I am looking this answer and maybe it's trivial but now can't understand one part of José Carlos Santos's answer. And if $n$ is odd, then $Q_{(n+1)}$ first decreases and then increases (it's trivial I know but I don't remember how can I proof that).
Asked
Active
Viewed 204 times
0
-
Because as mentioned in his answer, $Q_n$ is negative then becomes positive. – Kenta S Apr 02 '21 at 18:30
-
MathJax hint: to get multicharacter subscripts, put them in braces, so Q_{(n+1)} gives $Q_{(n+1)}$. What is it? It is never defined. – Ross Millikan Apr 02 '21 at 18:31
1 Answers
2
Note that$$Q_{n+1}'(x)=Q_n(x).$$So, if $n$ is odd, $Q_n$ has a single real root, at some point $x_0$. But, again because $n$ is odd, $\lim_{x\to\infty}Q_n(x)=\infty$ and $\lim_{x\to-\infty}Q_n(x)=-\infty$. So, $Q_n(x)<0$ if $x<x_0$ and $Q(x)>0$ if $x>x_0$. Therefore, $Q_{n+1}$ is decreasing on $(-\infty,x_0]$ and increasing on $[x_0,\infty)$.
J. W. Tanner
- 60,406
José Carlos Santos
- 427,504