I have the following recurrence relation, $$x_{i+1}=a\cdot x_i^{\frac{2-2\alpha}{3}}+x_i,$$ where $a>0, \alpha>0$, and $x_0>0$. My goal is to get an approximate the expression for $x_i$.
I tried to approximate the solution by setting $x_i=f(i)$ to get $$f(i+1)-f(i)=a\cdot f(i)^{\frac{2-2\alpha}{3}},$$ and then treating the difference like a derivative so that $$f'(i)=a\cdot f(i)^{\frac{2-2\alpha}{3}}.$$ Solving this gives me $x_i=(a\cdot i+C)^\frac{3}{2\alpha+1}$ where $C$ is a constant dependent on initial conditions. I think this would be good enough if I could show that the true value of $x_i$ is never lower than what I derived. I see from the original equation how for $\alpha=1$ my result is exact.
I see now that for $\alpha>1$, my approximation is too low which for my purpose is acceptable. I still have to take care of the case when $\alpha <1$.
Clearly, whatever $\alpha$ might be, if $x_0 > 0$ the sequence is increasing.
As a first approximation, if you disregard the troublesome term you have $x_n$ constant. Plug this into the recurrence:
$$ x_{n + 1} - x_n = x_0^{\frac{2 - 2 \alpha}{3}} $$
which gives:
$$ x_n = x_0 + n \cdot x_0^{\frac{2 - 2 \alpha}{3}} $$
We are underestimating $x_n$ here, so this is a lower bound. If not accurate enough, rinse and repeat. It gets ugly, perhaps estimating the sum by an integral is enough.