A simple zeroes polynomial which constructs another one

Question

Let $P(X) = (X-x_1)\ldots(X-x_N)$ be a complex polynomial with simple roots.

I define $$Q(X) = P(X-a)-bP(X),$$ with $a\in\mathbb{C}$ and $b\neq 1$ so that $Q(X)$ is also a polynomial of degree $N$.

Let me note $y_1,\ldots,y_N$ the roots of $Q(X)$. I would like to prove that

• $Q$ has simple roots : $\forall i,j, \quad y_i \neq y_j$
• one cannot shifts from a root to another by the use of $a$ : $\forall i,j,\quad y_i \neq y_j \pm a $

I am not sure this is the case for all $a$ and $b$, but in the context of my problem it seems $Q$ has to verify these conditions.

I thought it would be a simple exercise, but I keep struggling on it.

Any help is much appreciated ! :)

Answer 1

It's not true.

As a counterexample, letting

• $P(x)=(x+2)(x-2)$$\\[4pt]$
• $a=5$$\\[4pt]$
• $b=-4$

we get

$\;\;\;Q(x)=5(x+1)^2$.

Staying with the case $n=2$, let $P(x)=(x-r)(x-s)$ with $r\ne s$.

Then for $b\ne 1$, the polynomial $$Q(x)=P(x-a)-bP(x)$$ has simple roots if and only if either$\;b=0\;$or $$a^2\ne -\frac{(b-1)^2(r-s)^2}{4b}$$

Let's try an example with $n=3$ . . .

Let $P(x)=(x-1)x(x+1)$.

Then for $b\ne 1$, the polynomial $$Q(x)=P(x-a)-bP(x)$$ has simple roots if and only if $$ 4b^4+(36a^2-16)b^3+(-27a^6+108a^4-72a^2+24)b^2+(36a^2-16)b+4 $$ is nonzero.

The results of that example suggest that for the general case, trying to find usable necessary and sufficient conditions on $a,b$ for $Q(x)$ to have simple roots is not likely to succeed.

As a more reasonable goal, one might try to find sufficient conditions for $Q(x)$ to have simple roots, expressed in terms of inequalities relating $|a|,|b|$.

For example, if $a$ is fixed, then $Q(x)$ will have simple roots if

• $b\;$is sufficiently close to $0$.

  or

• $|b|\;$is sufficiently large.

Answer 2

If you draw $P(X)$ and $bP(X+a)$ you will see that, if $a$ is quite less than the minimum separation between the zeros and $b$ enough less than $1$ as not to alter the relative position of the min/max, then the sum will have simple zeros which shall between $x_k \pm a$.