofsey
190k14216347
answered Dec 9 '18 at 3:57
Eric WofseyEric Wofsey
190k14216347
answered Dec 9 '18 at 3:57
Eric WofseyEric Wofsey
190k14216347
answered Dec 9 '18 at 3:57
Eric WofseyEric Wofsey
190k14216347
190k14216347
$begingroup$
Beat me to it by 8 s! +1
$endgroup$
– Macavity
Dec 9 '18 at 3:58
$begingroup$
Generalizing slightly, you've shown for any $c,d in mathbb{C}$ and $p in mathbb{C}[x]$ we have $|p^{-1}(c)| + |p^{-1}(d)| geq 1 + deg p$. Very nice!
$endgroup$
– Jair Taylor
Dec 9 '18 at 18:08
add a comment |
$begingroup$
Beat me to it by 8 s! +1
$endgroup$
– Macavity
Dec 9 '18 at 3:58
$begingroup$
Generalizing slightly, you've shown for any $c,d in mathbb{C}$ and $p in mathbb{C}[x]$ we have $|p^{-1}(c)| + |p^{-1}(d)| geq 1 + deg p$. Very nice!
$endgroup$
– Jair Taylor
Dec 9 '18 at 18:08
$begingroup$
Beat me to it by 8 s! +1
$endgroup$
– Macavity
Dec 9 '18 at 3:58
$begingroup$
Beat me to it by 8 s! +1
$endgroup$
– Macavity
Dec 9 '18 at 3:58
$begingroup$
Generalizing slightly, you've shown for any $c,d in mathbb{C}$ and $p in mathbb{C}[x]$ we have $|p^{-1}(c)| + |p^{-1}(d)| geq 1 + deg p$. Very nice!
$endgroup$
– Jair Taylor
Dec 9 '18 at 18:08
$begingroup$
Generalizing slightly, you've shown for any $c,d in mathbb{C}$ and $p in mathbb{C}[x]$ we have $|p^{-1}(c)| + |p^{-1}(d)| geq 1 + deg p$. Very nice!
$endgroup$
– Jair Taylor
Dec 9 '18 at 18:08
add a comment |
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$begingroup$
A polynomial is determined by its zeroes, counted by multiplicity, along with one more constant factor. Knowing $f^{-1}(0)$ determines the zeros, but not their multiplicity.
$endgroup$
– Jair Taylor
Dec 9 '18 at 1:46
1
$begingroup$
@JairTaylor but different multiplicities will tend to put the solutions of $f(x)=1$ at different points.
$endgroup$
– Empy2
Dec 9 '18 at 2:05
$begingroup$
@Empy2 Ah, you are right. I see now based on the answer below.
$endgroup$
– Jair Taylor
Dec 9 '18 at 17:48