Formula for the pseudofrequency using approximations

Question

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

Answer 1

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.

Answer 2

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 $1/n^b$.