I am new to Asymptotic analysis so please bear with me and i apologize if the following question is not well formed or is trivial. I am trying to figure out Asymptotic behavior of the following two functions.
$$\phi_1(x) = 1 – \cos\left(1-\cos\left(x\right)\right)$$ and $$\phi_2(x) =x\sin\left(\sin x\right) -\sin^2 x$$
when x is small.
Now i am aware of the following two limits
$$\lim _{x\rightarrow 0}\dfrac {\sin x} {x}=1$$ and $$\lim _{x\rightarrow 0}\dfrac {1-\cos x} {x}=0$$
I suspect that $\phi_1(x)$ and $\phi_2(x)$ are of the fourth and the sixth order respectively, but i am unsure how to use the insight from the limit expressions to show this. Any help would be much appreciated.
Best Answer
I think one way to proceed is to use the Maclaurin expansions of $\sin$ and $\cos$: $\sin x \approx x - \frac{x^3}6$ and $\cos x \approx 1 - \frac{x^2}2$ when $x$ is small. If you plug in these approximate polynomials into your expressions, and then apply the sum-of-arguments formulas, polynomials will come out that will tell you the approximate values.
We would normally start by disregarding the second term of each polynomial, taking $\sin x \approx x$ and $\cos x \approx 1$, but if we do that here we get $\phi_1(x)\approx 0$ and $\phi_2(x)\approx 0$, which is correct, but less accurate than we want, and indicates that we disregarded too much.