# continued fraction (trivial?) substraction

This is my first post!

I have been working with continued fractions, trying to perform some math on them, but I have come to a point where I'm not sure if what i'm doing is right or wrong. Any help is welcome!

So I have two functions, say $f(n)$ and $g(n)$, one with a FINITE general continued fraction $$f(n)=\frac{b_0}{a_1-\frac{b_1}{a_2-\frac{b_2}{a_3-\frac{b_3}{\frac{ _\ddots}{a_n-b_n}}}}}$$ for which I have proven that $$\lim_{n \to \infty}f(n)=\frac{b_0}{a_1-\frac{b_1}{a_2-\frac{b_2}{a_3-\frac{b_3}{ _\ddots}}}}$$ that is, it just adds more and more $a_m+b_m$ terms on the bottom to infinity. And the second one with an INFINITE general continued fraction $$g(n)=\frac{c_0(n)}{a_1-\frac{c_1(n)}{a_2-\frac{c_2(n)}{a_3-\frac{c_3(n)}{ _\ddots}}}}$$ for which I have proven that $$\lim_{n \to \infty}g(n)=\frac{c_0}{a_1-\frac{b_1}{a_2-\frac{b_2}{a_3-\frac{b_3}{ _\ddots}}}}$$ that is, other than the top numerator the rest is the same as for $f(n)$.

I tried to perform $\lim_{n \to \infty}(f(n)-g(n))$ as $$\lim_{n \to \infty}(f(n))-\lim_{n \to \infty}(g(n))=\frac{a_0}{a_1-\frac{b_1}{a_2-\frac{b_2}{a_3-\frac{b_3}{ _\ddots}}}}-\frac{c_0}{a_1-\frac{b_1}{a_2-\frac{b_2}{a_3-\frac{b_3}{ _\ddots}}}}=\frac{a_0-c_0}{a_1-\frac{b_1}{a_2-\frac{b_2}{a_3-\frac{b_3}{ _\ddots}}}}$$

is this even possible? my answer is not what I expected, so I'm guessing this substraction it's not as trivial as this... any comment?

By the way and on a more general sense, I have not very advanced programming skills or knowledge, could you provide some reference where I can learn some algorithms to calculate several decimal places, with a known error, for this kind of continued fractions? So far I have done it the direct way and with my calculator's decimal place limit (...!), how can I get better results?

Thanks very much!

For two irrational numbers, whose continued fractions must converge, we can write $$\frac{a_0}{a_1-\frac{b_1}{a_2-\frac{b_2}{a_3-\frac{b_3}{ _\ddots}}}}-\frac{c_0}{a_1-\frac{b_1}{a_2-\frac{b_2}{a_3-\frac{b_3}{ _\ddots}}}}$$ as $$a_0\frac{1}{a_1-\frac{b_1}{a_2-\frac{b_2}{a_3-\frac{b_3}{ _\ddots}}}}-c_0\frac{1}{a_1-\frac{b_1}{a_2-\frac{b_2}{a_3-\frac{b_3}{ _\ddots}}}}$$ that is $$(a_0-c_0)\frac{1}{a_1-\frac{b_1}{a_2-\frac{b_2}{a_3-\frac{b_3}{ _\ddots}}}}$$ is valid, corresponding to my original question.
• I changed all the $x$ to $n$, since obviously there is no $x$ anywhere in these functions – Yuriy S Dec 16 '16 at 8:50
• The question is interesting, however it would be better and have more of a chance to be answered here if you just provided the explicit expressions for $f(n)$ and $g(n)$ – Yuriy S Dec 16 '16 at 8:55
• It is correct in general that if $\lim f(n)$ and $\lim g(n)$ exist, then $\lim [f(n)-g(n)]$ exists and is equal to $\lim f(n) - \lim g(n)$. – Patrick Stevens Dec 16 '16 at 9:01