How prove this Polynomial $g(x)=\sum_{i=1}^{n}a^m_{i}x^i$have only real roots?

Question 1:

let Polynomial $f(x)=\displaystyle\sum_{i=0}^{3}a_{i}x^i,$ have three real numbers roots,where $a_{i}>0,i=1,2,3$.

show that: $$g(x)=\sum_{i=0}^{3}a^m_{i}x^i$$ have only real roots,where $m\in R,m\ge 1$

My try:

case (1):

suppose that $f(x)$ has a zero of multiplicity 3, then we assume

$$f(x)=(x+p)^3=x^3+3x^2p+3xp^2+p^3$$ then $$g(x)=x^3+(3p)^mx^2+(3p^2)^mx+(p^3)^{m}=(x+p^m)[x^2+(3^mp^m-p^m)x+p^{2m}]$$

then

$$h(x)=x^2+p^m(3^m-1)x+p^{2m}\Longrightarrow \Delta =(p^m(3^m-1))^2-4p^{2m}>0$$ so this case $g(x)$ have only three real roots.

for case (2):

let $f(x)=(x+p)^2(x+q)$,

I can't prove it,

and the case (3):

$$f(x)=(x+p)(x+q)(x+r)$$ and this case I can't prove it too.

I hope someone can help solve this nice problem ;

Thank you very much!

Solution 1:

As I explained in my comment, it is enough to do case (2).

Let $x,y \ge 0$. Then $(X+x)(X+x)(X+y) = X^3+(2x+y)X^2 + (x^2+2xy)X + x^2y$. The discriminant of $X^3+bX^2+cX+d$ is, according to wikipedia, $\Delta(b,c,d) = b^2c^2-4c^3-4b^3d-27d^2+18bcd$.

The derivative of $m \mapsto \Delta(b^m,c^m,d^m)$ at $m=1$ is $$\delta(b,c,d) \\ = (2b^2c^2-12b^3d+18bcd)\log b + (2b^2c^2-12b^3d+18bcd)\log c + (18bcd-4b^3d-54d^2)\log d $$

Plugging $b = 2x+y,c = x^2+2xy,d = x^2y$ we compute and obtain $$ \delta(x,y) = 4x^6(\frac yx -1)^3\left(-(2+ \frac yx)\log \frac {(2x+y)^2}{x^2+2xy} + \frac yx \log \frac {(2x+y)^3} {x^2y}\right)$$

$\delta(x,y)/4x^6$ is actually a function of $t = \frac yx$ and has the same sign as $\delta(x,y)$ so it is enough to study the sign of the function $$g(t) = (t-1)\left(-(2+t)\log \frac{(2+t)^2}{1+2t} + t \log \frac {(2+t)^3}t\right) \\ = (t-1)\left((t-4)\log(2+t) + (2+t)\log(1+2t) - t\log t\right) $$

Letting $g(t) = (t-1)h(t)$, we compute $h'(t) = \log(9+(t- t^{-1})^2) + 2\frac{(t-1)^2}{2t^2+5t+2} \ge \log 9 > 0 $. Since $h(1) = -3\log3+3\log3-\log1 = 0$, we have $h(t)>0$ for $t>1$, $h(t)<0$ for $t<1$, and $h(1)=0$.

Going back to $g$ we learn that $g(t)>0$ except when $t=1$, which means that $\delta(x,y) > 0$ except when $x=y$, which is what we wanted.

Solution 2:

It is really a tough question. A generalization of this problem follows from a theorem in the famous book by Polya and Szego(Book II, Chapter 5, Problem 155), which states:

If

$$a_0+a_1 x+\cdots+a_n x^n$$

and

$$b_0+b_1 x+\cdots+b_n x^n$$

have only real zeros, while all the zeros of the latter have the same sign, then

$$a_0 b_0+a_1 b_1 x+\cdots+a_n b_n x^n$$

has real zeros only.

The proof is complicated, so borrowing a copy from library is recommended.