Finding the upper bound on an expression.

Question

I am trying to find the smallest $N_0$ such that

$$\max_{n\leq N}\frac{1}{4NL}\bigg(\bigg(1+\frac{h}{N}\bigg)^n-1\bigg) \leq 10^{-4}$$ for $N \geq N_0$

I guess that this is equivalent to solving$$ \frac{1}{4NL}\bigg(\bigg(1+\frac{h}{N}\bigg)^N-1\bigg) \leq 10^{-4}$$

I can not seem to get to grips with it though, I have considered using a binomial expansion and writing it as $$\frac{1}{4NL}\sum_{k=1}^NC^N_k\bigg(1+\frac{h}{N}\bigg)^k\leq 10^{-4}$$

again, I am a stumped here.

Some suggestions please? even some improvements on the title would be great, thanks.

Answer 1

An inequality which can be useful in situations such as this one is derived from the Taylor expansion of the natural exponential function about the point $0$. We have $$ 1 + x \leq \exp(x),$$ for all $x \ge 0$.

Using this inequality, we can estimate $$ \left( 1 + \frac{h}{N}\right)^N \leq \left( \exp\left(\frac{h}{N}\right) \right)^N = \exp(h).$$

This will certainly allow you to find an $N_0$, such that $$ \frac{1}{4NL} \left [ \left( 1 + \frac{h}{N}\right)^N - 1 \right] \leq \tau,$$ for all $N \ge N_0$. However, it will not give you the smallest value of $N_0$.

Normally, I would just examine my choices, i.e. $\{1,2,\dotsc,N_0\}$ using either brute force, i.e. one at a time starting from the last, or using bisection.