How to linearize the equation

Question

I have an equation that I read from a book as $$A = R^2 \arcsin{\left(\frac{a}{R}\right)} - a\left(R-b\right) (1)$$ And it says: compared with R, b is very small, thus, the equation above can be further linearized about $$\frac{a}{R}$$ as: $$A = ab \left(1 + O\left(\frac{b}{R}\right)\right) (2)$$ I don't know the meaning of the sign if it is a function with its argument in parentheses $$O\left(\frac{b}{R}\right)$$ as well as how to convert the equation (1) to (2). Can you help me understand the problem we are facing with the project? Thank you very much!

Answer 1

$$A=R^2 \sin ^{-1}\left(\frac{a}{R}\right)-a (R-b) \quad\text{with}\quad a = \sqrt{R^2 - (R-b)^2}$$ Let $R=\frac b x$ and write $$A=\frac b {x^2}\left((x-1) x \sqrt{-\frac{b^2 (x-2)}{x}}+b \sin ^{-1}\left(\frac{x \sqrt{-\frac{b^2 (x-2)}{x}}}{b}\right)\right)$$ Now, using Taylor series for small $x$ , assuming $x>0$ and $b>0$ $$A=\frac{4 \sqrt{2}\, b^2}{3 \sqrt{x}}-\frac{\sqrt{2}}{5} b^2 \sqrt{x}+O\left(x^{3/2}\right)$$ Back to $x=\frac b R$ $$A=\frac{4\sqrt{2R}}{3} b^{3/2} \left(1-\frac{3 b}{20 R}+O\left(\frac {b^2}{R^2}\right) \right)=\frac{4\sqrt{2R}}{3} b^{3/2}+O\left(\frac {b}{R}\right) $$

Using $b=1$ and $R=10$ would give $A=\frac{8 \sqrt{5}}{3}=5.96285$ to be compared with the exact value $5.87259$; this corresponds to a relative error of $1.5$%.

Answer 2

The $O$ is for what's called the big $O$ notation, part of the family of the Landau notations.
$f(x) = \underset{x \to a}{O}(g(x))$ means that there exists $M$ such that for all $x$ close enough to $a$, $|f(x)| \leq M |g(x)|$ (with $a$ being able to be $\pm\infty$). $O$, $o$ and $\sim$ are fairly widely used notations, so do not hesitate to learn how to use them if you're interested.

As for the linearisation of the equation, I'm guessing they used Taylor's theorem on $\arcsin$ at $0$ (so for big $R$, or small $a$). Since they only wrote $O$ without anything under it I'm having a hard time seeing if they meant to make $b$ tend to $0$ or $R$ to infinity, but essentially the $1$ is from $-a(...-b)$, and the $O$ is the rest.