Closed form for continued fraction

Question

I am working with a recursive algorithm and I have realized that in each step it computes something equivalent to:

$A_{0}=\frac{a}{b}$

$A_{1}=a/(b-\frac{a^{2}}{b})$

$A_{2}=a/(b-\frac{a^{2}}{b-\frac{a^{2}}{b}})$

$\vdots$

$A_{n+1}=a/(b-aA_{n})$

And so on. I was wondering if there could exist some closed form for $A_{n}$ to avoid unnecessary iterations

Answer 1

The closed form does exist, I found it with Mathematica and it's amazing, to use in a program I mean.

I found it with this function

RSolve[{A[n + 1] == a/(b - a A[n]), A[0] == a/b}, A[n], n]

Set: $$ \begin{cases} D=\sqrt{b^2-4 a^2}\\ x=\frac{b-D}{a}\\ y=\frac{b+D}{a}\\ \end{cases} $$ Then $$A_n=\frac{a (b+D ) y^n-a (b-D ) x^n}{x^n \left(2 a^2-b (b-D )\right)+y^n \left(b (b+D )-2 a^2\right)}$$

You will surely find a way to optimize better this formula in order to use it in a program.

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Edit

This is the formula as output by Mathematica $$A_n=\frac{a \left(b \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n+\sqrt{b^2-4 a^2} \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n+(-b) \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n+\sqrt{b^2-4 a^2} \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n\right)}{-2 a^2 \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n+b^2 \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n+b \sqrt{b^2-4 a^2} \left(\frac{\sqrt{b^2-4 a^2}+b}{a}\right)^n-b^2 \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n+2 a^2 \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n+b \sqrt{b^2-4 a^2} \left(-\frac{\sqrt{b^2-4 a^2}-b}{a}\right)^n}$$