Find a closed formula for the following recursive function

Question

How can I express a closed-form formula for the following equation? $$f(n)=f(n-1)+\frac{C}{f(n-1)} $$ Where $C>0$ and $f(0)=\sqrt{C}$.

Answer 1

Your $f(n) = \sqrt{C} g(n)$, where $g(n)$ is obtained by the recursion $$ g(n) = g(n-1) + 1/g(n-1)$$ with $g(0)=1$. The numerators are OEIS sequence A073833, denominators A073834, and $g(n)$ is the fractional chromatic number of the Mycielski graph $M_{n+1}$. There seems to be no closed-form formula, though.

Answer 2

This is the best I can do. It seems like \begin{align} \frac{f(n)-f(n-1)}{n-(n-1)} = \frac{C}{f(n-1)}. \end{align} Let us solve the differential equation \begin{align} f' = \frac{C}{f} \ \ \implies \ \ \frac{d}{dx}(f)^2 = C \end{align} which means $f^2 = Cx$ or $f(x) = \sqrt{C}\sqrt{x}$. Observe \begin{align} \sqrt{C}\sqrt{n}-\sqrt{C}\sqrt{n-1} = \frac{\sqrt{C}}{\sqrt{n}+\sqrt{n-1}} \end{align} then it follows \begin{align} \frac{\sqrt{C}}{2f(n)}=\frac{\sqrt{C}}{2\sqrt{n}}\leq f(n)-f(n-1) \leq \frac{\sqrt{C}}{2\sqrt{n-1}} = \frac{\sqrt{C}}{2f(n-1)} \end{align}