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I have come across continued fractions approximation but I am unsure what the steps are. For example How would you express the following rational function in continued-fraction form: $${x^2+3x+2 \over x^2-x+1}$$

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I think continued fractions are used for constants such as $\sqrt 2 $, you are probably referring to something like Pade approximant. – glebovg Oct 22 '12 at 18:31
Well it is related to Pade approximant. But from what I understand its a simplification of the pade approximant – math101 Oct 22 '12 at 18:35
up vote 4 down vote accepted

This might be what you are looking for: $$ \begin{align} \frac{x^2+3x+2}{x^2-x+1} &=1+\cfrac{4x+1}{x^2-x+1}\\[4pt] &=1+\cfrac1{\frac14x-\frac5{16}+\cfrac{\frac{21}{16}}{4x+1}}\\[4pt] &=1+\cfrac1{\frac14x-\frac5{16}+\cfrac1{\frac{64}{21}x+\frac{16}{21}}} \end{align} $$ 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: $$ 1+\cfrac1{\frac14x-\frac5{16}}=\frac{4x+11}{4x-5} $$ That is, we can write the polynomial GCD (a constant since they are relatively prime) as $$ (4x+11)(x^2-x+1)-(4x-5)(x^2+3x+2)=21 $$

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The answer in the back of the book is:$$1+{4 \over x- \frac{5}{4}} +{ \frac{21}{16} \over x + \frac{1} {4}}$$ – math101 Oct 22 '12 at 19:50
ohhhh I seee it now. I was getting confused as to how you remove that staircase. Now I seee. Thanks for your help. I really appreciate it – math101 Oct 22 '12 at 19:52

Start out writing $$\frac{x^2+3x+2}{x^2-x+1}=\frac{1}{\frac{(x^2+3x+2)-(4x+1)}{x^2+3x+2}}=\frac{1}{1-\frac{4x+1}{x^2+3x+2}}$$ and then iterate doing the same with the fraction in the denominator.

EDIT: Complete solution:

By polynomial long division we have $\frac{x^2+3x+2}{4x+1}=\frac{x}{4}+\frac{11}{16}+\frac{21}{16}\frac{1}{4x+1}$. Hence the above expression equals

$$\huge \frac{1}{1-\frac{1}{x/4+11/16+\frac{1}{40x/21+10/21}}}. $$

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hahah I must be doing something wrong cuz it seems like I am just going on for forever – math101 Oct 22 '12 at 18:57
Yes, it should stop if you're doing it correctly. Just do polynomial long division. – Your Ad Here Oct 22 '12 at 19:04
Can you please show me the complete process. This is my first time touching this topic and I quite confused – math101 Oct 22 '12 at 19:11

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