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Let $u$ be a posive irrational number and let

$$u=a_0+ \frac{1}{a_1+\frac{1}{a_2 \cdots} }$$

be its continued fraction expansion.

Consider the second-order finite continued fraction expansion of $u$:

$\frac{p_2}{q_2}:=a_0+ \frac{1}{a_1+\frac{1}{a_2}}$ with $p_2$ and $q_2$ being positive coprime integers.

I wonder if we have the following inequality and how to prove it:

$$|p_2-q_2 u|<\|u\|,$$

where $\|u\|$ denotes the distance from $u$ to its nearest integer.


Please also note that I am not asking how to prove $|\frac{p_2}{q_2}-u|<\|u\|$, which is simpler.


If the above is true (I have verified it with many examples), please feel free to use any very basic properties of continued fractions to prove it.

But please do not use any general fact about the optimality of the rational approximation with continued fractions to prove this because I actually want to use the above fact as one of the steps to prove optimality of the rational approximation with continued fractions.

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2 Answers 2

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Hint:

The Continued Fraction expansion (and approximation) is tied to the Stern-Brocot tree.

In turn, the properties of the Stern-Brocot tree and related Farey sequence, ensure that a truncation of the CF, that is a truncated path on the tree, represents the best rational approximation (with the denominator being less than a fixed threshold).

The above is for the general case of taking $n$- terms of the CF.

A more simple reply to your specific case is that, being $$ u = \left\lfloor u \right\rfloor + \left\{ u \right\}\quad \left| {\;0 \le \left\{ u \right\} < 1} \right. $$ the splitting of $u$ into the integral and fractional part, then $$ a_{\,0} = \left\lfloor u \right\rfloor $$ and $$ \eqalign{ & \left\{ u \right\} = {1 \over {{1 \over {\left\{ u \right\}}}}} = {1 \over {\left\lfloor {{1 \over {\left\{ u \right\}}}} \right\rfloor + \left\{ {{1 \over {\left\{ u \right\}}}} \right\}}} = {1 \over {a_{\,1} + {1 \over {{1 \over {\left\{ {1/\left\{ u \right\}} \right\}}}}}}} = \cr & = {1 \over {a_{\,1} + {1 \over {\left\lfloor {{1 \over {\left\{ {1/\left\{ u \right\}} \right\}}}} \right\rfloor + \left\{ {{1 \over {\left\{ {1/\left\{ u \right\}} \right\}}}} \right\}}}}} = {1 \over {a_{\,1} + {1 \over {a_{\,2} + r_{\,2} }}}} \cr} $$

So you can conclude by yourself.

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    $\begingroup$ Thanks but it's a little hard for my to understand this. Is it possible to give a direct proof for my question? My question is just about the order 2 case... what is the relation between p_2, q_2 and Stern-Brocot tree specifically? $\endgroup$
    – No One
    Commented Jan 1, 2021 at 20:37
  • $\begingroup$ @NoOne: don't know your level of knowledge about CF, Stern-Brocot, Euclidean Algorithm inter-relation. I addeda trace of how the pattern begins, hope it is clear for you. $\endgroup$
    – G Cab
    Commented Jan 1, 2021 at 23:04
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cannot comment but some clarification would be fitting;

What is meant by ||[blah]|| in this context (supposing uϵ[blah]), or more to the point Why/how would it be equal to n minus [blah] given $nϵ{N},0<|u|<1$?

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