Example 6 Find the simplest fraction (with the smallest denominator) that approximates , with an error .
Solution
In Example 4, we have found . We first list to find , , and then estimate the error according to formula (16).
\begin{tabular}{|c|rrrrrrrrrr|}
\hline & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline & 2 & 1 & 4 & 1 & 4 & 1 & 4 & 1 & 4 & 1 \\
& 2 & 3 & 14 & 17 & 82 & 99 & 478 & 577 & 2786 & 3363 \\
& 1 & 1 & 5 & 6 & 29 & 35 & 169 & 204 & 985 & 1189 \\
\hline
\end{tabular}
From the table above, when , by formula (16) we get
Therefore, the required convergent fraction is .
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