Good morning everyone, I encountered a calculation procedure (structural dimensioning) that proceeds by trial and error in a recursive manner, which converges to a value. From various steps I arrived at the following nested infinite radical which unfortunately I cannot resolve further with my skills.
$$x_{1} = \frac{a}{1+k \sqrt{a}}$$ $$x_{2} = \frac{a}{1+k \sqrt{x_{1}}}=\frac{a}{1+k \sqrt{\frac{a}{1+k \sqrt{a}}}}$$ $$x_{3} = \frac{a}{1+k \sqrt{x_{2}}}=\frac{a}{1+k \sqrt{\frac{a}{1+k \sqrt{x_{1}}}}}=\frac{a}{1+k \sqrt{\frac{a}{1+k \sqrt{\frac{a}{1+k \sqrt{a}}}}}}$$ $$...$$ $$ k, a \in \Re \qquad k > 0 \qquad a >0$$
What I would like to know is:
$$ \lim_{n\to +\infty}{x_n} = ? $$
Since I don't need an exact result, which may be too complicated to quantify, I would also like a formula with an approximation of $$ \pm 1‰$$
Thank you
The recurrence relation is $x_{n+1} = \frac{a}{1+k \sqrt{x_n}}$
If we assume a fixed point $L$ exists, it will satisfy $L = \frac{a}{1+k \sqrt L}$:$$ \begin{align} L+kL\sqrt{L} &= a\\ k (\sqrt L)^3 + (\sqrt L)^2 -a & = 0 \end{align} $$
This is a cubic in terms of $\sqrt L$. This is a special case of cubic (no linear term). Because the constant term is negative, this will have exactly one positive root. Since we're solving for $\sqrt L > 0$, that's the solution we want. I had Mathematica evaluate it for me (you can read up on the cubic formula if you're interested); then I simplified it a bit by hand: $$ \begin{align} \sqrt L &= \frac{\sqrt[3]{27 a k^2+\sqrt{\left(2-27 a k^2\right)^2-4}-2}}{3 \sqrt[3]{2} k}+\frac{\sqrt[3]{2}}{3 k \sqrt[3]{27 a k^2+\sqrt{\left(2-27 a k^2\right)^2-4}-2}}-\frac{1}{3 k} \\ A &= ak^2 > 0 \\ \sqrt L &= \frac{1}{3k} \left( \frac{\sqrt[3]{27A+\sqrt{\left(2-27 A\right)^2-4}-2}}{\sqrt[3]{2} }+\frac{\sqrt[3]{2}}{\sqrt[3]{27 A+\sqrt{\left(2-27 A\right)^2-4}-2}}-1 \right) \\ B &= 27A-2 = 27 ak^2 - 2 \\ \sqrt L &= \frac{1}{3k} \left( \frac{\sqrt[3]{B+\sqrt{B^2-4}}}{\sqrt[3]{2} }+\frac{\sqrt[3]{2}}{\sqrt[3]{B+\sqrt{B^2-4}}}-1 \right) \end{align} $$ If $|B|<2$, $B^2-4<0$ so $\sqrt{B^2-4}$ doesn't work/produces a complex result. That can happen if $27 ak^2 < 4$. There's still a positive root to the equation in that case, but this formula Mathematica produced is not directly applicable in that case. You can read more details here