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












2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














2












2








2


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 4:06







user521337

















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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















4












$begingroup$

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 $f=g$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3031906%2ftwo-polynomials-with-complex-coefficients-whose-level-sets-at-0-and-1-are-th%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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 $f=g$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















4












$begingroup$

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 $f=g$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














4












4








4





$begingroup$

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 $f=g$.






share|cite|improve this answer









$endgroup$



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 $f=g$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3031906%2ftwo-polynomials-with-complex-coefficients-whose-level-sets-at-0-and-1-are-th%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
Read more