Proof that Qm(x) and Pn(x) have the same sign

Question

I have difficulty in this exercise, is from the book Apostol Calculus Vol 2, I really want to know is there is a technique that can show that the two polynomials share the same roots.

Thanks.

Answer 1

In Apostol's book it is said (just before exercise $8$) that $P_n$ is the Legrendre polynomial of degree $n$ (assume $n\ge$ since the case $n=0$ case is trivial $P_1=Q_0=1).$

If you have solved the part $a)$ of this exercise then you know that every root of $P_n$ is simple. So, if $x_1,\cdots, x_m$ are the roots in $(-1,1)$ then $P_n(x)=r(x)(x-x_1)(x-x_2)\cdots (x-x_m)$ where $r(x)$ is a polynomial of degree $n-m,$ and that doesn't vanish on $(-1,1).$ So, to solve the exercise, we only need to show that $r(x)>0$ on $(-1,1).$ But, since $P_n(1)=1$ we can conclude that $r(1)(1-x_1)(1-x_2)\cdots (1-x_m)>0,$ from where, using that $(1-x_1)(1-x_2)\cdots (1-x_m)>0,$ we get $r(1)>0.$