I have the function $ F(x) $ where $ x >> a $ and I have derived two linear approximations of $ F(x) $:
- $ L_{1}(x) $, where I take that $ \frac{\left(a^{2}-2xa\right)}{x^{2}} $ is way way less than 1 (and therefore, $ (1+x)^{m} \approx 1+mx $ )
- $ L_{2}(x) $, where I say that $ a^2 $ is negligible as is very small compared to $ x^2 $ and after discarding it I take the binomial approximation in the same way I did in $ L_{1} $
I'm quite surprised that discarding $ a^2 $ and doing the approximation leaves me with a better function (when x is very big) than if I don't do it. Is there any explanation for it?
$$ F\left(x\right)=\left(1+\frac{\left(a^{2}-2xa\right)}{x^{2}}\right)^{\left(\frac{1}{2}\right)}\left\{x>0\right\} \\ L_1\left(x\right)=\left(1+\frac{1}{2}\frac{\left(a^{2}-2xa\right)}{x^{2}}\right)\left\{x>0\right\} \\ L_2\left(x\right)=\left(1+\frac{1}{2}\frac{\left(-2xa\right)}{x^{2}}\right)\left\{x>0\right\} $$
$F(x) = |\frac{x-a}{x}|$ and $L_2(x) = \frac{x-a}{x}$, so $L_2$ is not an approximation of $F$, it is equal to $F$ for $x \geq a$.
Basically what you have done in $L_2$ gives you the equality thanks to the simplification of the square with the square root. Indeed if instead you use for example $$ F'\left(x\right)=\left(1+\frac{\left(a^{2}-2xa\right)}{x^{2}}\right)^{\color{red}{3}} $$ where there is no simplification involved, you will see that $L'_1$ is better than $L'_2$ for large $x$.