What does the $O$ mean in "$O(v^4)$"?

Question

I am currently studying General Relativity and I am using the book by Schutz.

I encountered a problem when I was reading about the gravitational redshift experiment. Here is the website (it is not the book, but it depicts the exact same experiment):gravitational redshift experiment

I cannot figure out the meaning of the $O(v^4)$ term in the equation of the total energy of the particle when it reaches the ground, as measured by the observer on ground:

$$ E=mc^2+\frac{1}{2}mv^2+O(v^4)=mc^2+mgh+O(v^4)$$

I don’t know what $O(v^4)$ is supposed to mean. What is the context of it? I’ve done some searches on Google but I can’t seem to find any explanations. Any help would be so great!

Answer 1

$O(x^n)$ stands for big O notation. It means that you wrote all terms that are proportional to $x$ up to the ones that are proportional to $x^{n-1}$ and that there are some more terms proportional to $x$ to a higher power $x^n,x^{n+1},\cdots$ that are not written explicitly. It usually indicates that higher terms are negligible.

In the formula that you have written above, there is no term $v^3$. The additional terms are proportional to $v^4$ or higher powers of $v$.