sufficient condition for a polynomial to have roots in $[0,1]$

Question is to check :

which of the following is sufficient condition for a polynomial

$f(x)=a_0 +a_1x+a_2x^2+\dots +a_nx^n\in \mathbb{R}[x] $ to have a root in $[0,1]$.

• $a_0 <0$ and $a_0+a_1+a_2+\dots +a_n >0$
• $a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots +\frac{a_n}{n+1}=0$
• $\frac{a_0}{1.2}+\frac{a_1}{2.3}+\dots+\frac{a_n}{(n+1).(n+2)} =0$

First of all i tried by considering degree $1$ polynomial and then degree $2$ polynomial and then degree $3$ polnomial hoping to see some patern but could not make it out.

And then, I saw that $a_0= f(0)$ and $f(1)=a_0+a_1+a_2+\dots +a_n$.

So, if $f(0)<0$ and $f(1)>0$ it would be sufficient for $f$ to have root in $[0,1]$

In first case we have $a_0 <0$ i.e., $f(0)<0$ and $f(1)>1>0$.

So, first condition should be implying existence of a root in $[0,1]$

for second case, let $f(x)$ be a linear polynomial i.e., $f(x)=a_0+a_1x$

Now, $a_0+\frac{a_1}{2}=0$ implies $0\leq x=\frac{-a_0}{a_1}=\frac{1}{2}< 1$ So, this might be possibly give existence in case of linear polynomials.

Now, $\frac{a_0}{1.2}+\frac{a_1}{2.3}=0$implies $0\leq x=\frac{-a_0}{a_1}=\frac{1}{3}< 1$ So, this might be possibly give existence in case of linear polynomials.

So, for linear polynomials all the three conditions imply existence of a root in $[0,1]$.

But, i guess this can not be generalized for higher degree polynomial.

I think there should be some "neat idea" than checking for roots and all.

I am sure about first case but I have no idea how to consider the other two cases.

please provide some hints to proceed further.

Solution 1:

For the second case consider the polynomial $$ F(x)=a_0x+\frac12a_1x^2+\frac13a_2x^3+\cdots+\frac1na_{n-1}x^{n}+\frac1{n+1}a_nx^{n+1} $$ and then use Rolle's theorem.

For the third case consider some other polynomial (which?) and then use two times Rolle's theorem.

Solution 2:

Hints:

• $a_0 <0$ and $a_0+a_1+a_2+\dots +a_n >0$ means that $f(0)<0$ and $f(1)>0$.

• $a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots +\frac{a_n}{n+1}=0$ means that $F(1)=0$, where $F(x)=\int_0^x f(t) \, dt$. Note that $F'=f$ and $F(0)=0$ and recall Rolle's theorem.

Can you think of what the third condition means in this context?

Solution 3:

for third case we consider polynomial

$F(x)=\frac{a_0}{1.2}x^2+\frac{a_1}{2.3}x^3+\dots + \frac{a_n}{(n+1)(n+2)}a_nx^{n+2}$

we now assume third condition i.e., $\frac{a_0}{1.2}+\frac{a_1}{2.3}+\dots+\frac{a_n}{(n+1).(n+2)} =0$

In that case, for polynomial $F(x)$ we would then have $F(0)=0$ and $F(1)=0$ (with given condition)

So, by rolle's theorem we have a root for $F'(x)$ in $[0,1]$

i.e., we have a root for $F'(x)=\frac{a_0}{1}x+\frac{a_1}{2}x^2+\dots+ \frac{a_n}{n+1}x^{n+1}$ in $[0,1]$ say at $c\in [0,1]$

Now, for $F'(x)$ we have two zeros.. i.e., $F'(0)=0$ and $F'(c)=0$

Now, i will use rolle's theorem again i.e, i have root for $F''$ in $[0,c]$

where $F''(x)=a_0+a_1x+a_2x^2+\dots+a_nx^n$

to conclude, i now set $F''(x)=f(x)$ and with given condition,

i have a root in $[0,c] $ for some $c\in [0,1]$ particularly, it has a zero in $[0,1]$

i.e., $f(x)=a_0+a_1x+a_2x^2+\dots+a_nx^n$ has a root in $[0,1]$

To conlcude, with above answer and my previous observation of second case,

$f(x)=a_0 +a_1x+a_2x^2+\dots +a_nx^n\in \mathbb{R}[x] $ have a root in $[0,1]$ in all three following cases:

• $a_0 <0$ and $a_0+a_1+a_2+\dots +a_n >0$
• $a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots +\frac{a_n}{n+1}=0$
• $\frac{a_0}{1.2}+\frac{a_1}{2.3}+\dots+\frac{a_n}{(n+1).(n+2)} =0$

P.S : This is completely for the sake of my reference and all the credit goes to above two users who have helped me to go through this idea.