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I want to prove that following polynomial $f_p$ has distinct zeros

$$ f_p(x) = \begin{cases} x^{2p+2} - abx^{2p} - (a+b)x^p-1, &\text{if $p$ is even}\\ x^{2p+2} - abx^{2p} - 2x^{p+1} +1, & \text{if $p$ is odd} \end{cases} $$

where $a$, $b$ are positive real numbers and $p>0$ is an integer. How can I prove this polynomial has distinct zeros for any arbitrary $a$,$b$ and $p$.

Thanks in advance.

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    $\begingroup$ I don't have a clue. But $p$ has a zero of multiplicity (at least) two at $x$ iff $p(x) = p'(x) = 0$. However, I cannot see how this should help. $\endgroup$ – Friedrich Philipp Mar 15 '16 at 19:30
  • $\begingroup$ @FriedrichPhilipp, $\mathbb R[x]$ is a Euclidean ring, with the degree of a polynomial as the size function. A polynomial $f_p(x)$ has a repeated root if and only if $\gcd(f_p(x), f_p'(x)$ has degree at least one; that is, not a constant. $\endgroup$ – Will Jagy Mar 15 '16 at 20:06
  • $\begingroup$ How does this differ from what I wrote? $\endgroup$ – Friedrich Philipp Mar 15 '16 at 21:10
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Standard way of doing it is creating an auxiliary polynomial in your case of $2p$ degree. Multiply this one with a hypothetical factor $(x-c)^2$. Finally compare coefficients and prove (or disprove) that such system does not have a solution. Try with smaller $p$ first until you realize what is going on then generalize.

$$(\sum\limits_{k=0}^{2p}a_{k}x^{k})(x-c)^2 = \sum\limits_{k=2}^{2p+2}a_{k-2}x^k - \sum\limits_{k=1}^{2p+1}2a_{k-1}cx^k + \sum\limits_{k=0}^{2p}a_{k}c^2x^k=$$ $$a_{2p}x^{2p+2}+(a_{2p-1}-2a_{2p}c)x^{2p+1}+\sum\limits_{k=2}^{2p}(a_{k-2}-2a_{k-1}c+a_{k}c^2)x^k+(-2a_{0}c+a_{1}c^2)x +a_{0}c^2$$

Now you have $a_{0}c^2=1$ or $a_{0}c^2=-1$. Obviously this resolves $a_{1}$ as well. On the other side you have $a_{2p}=1$ and $a_{2p-1}-2a_{2p}c=0$ which resolves $a_{2p-1}$

As you will realize yourself soon enough $a_{k-2}-2a_{k-1}c+a_{k}c^2=0$ up to $p$ (and there is another one after it up to $2p-1$) is actually a recursion $a_{k}=-\frac{a_{k-2}}{c^2}+\frac{2a_{k-1}}{c}$ which you can solve for the given initial condition. Since you have $a_{p}$ and $a_{2p}$ you can find the condition for these two and compare if they can satisfy recursion up to $p$ and the following one from $p$ to $2p-1$ together with the initial conditions for $a_{1}$ and finishing condition for $a_{2p-1}$.

That is the plan that should either give you a possible solution or give you the reason for not having a solution.

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  • $\begingroup$ Thank you for answer but Isn't it $(\sum\limits_{k=0}^{2p}a_{k}x^{k})(x-c)^2 = \sum\limits_{k=2}^{2p+2}a_{k-2}x^k - \sum\limits_{k=1}^{2p+1}2a_{k-1}cx^k + c^2\sum\limits_{k=0}^{2p}a_{k}x^k$ ? $\endgroup$ – drxy Mar 15 '16 at 21:33
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    $\begingroup$ Fixed, but the solution is giving you just a general picture, you should check everything at your desk, here, it is not that important. $\endgroup$ – user318107 Mar 15 '16 at 21:46
  • $\begingroup$ Thank you for the solution, I will try. $\endgroup$ – drxy Mar 15 '16 at 21:50
  • $\begingroup$ There are many other ways you can try in order to check if there is a solution given by the system of coefficients. Gauss, matrix, trying some small values for p and then using induction and so on. Some of them might be simpler. The core is to work around the system of coefficients as you would for any system of equations, and this one is very symmetrical. Find its crucial weakness. Try small p's and you will be quickly confident enough to attack the general statement. $\endgroup$ – user318107 Mar 15 '16 at 21:55
  • $\begingroup$ I have tried many times but I got some unsatisfactory results for small $p$ values. For example, I have tried to find the zeros of the polynomial for $p=1,2,3$. I saw that all zeros are distinct up to $p=3$. But I couldn't guarantee this for the higher values. I will continue to try figure out how to achieve this. Thank you for your helpful suggestions. $\endgroup$ – drxy Mar 15 '16 at 22:04

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