I would rather this not be solved for me , as its a homework question , I just want some clarification on whether I'm understanding it correctly.
Say we have the continued fraction $\alpha=[3,\bar{2},\bar{4},\bar{5}]$.
I know this is the root of the quadratic polynomial $P(x)=9x^2-11x-69$, and also that written as a quadratic irrational $\alpha=\tfrac{11+\sqrt{2605}}{18}$.
In the question it says find bounds for $|P'(x)|, x\in(\alpha-1,\alpha+1)$.
Which I did ,and they're: $-18+ \sqrt{2605}\leq|P'(x)|\leq 18+\sqrt{2605}$.
The question then goes on to say , Hence or otherwise find a constant C s.t.
$|\alpha-\tfrac{p}{q}|>\tfrac{1}{Cq^2}$.
This is where I start to become unsure, unfortunately I wasn't able to attend the lecture on this due to health reasons.
Here are my questions:
1) understanding the terms :
Should we here say $\alpha=3+\tfrac{1}{2+\tfrac{1}{4+\tfrac{1}{5+...}}}$, $\tfrac{p}{q}=\tfrac{11+\sqrt{2605}}{18}$ and so then $q=18$?
2) the connection between having found the bounds of $P'(x)$ and the question itself:
Could anyone explain to me the theory linking these two parts of the question? Or alternatively perhaps a link, if the theory is too complicated to write all the way out
There is a number $M$ such that $|P'(x)|<M$ for $\alpha-1<x<\alpha+1$ (as you know, since you have actually found such a number). We may assume $p/q$ lies between $\alpha-1$ and $\alpha+1$, and is closer to $\alpha$ than is the other root of $P$. So $P(p/q)\ne0$, so $|P(p/q)|=|9p^2-11pq-69q^2|q^{-2}\ge q^{-2}$. Now, $$P(p/q)=P(p/q)-P(\alpha)=((p/q)-\alpha)P'(x)$$ for some $x$ between $p/q$ and $\alpha$ (by the Mean Value Theorem). So $$|(p/q)-\alpha|=|P(p/q)|/|P'(x)|>(Mq^2)^{-1}$$
More generally, this argument can be used to prove Liouville's Theorem, q.v., and indeed I have cribbed the above from the proof of Liouville in Hardy and Wright.