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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.

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@MarkDominus Thanks so much i had not thought of that, but ofcourse that makes so much sense. –  Hardy Jun 4 '12 at 18:09
    
@MarkDominus, you would have to prove convergence of the MacLaurin series, which involves exactly these limits. It gives a hint though. –  akkkk Jun 4 '12 at 18:09
    
The question is not to prove the two limits, so that is not a problem with using that approach for this problem. –  MJD Jun 4 '12 at 18:10
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1 Answer

up vote 1 down vote accepted

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.

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@Hardy: Did this work? I didn't actually try it myself. –  MJD Jun 4 '12 at 18:24
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