Computational Problems, Methods and Results in Algebraic by Horst G Zimmer

By Horst G Zimmer

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Example text

An ] You tell them the numerator and denominator of this fraction. There are a couple of traps, one theoretical, one practical. First of all, as we have seen, the greatest common divisor of the numerator and denominator of a continued fraction is 1. So if the original numbers r and s have greatest common divisor d > 1, then you will find r/d and s/d instead of r and s. There is nothing that can be done about this; you have to bluff your way out of it as well as you can. The practical problem is caused by rounding errors.

An−2 ]. Proof (a) holds because in Euler’s formula the reversed sequence obviously gives exactly the same result. Then (b) is straightforward: [a0 , . . , an ] = [an , . . , a0 ] = an [an−1 , . . , a0 ] + [an−2 , . . , a0 ] = [a0 , . . , an−1 ]an + [a0 , . . , an−2 ]. 3 = = = = = 3 [3]1 + 1 = 4 [3, 1]4 + [3] = 19 [3, 1, 4]1 + [3, 1] = 23 [3, 1, 4, 1]6 + [3, 1, 4] = 157. The convergents of a finite continued fraction Let a0 , a1 , . . , an be positive integers. We define the convergents of the continued fraction [a0 ; a1 , .

We begin with a lemma about purely periodic continued fractions. 2 If y = [a0 ; a1 , . . , an−1 , an ], then −1/y = [an ; an−1 , . . , a1 , a0 ]. Proof y = [a0 ; a1 , . . , an , y] = ypn + pn−1 , yqn + qn−1 so qn y2 + (qn−1 − pn )y + pn−1 = 0. Let z = [an ; an−1 , . . , a0 ]. Then z[an , . . , a0 ] + [an , . . , a1 ] z[an−1 , . . , a0 ] + [an−1 , . . , a1 ] zpn + qn = , zpn−1 + qn−1 z = [an ; an−1 , . . , a0 , z] = where we use the fact that [a0 , . . , an ] = [an , . . , a0 ]. So pn−1 z2 + (qn−1 − pn )z − qn = 0.

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