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

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!

$$O(x^n)$$ 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 $$v$$.