Two polynomials with complex coefficients whose level sets at $0$ and $1$ are the same












2














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







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  • 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.
    – Jair Taylor
    Dec 9 at 1:46






  • 1




    @JairTaylor but different multiplicities will tend to put the solutions of $f(x)=1$ at different points.
    – Empy2
    Dec 9 at 2:05










  • @Empy2 Ah, you are right. I see now based on the answer below.
    – Jair Taylor
    Dec 9 at 17:48
















2














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







share|cite|improve this question
























  • 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.
    – Jair Taylor
    Dec 9 at 1:46






  • 1




    @JairTaylor but different multiplicities will tend to put the solutions of $f(x)=1$ at different points.
    – Empy2
    Dec 9 at 2:05










  • @Empy2 Ah, you are right. I see now based on the answer below.
    – Jair Taylor
    Dec 9 at 17:48














2












2








2


1





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







share|cite|improve this question















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






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share|cite|improve this question













share|cite|improve this question




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edited Dec 9 at 4:06

























asked Dec 9 at 1:41









user521337

7951314




7951314












  • 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.
    – Jair Taylor
    Dec 9 at 1:46






  • 1




    @JairTaylor but different multiplicities will tend to put the solutions of $f(x)=1$ at different points.
    – Empy2
    Dec 9 at 2:05










  • @Empy2 Ah, you are right. I see now based on the answer below.
    – Jair Taylor
    Dec 9 at 17:48


















  • 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.
    – Jair Taylor
    Dec 9 at 1:46






  • 1




    @JairTaylor but different multiplicities will tend to put the solutions of $f(x)=1$ at different points.
    – Empy2
    Dec 9 at 2:05










  • @Empy2 Ah, you are right. I see now based on the answer below.
    – Jair Taylor
    Dec 9 at 17:48
















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.
– Jair Taylor
Dec 9 at 1:46




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.
– Jair Taylor
Dec 9 at 1:46




1




1




@JairTaylor but different multiplicities will tend to put the solutions of $f(x)=1$ at different points.
– Empy2
Dec 9 at 2:05




@JairTaylor but different multiplicities will tend to put the solutions of $f(x)=1$ at different points.
– Empy2
Dec 9 at 2:05












@Empy2 Ah, you are right. I see now based on the answer below.
– Jair Taylor
Dec 9 at 17:48




@Empy2 Ah, you are right. I see now based on the answer below.
– Jair Taylor
Dec 9 at 17:48










1 Answer
1






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4














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 $f=g$.






share|cite|improve this answer





















  • Beat me to it by 8 s! +1
    – Macavity
    Dec 9 at 3:58










  • 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!
    – Jair Taylor
    Dec 9 at 18:08











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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 $f=g$.






share|cite|improve this answer





















  • Beat me to it by 8 s! +1
    – Macavity
    Dec 9 at 3:58










  • 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!
    – Jair Taylor
    Dec 9 at 18:08
















4














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 $f=g$.






share|cite|improve this answer





















  • Beat me to it by 8 s! +1
    – Macavity
    Dec 9 at 3:58










  • 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!
    – Jair Taylor
    Dec 9 at 18:08














4












4








4






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 $f=g$.






share|cite|improve this answer












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 $f=g$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 at 3:57









Eric Wofsey

179k12204331




179k12204331












  • Beat me to it by 8 s! +1
    – Macavity
    Dec 9 at 3:58










  • 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!
    – Jair Taylor
    Dec 9 at 18:08


















  • Beat me to it by 8 s! +1
    – Macavity
    Dec 9 at 3:58










  • 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!
    – Jair Taylor
    Dec 9 at 18:08
















Beat me to it by 8 s! +1
– Macavity
Dec 9 at 3:58




Beat me to it by 8 s! +1
– Macavity
Dec 9 at 3:58












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!
– Jair Taylor
Dec 9 at 18:08




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!
– Jair Taylor
Dec 9 at 18:08


















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