Solving a particular inequation

Question

I don't know the method to solve the following inequation

$$C_1 (1+ \sqrt{a})^{n-1} + C_2(1-\sqrt{a})^{n-1} \geq k$$

where $C_1$, $C_2$, and $a$ are constants. I would like to express a lower bound for $n$ as a function of $k$.

Answer 1

This is not simple. The addition doesn't play nicely with the powers so you will need to find a numeric solution. The best approach will depend upon the constants you have. We set $m=n-1$ to save typing.

The easiest case is when $k$ is rather large, especially where $C_2$ is not much greater than $C_1$. In that case the first term will dominate. You can ignore the second to get a starting value, then iterate. Set $$C_1(1+\sqrt a)^{m_0}=k \\m_0=\frac {\log k - \log C_1}{\log (1+\sqrt a)}\\m_{i+1}=\frac {\log (k-C_2(1-\sqrt a)^{m_i}) - \log C_1}{\log (1+\sqrt a)}$$ and iterate to convergence

Another easy case is when $a$ is small and $m$ is not too large. You can use the binomial expansion then and write $$(1+\sqrt a)^m \approx 1+m\sqrt a\\C_1 (1+ \sqrt{a})^{m} + C_2(1-\sqrt{a})^{m}\approx C_1+C_2 +(C_1-C_2)m\sqrt a$$ and you have a linear equation to solve for $m$. Again, you can use that as a starting value for iteration if it is not accurate enough. The test for accuracy is $m \sqrt a \ll 1$

If $C_2$ is rather large compared with $C_1$ you can have two ranges of $m$. An example is in the plot below where I took $C_1=1, C_2=100, \sqrt a=0.1$ You can see that for $k=40$ you have two regions of solution. The minimum value for the higher region will be found easily by the first approach. The root around $10$ will be harder to find.