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The sequence $Q_n(x)$ of polynomials is defined by $Q_1(x) = 1+x, Q_2(x)=1+2x$, and for $m \ge 1$,

$$Q_{2m+1}(x)=Q_{2m}(x)+(m+1)xQ_{2m-1}(x)$$ $$Q_{2m+2}(x)=Q_{2m+1}(x)+(m+1)xQ_{2m}(x)$$

Let $x_n$ be the largest real solution of $Q_n(x)=0$. Prove that {$x_n$} is an increasing sequence and that $\lim_{x \to \infty}x_n=0$

Attempt: Obviously {$x_n$} is bounded above by zero, as $Q_n(x)$ are polynomials with positive coefficients and a constant term 1. I just have trouble showing {$x_n$} is strictly increasing. I'm guessing that I need to apply the squeeze principle, but I have no idea how to do that.

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    BTW, it's not immediately clear to me why all these polynomials would _have any real roots_ in the first place. – 2017-02-21
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    @IvanNeretin looks true from the first few cases. I tried induction but didn't get it to work – 2017-02-21
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    Yeah, I see that it looks true, but why must it always be that way? – 2017-02-21
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    @IvanNeretin Well the odd degree polynomials certainly do. I don't know about the even degree ones. But isn't that part of what we're trying to prove? – 2017-02-21
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    Yes, that's right, and I'm stuck. – 2017-02-21
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    Hint: Use induction (starting with $n=1$ and $n=2$) and plug the $x_n$ into the recursive formula for $Q_{n+1}$. What can you say about the sign of $Q_{n+1}(x_n)$? – 2017-02-21

1 Answers 1

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With $c_n=\lfloor{\frac n2}\rfloor + 1$ we can rewrite the recursion as one equation:

$$Q_{n+1}(x) = Q_n(x)+c_nxQ_{n-1}(x)$$

We will prove by induction that $x_n$ exists (that means that $Q_n(x)$ has at least one real root) and that $x_n < x_{n+1}$ holds.

First, as the OP noted, $Q_n(x) \ge 1$ for $x \ge 0$, so any root of any $Q_n(x)$ must be negative, hence $x_n < 0$ (if it exists). Obviously, as the start of our induction, we see that $x_1 =-1 < -\frac 12 = x_2$.

Let's assume that $x_{n-1} < x_n$ for some $n \ge 2$. We then have

$$ Q_{n+1}(x_n) = Q_n(x_n) + c_n x_n Q_{n-1}(x_n) = c_n x_n Q_{n-1}(x_n), $$ as $Q_n(x_n)= 0$ by definition of $x_n$. In the product on the right, we have $c_n > 0, x_n <0$ and $Q_{n-1}(x_n) > 0$. The last inequality is due to the fact that by induction hypothesis $x_n$ is larger than the largest root $x_{n-1}$ of $Q_{n-1}(x)$ and since $Q_{n-1}(0)=1$ a negative value of $Q_{n-1}(x_n)$ would imply a root of $Q_{n-1}(x)$ in the interval $(x_n,0)$, contradicting the definition of $x_{n-1}$.

So we finally get $$Q_{n+1}(x_n) < 0,$$ which means that there is a root of $Q_{n+1}(x)$ in the interval $(x_n,0)$, proving $x_{n+1}$ is defined and that $x_{n+1} > x_n$ holds. This concludes the prove by induction.$\blacksquare$

We now know that ${x_n}$ is strictly increasing and bounded above by $0$, so it converges: $$\lim x_n= l\le0$$.

We want to prove $l=0$, so let's assume to the contrary that $l < 0$. We consider the values $Q_n(l)$ for all $n$ . First, since $x_n < l$ for all $n$, we have $Q_n(l) >0$ as in the proof above. Second, we have $$Q_{n+1}(l) = Q_n(l)+c_n l Q_{n-1}(l) < Q_n(l),$$ as in the term $c_n l Q_{n-1}(l)$ the factors $c_n$ and $Q_{n-1}(l)$ are positive and $l$ is negative.

But with $l < 0$ there is an $N$ such that $c_Nl < -1$ (remember that $c_n$ is roundabout $\frac n2$). This gives

$$Q_{N+1}(l) = Q_N(l)+c_N l Q_{N-1}(l) < Q_N(l) - Q_{N-1}(l) < 0,$$

in contradiction to $Q_n(l) >0$ for all $n$.

Thus we have proved that $$\lim x_n= 0.$$