# 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!

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Do you mean the limits of summation of $f$ and $g$ to be different? –  Igor Rivin Nov 20 '13 at 5:06
Hello,what you mean? Thank you –  math110 Nov 20 '13 at 5:09
It appears that the definition of $g(x)$ always has the root $x=0$ according to the summation, but the written-out function seems to not have this property... –  abiessu Nov 20 '13 at 5:43
@sorry,I have edit,this $g(x)$ sum is $i=0$ to $i=3$ –  math110 Nov 20 '13 at 14:38
You don't need to do case (3) : when you change a real polynomial continuously, real roots won't jump into complex conjugate pairs of roots unless you have a double real root somewhere. So there is no doubt that there still are $3$ real roots until you find some $m$ where the polynomial has a double real root, and then we can simply use case (2). So it is enough to show the result for cases (1) and (2). Maybe you can try looking at the derivative of the function $m \mapsto \Delta(f_m)$ at $m=1$ and try to prove it's strictly positive (this would be enough !), but it doesn't look good. –  mercio Nov 20 '13 at 17:23

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.

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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.

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Thank you,No But my problem $m\in R$,This generalization only can prove $m\in N^{+}$. –  math110 Nov 20 '13 at 16:37
@math110: This generalization reduces your problem to the case $1\leq m<2$. –  y zhao Nov 20 '13 at 16:46
@yzhao : to be fair, any proof for $1 \le m \le 1 + \varepsilon$ (with $\varepsilon$ independant of $f$ of course) is enough. –  mercio Nov 20 '13 at 16:56
and How prove this case $1\le m<2$? –  math110 Nov 20 '13 at 18:15
Hmm, for fourth or higher degree equations, counterexamples exist. –  y zhao Nov 23 '13 at 14:13

For the first question, this is really a question about the derivative. $f$ has three real zeros if and only if its derivative has two zeros, and $f$ is positive at the smaller critical point, and negative at the bigger critical point [by the way, I am dividing through by $a_3$ to make the polynomial monic -- this does not change the question). This can be checked for $g$ by tedious computation, which I leave to you...

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No,This methods is not usefull,But Thank you all the same –  math110 Nov 20 '13 at 14:32