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2022-07-01

Continued Fractions Approximation
$\frac{{x}^{2}+3x+2}{{x}^{2}-x+1}$

### Answer & Explanation

Alisa Jacobs

$\begin{array}{rl}\frac{{x}^{2}+3x+2}{{x}^{2}-x+1}& =1+\frac{4x+1}{{x}^{2}-x+1}\\ & =1+\frac{1}{\frac{1}{4}x-\frac{5}{16}+\frac{\frac{21}{16}}{4x+1}}\\ & =1+\frac{1}{\frac{1}{4}x-\frac{5}{16}+\frac{1}{\frac{64}{21}x+\frac{16}{21}}}\end{array}$
At each stage, we are doing a polynomial division instead of an integer division, but otherwise, the process is the same as with continued fractions with integers.
We can get the Bezout polynomials by truncating the continued fraction:
$1+\frac{1}{\frac{1}{4}x-\frac{5}{16}}=\frac{4x+11}{4x-5}$
That is, we can write the polynomial GCD (a constant since they are relatively prime) as
$\left(4x+11\right)\left({x}^{2}-x+1\right)-\left(4x-5\right)\left({x}^{2}+3x+2\right)=21$

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