5
$\begingroup$

I want to prove that the polynomial

$$ f_p(x) = x^{2p+2} - abx^{2p} - 2x^{p+1} +1 $$

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

Thanks in advance.

$\endgroup$
  • $\begingroup$ Perhaps proving that $\gcd(f,f')=1$ is feasible. $\endgroup$ – lhf Apr 13 '16 at 12:57
  • $\begingroup$ We tried to prove it with your way but didn't achieve. Could you please prove $gcd(f,f')=1$? $\endgroup$ – drxy Apr 13 '16 at 12:58
  • 6
    $\begingroup$ Why do you have a coefficient $ab$? Why not just call it $c$? Is it a typo? $\endgroup$ – M. Vinay Apr 13 '16 at 13:14
9
$\begingroup$

Let $c$ denote $ab$. Note that \begin{equation*} f(x) = (x^{p+1}-1)^2 - cx^{2p} \end{equation*} and \begin{equation*} f'(x) = 2(p+1)x^p(x^{p+1}-1) - 2pcx^{2p-1} \end{equation*}

Then, $f(x) = 0 \iff$ \begin{equation*} c = \dfrac{(x^{p+1}-1)^2}{x^{2p}} = \varphi(x)\ (\text{say}) \end{equation*} and $f'(x) = 0 \iff$ \begin{align*} c & = \dfrac{(p+1)x^p(x^{p+1}-1)}{px^{2p-1}} = \dfrac{(p+1)x}{p}\dfrac{x^{p+1}-1}{x^p} \iff\\ c & = \left(\dfrac{p+1}{p}\right)x \sqrt{\varphi(x)}. \end{align*}

Thus, $f(x)$ and $f'(x)$ vanish for the same $x$ if and only if for some root $x$ of $f(x)$, \begin{align*} c & = \left(\dfrac{p+1}{p}\right)x \sqrt c \iff\\ x & = \dfrac{p\sqrt c}{p+1}. \end{align*}

Thus, for every $p$ and $c = ab$, if such an $x$ is a root, it is a multiple root.

Now, when does such a root $x$ exist? Let $x = t$ be one such. Then $c = \left(\dfrac{p+1}{p} \right)^2 t^2$. Then, since $f(t) = 0$, we have (from the original form of the equation): \begin{align*} t^{2p+2} - \left(\dfrac{p+1}{p}\right)^2 t^{2p + 2} - 2t^{p+1} + 1 = 0\\ -\dfrac{(2p + 1)}{p^2}t^{2p + 2} - 2t^{p+1} + 1 = 0\\ (2p + 1)t^{2(p + 1)} + 2p^2 t^{p + 1} - p^2 = 0. \end{align*}

When treated as a quadratic equation in $t^{p+1}$, the discriminant is \begin{equation*} 4p^4 + 4p^2(2p + 1) = 4p^2(p + 1)^2, \end{equation*} and therefore, the solutions are \begin{equation*} t^{p+1} = -p, \dfrac{p}{2p + 1}. \end{equation*} That is, \begin{equation*} t = (-p)^{\frac 1 {p + 1}}, \left(\dfrac p {2p + 1} \right)^{\frac 1 {p + 1}}. \end{equation*}

But substituting the same $c$ in $f'(t) = 0$, we get \begin{align*} & 2(p+1)t^p(t^{p+1}-1)-2p\left(\dfrac{p+1}{p}\right)^2t^{2p+1}=0\\ & p(t^{p+1}-1)-(p+1)t^{p+1}=0 \implies\\ & t = (-p)^{\frac{1}{p+1}} \end{align*}

Thus, only the first of the previous two solutions satisfies both equations.

Then, $c = \left( \dfrac{p+1}{p} \right)^2 t^2$ gives us \begin{equation*} \boxed{c= \dfrac{(p+1)^2}{(-p)^{\frac{2p}{p+1}}}}. \end{equation*}

Thus, the equation has multiple roots exactly when $c$ and $p$ are related as above.

Note that for odd values of $p$, $c$ will be a real number if and only if $p$ is of the form $4k + 1$, and then, $c < 0$. If, as stated in the question, $c = ab$ is a positive real number, the equation will have distinct roots.


Example

For $p = 1$, $f(x) = x^4 - cx^2 - 2x^2 + 1$ and $f'(x) = 4x^3 - 2cx - 4x$.

Then, $f(x) = 0$ and $f'(x) = 0$ imply that $c = \left(\dfrac{x^2 - 1}{x}\right)^2$ and $c = 2x\left(\dfrac{x^2-1}{x}\right)$ respectively. Thus, if $x$ is a multiple root, then $x = \dfrac{\sqrt c}{2}$.

Taking $t$ to be such a root, so that $c = 4t^2$, and substituting in $f(t) = 0$, we get \begin{align*} t^4 - 4t^4 - 2t^2 + 1 = 0\\ 3t^4 + 2t^2 - 1 = 0. \end{align*} Thus, $t^2 = -1, \dfrac 1 3$, of which only the first one satisfies $f'(t)=0$. Thus, $c = -4$.

For $c = 4$, $x^4 + 2x^2 + 1 = 0$ has roots $\pm i, \pm i$.

$\endgroup$
  • 1
    $\begingroup$ I think an example would be useful. Perhaps p=1 or 3. $\endgroup$ – marty cohen Apr 13 '16 at 19:14
  • $\begingroup$ @martycohen Added. Thanks, working out the example also brought out an error in the derivation, which I've fixed now. $\endgroup$ – M. Vinay Apr 14 '16 at 13:46
  • $\begingroup$ Since $a$ and $b$ are positive real numbers, $c=ab$ must be positive. But in your answer $c=-4$. So does it mean there is no multiple root? $\endgroup$ – drxy Apr 14 '16 at 20:43
  • $\begingroup$ @drxy Yes, for real $c > 0$, the equation has distinct roots (I suppose that's what you wish to prove). I've updated my answer with this information. $\endgroup$ – M. Vinay Apr 15 '16 at 2:32
  • $\begingroup$ @M.Vinay thank you for invaluable response. If my polynomial turns into $ f_p(x) = x^{2p+2} - abx^{2p} - (a+b)x^{p} -1 $, where p is even, then have you got any idea that $f_p(x)$ has distinct roots? $\endgroup$ – drxy Apr 16 '16 at 9:37
8
$\begingroup$

The claim is false. Set $ab:=(\frac{729}{16})^\frac{1}{3}$ and $p:=2$

Then $f(x)=x^6-abx^4-2x^3+1$ has a double root at $x=-2^{\frac{1}{3}}$

$\endgroup$
  • $\begingroup$ Sorry, I forgot that $p$ is odd. $\endgroup$ – drxy Apr 13 '16 at 18:15
  • $\begingroup$ @drxy Even for odd $p$, it's possible to get multiple roots. $\endgroup$ – M. Vinay Apr 14 '16 at 13:48
  • $\begingroup$ @M.Vinay If it is possible, can you give me some example for odd $p$. $\endgroup$ – drxy Apr 14 '16 at 14:04
  • $\begingroup$ @drxy I have done so in my answer (both the general case describing exactly when [for what value of $c$] the polynomial will have multiple roots, as well as an example, with $p = 1$). $\endgroup$ – M. Vinay Apr 14 '16 at 14:05
  • $\begingroup$ Oh, I have just saw it. So many thanks @M.Vinay $\endgroup$ – drxy Apr 14 '16 at 14:06
4
$\begingroup$

With the help of Mathematica the discriminant is given by:

$$\Delta = \left \{ \begin{array}{cc} (4c)^{p+1} \left[ p^{p} c^{(p+1)/2}+(p+1)^{p+1} \right]^{2} & \text{odd } p \\[5pt] (4c)^{p+1} [(p+1)^{2(p+1)}-p^{2p}c^{p+1}] & \text{even } p \end{array} \right. \\$$

$\Delta \neq 0 \,$ if $\, \left \{ \begin{array}{ll} \text{odd } p=4n-1 , & c\neq 0 \\[5pt] \text{odd } p=4n-3, & c^{(p+1)/2} \neq 0, \displaystyle -\frac{(p+1)^{p+1}}{p^{p}} \\[5pt] \text{even } p, & c^{p+1} \neq 0, \displaystyle \frac{(p+1)^{2(p+1)}}{p^{2p}} \end{array} \right.$

$\endgroup$
  • $\begingroup$ Discriminant[x^(2n+2)- a x^(2n) - 2x^(n+1)+1 ,x] did not work for me... $\endgroup$ – lhf Apr 13 '16 at 14:14
  • 1
    $\begingroup$ @lhf Table[{p, Discriminant[x^(2 p + 2) - c x^(2 p) - 2 x^(p + 1) + 1, x] // Simplify}, {p, 1, 10}] // TraditionalForm $\endgroup$ – Ng Chung Tak Apr 13 '16 at 14:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.