# The polynomial $p(x)=a_0+a_1x+a_2x^2+\cdots +a_nx^n$ has a zero in $[0,1]$ under a given condition

Show that the polynomial $p(x)=a_0+a_1x+a_2x^2+\cdots +a_nx^n$ has a zero in $[0,1]$ when it is given that , $$\frac{a_0}{1.2}+\frac{a_1}{2.3}+\cdots+\frac{a_n}{(n+1)(n+2)}=0.$$

First we consider the function , $\displaystyle f(x)=\frac{a_0}{1.2}x^2+\frac{a_1}{2.3}x^3+\cdots+\frac{a_n}{(n+1)(n+2)}x^{n+2}$. Then , $f(0)=0$ and $f(1)=0$. Then by Rolle's theorem , $\exists$ $y\in (0,1)$ such that $f'(y)=0$ which gives , $\displaystyle a_0y+\frac{a_1}{2}y^2+\cdots +\frac{a_n}{n+1}y^{n+1}=0$.

Next consider , $\displaystyle g(z)=a_0z+\frac{a_1}{2}z^2+\cdots +\frac{a_n}{n+1}z^{n+1}$. Then , $g(0)=0$ and $g(y)=0$. So by Rolle's theorem , $\exists$ $w\in (0,y)\subset (0,1)$ such that $g'(w)=p(w)=0.$ Hence the proof is complete.

Is this proof is correct ? I think it is correct. But I am looking for an another proof so that I can avoid two times consideration of functions such as $f$ and $g$.

• The proof is correct, well described. – André Nicolas Nov 9 '15 at 7:11
• @ André Nicolas) Please see the next question. – Empty Nov 9 '15 at 7:12
• I can't think of a way to avoid differentiating twice, or integrating twice. – André Nicolas Nov 9 '15 at 7:16
• – Arnaud D. Sep 4 '18 at 12:04

The proof is correct (but note it is Rolle, not Roll). You can also use the fact that $$\int_0^1\int_0^x p(t)dt=\sum_{k=0}^n\frac{a_k}{(k+1)(k+2)}=0$$
Now if you suppose that $p(t)\not = 0$ on $[0,1]$, then (as $p$ is continuous)there exists $m$ such that $p(t)\geq m>0$ for all $t\in [0,1]$ (or $p(t)\leq -m<0$), and this show that the multiple integral is $\geq m/2$, a contradiction.
As $\displaystyle \frac{1}{(k+1)(k+2)}=\frac{1}{k+1}-\frac{1}{k+2}$, we get $$\int_0^1(1-t)p(t)dt=\sum_{k=0}^n\frac{a_k}{k+1}-\sum_{k=0}^n\frac{a_k}{k+2} =0$$ and we finish easily.