Formula for the pseudofrequency using approximations We know that $w_1=\frac{\sqrt{4w_{0}^2-b^2}}{2}$ and $(1+u)^{\alpha}=1+\alpha u +  \mathrm{O}\left(u^2\right)$
I need to show that $\frac{w_1}{w_0}=1+\frac{\lambda}{N^2}+\mathrm{O}\left(\frac{1}{N^3}\right)$ such that $N=\frac{w_0}{b}$
We have:
$$\frac{w_1}{w_0}=\frac{\sqrt{4w_{0}^2-b^2}}{2w_0}=\sqrt\frac{{4w_{0}^2-b^2}}{4w_{0}^2}= \left(1-\frac{b^2}{4w_{0}^2}\right)^{1/2}= 1-\frac{1}{8N^2}+\mathrm{O}\left(\frac{1}{16N^4}\right) $$
But I am unable to get the expression $\mathrm{O}\left(\frac{1}{N^3}\right)$
What am I doing wrong?
Thank you
 A: You are doing nothing wrong.
Since $1/N^4$ is of smaller order than $1/N^3$,
$1/N^4 = O(1/N^3)$.
More precisely,
if $f(N) = O(1/N^4)$,
then $f(N) = O(1/N^3)$.
In addition, $f(N) = o(1/N^3)$,
a stronger result.
A: Expanding on my previous answer,
you need to understand that
big-o (and little-o) notation
refers to $sets$ of functions.
$O(f(n))$ means the set of all functions
$g$ such that
$g(n)/f(n)$
is bounded.
Similarly,
$h(n)+O(f(n))$
means the set of all functions
$g$ such that
$(g(n)-h(n))/f(n)$
is bounded.
That is why you can argue that,
if $f(n) = a(n) + O(g(n))$
and $g(n)/h(n) \to 0$ or
$g(n)/h(n)$ is bounded,
then $f(n) = a(n) + O(h(n))$.
So $1-\frac{1}{8n^2}+\mathrm{O}(\frac{1}{n^4})$
is the set of functions $g$
such that
$\frac{g-(1-1/(8n^2))}{1/n^4}$
is bounded (you can always ignore a constant
multiplying a big-o function),
and
$1-\frac{1}{8n^2}+\mathrm{O}(\frac{1}{n^3})$
is the set of functions $g$
such that
$\frac{g-(1-1/(8n^2))}{1/n^3}$
is bounded.
Since $1/n^4$ is smaller than
$1/n^3$ (any constants in front of these
do not matter as $n$ gets large),
any function that is
$O(1/n^4)$
is
$O(1/n^3)$.
More generally,
if $a > b$,
any function that is
$O(1/n^a)$
is
$O(1/n^b)$
since $1/n^a$ is smaller than
$1/n^b$.
