What is a/O(a)? I have an algorithm where one parameter is defined as $p = O(a^{2}\ln(b/c))$, where $a,b,c$ are positive numbers.
In the proof of the convergence rate one step goes as follows:
$$16a\sqrt{\frac{\ln(b/c)}{p}}+\frac{1}{16} \leq \frac{1}{4}$$ 
What exactly do i substitute for the value of p to arrive at $\frac{1}{4}$? If I use $p = a^{2}\ln(b/c)$ I get: $16 + \frac{1}{16} \leq \frac{1}{4}$, which is obviously not correct.
Any answer is appreciated!
 A: Here's my analysis. I'm new here, and mathematics is more of a hobby for me. Still, this seems like a good place to start.
We're going for equality, so let's start with
$$16a\sqrt{\frac{\ln(b/c)}{p}}+\frac{1}{16} = \frac{1}{4}$$
subtract $1/16$ from each side to arrive at
$$16a\sqrt{\frac{\ln(b/c)}{p}} = \frac{3}{16}$$
square both sides to obtain the next equality. (Note that I'm not sure about this part of the whole business. You have stated that $a$, $b$, and $c$ are positive integers, but we don't know how $b$ and $c$ relate. For example, is $b < c$, or more importantly is $(b/c) \lt e$ ? Note that, having a background in physics, I've denoted $e$ as the base of $\ln$ , but considering you're working on an algorithm, I imagine your $e = 2$ and not necessarily Euler's constant.)
$$16^2 a^2 \frac{\ln{(b/c)}}{p} = \frac{9}{16^2}$$
Divide and multiply to get $p$ by itself, and you end up with
$$p = \frac{16^4 a^2 \ln{(b/c)}}{9}$$
I realize that it's very probable that this is a naïve answer, with the concerns about a negative radicand that I stated and concerns about $p$ being a function of $a$, $b$, and $c$, but you said

any answer is appreciated

so I figured I'd give it a shot.
Please comment on anything about my answer, no matter how wrong or naïve my thoughts might be. Rip it all to pieces - that will help me learn.
A: From the data you have given us, it is impossible to deduce $$16a\sqrt{\frac{\ln(b/c)}{p}}+\frac{1}{16} \leq \frac{1}{4}$$
For two reasons:  


*

*$p=O(\ldots)$ sets an upper bound on $p$, whereas your inequality reauires a lower bound on $p$ (because it is in the denominator).  

*In any case, the $O(\ldots)$ notation only tells us about the limiting behaviour as $a,b,c$ tend to infinity; whereas your inequality claims to be true for all $a,b,c$.


So you need to tell us the whole story. Or perhaps you have simply misunderstood the algorithm.
