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I have expanded $\lim_{n\to \infty} \arccos(\frac{n^2-1}{n^2+1})$ to $\arccos(1-\frac{2}{n^2})$ and now i dont know what to do. I wrote the function on walfram alpha and he told me that the result is $\frac{2}{n}+\frac{1}{3n^3}+O((\frac{1}{n})^6)$ you can explain to me how that result came out from where?

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    $\begingroup$ walfram alpha seems like a great guy. $\endgroup$
    – Mathgrad
    Dec 24, 2013 at 12:10
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    $\begingroup$ See this technique. $\endgroup$ Dec 24, 2013 at 13:32
  • $\begingroup$ @MhenniBenghorbal you helped me indirectly, to solve another doubt! thx! :) $\endgroup$
    – malloc
    Dec 24, 2013 at 13:39
  • $\begingroup$ @malloc: you are welcome. $\endgroup$ Dec 24, 2013 at 13:44

2 Answers 2

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Actually, for $x\in [0,+\infty[$,

$$\arccos \frac{1-x^2}{1+x^2}=2\arctan x$$

To see why, write $x=\tan \theta$ for $\theta \in[0,\pi/2[$, then $\theta=\arctan x$ and

$$\frac{1-x^2}{1+x^2}=\frac{1-\tan^2 \theta}{1+\tan^2 \theta}=\frac{\cos^2\theta -\sin^2\theta}{\cos^2\theta +\sin^2\theta}=\cos 2\theta$$

And since $2\theta \in [0,\pi[$, $\arccos(\cos2\theta)=2\theta$

Now, for $|x|<1$

$$\arctan x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}$$

And (don't forget to rename the sum variable)

$$\arccos \frac{n^2-1}{n^2+1}=\arccos \frac{1-1/n^2}{1+1/n^2}=2\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)n^{2k+1}}$$

By the way, since the LHS in the first equation is even, and the RHS is odd, you get, for any $x \in \Bbb R$,

$$\arccos \frac{1-x^2}{1+x^2}=2|\arctan x|$$

One way to "guess" the result in the first place, is to remember the half-angle formula: if $x=\tan (\theta/2)$, then $\cos \theta = \frac{1-x^2}{1+x^2}$.


Edit, for malloc

Here is a slightly different proof.

In the above, I apply the development of $\arctan$ on $]-1,1[$, and I let $x=1/n$.

But, still using

$$\arccos \frac{1-x^2}{1+x^2}=2\arctan x$$

And with $x=n$, we can also use the development of $\arctan$ at $\infty$. For this, you will need, for $x>0$:

$$\arctan x + \arctan \frac 1 x = \frac {\pi}2$$

Thus if $n=x >1$ (so that the development of $\arctan \frac 1x$ holds),

$$\arctan n = \frac {\pi}2-\arctan \frac 1 n = \frac{\pi}{2}-\sum_{k=0}^\infty (-1)^k \frac{(\frac{1}{n})^{2k+1}}{2k+1}$$

And finally (since $\arccos (-x)=\pi-\arccos(x)$),

$$\arccos \frac{n^2-1}{n^2+1}=\pi-\arccos \frac{1-n^2}{1+n^2}$$ $$=\pi-2\arctan n=\pi-2\left(\frac{\pi}{2}-\sum_{k=0}^\infty (-1)^k \frac{(\frac{1}{n})^{2k+1}}{2k+1}\right)$$

$$\arccos \frac{n^2-1}{n^2+1}=2\sum_{k=0}^\infty (-1)^k \frac{(\frac{1}{n})^{2k+1}}{2k+1}$$

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  • $\begingroup$ but the expantion of $2arctg(x)$ is $\pi-\frac{2}{n}+\frac{2}{3 n^3}-\frac{2}{5 n^5}+O((\frac{1}{n})^7)$ $\endgroup$
    – malloc
    Dec 24, 2013 at 13:22
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    $\begingroup$ @malloc Definitely not. $\arctan 0=0$, to begin with. Be careful with $\arcsin x + \arccos x = \pi/2$, maybe it's what hurts you here. $\endgroup$ Dec 24, 2013 at 13:24
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    $\begingroup$ I asked $2arctan(x)$, I checked! $\endgroup$
    – malloc
    Dec 24, 2013 at 13:44
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    $\begingroup$ To infinity. Of course it's not the same expansion! The one I give is at 0 (notice I wrote $|x|<1$). Now to get the expansion at infinity, just use $\arctan (x) + \arctan (1/x) = \mathrm{sign}(x) \pi/2$. Hence the $\pi$ lying in the result! $\endgroup$ Dec 24, 2013 at 13:54
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    $\begingroup$ @malloc Done... $\endgroup$ Dec 24, 2013 at 14:14
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$\lim_{n\to \infty}= arccos{\frac{n^2-1}{n^2+1}}=arccos\left({1-\frac{2}{n^2+1}}\right)$

So as $n\to\infty$ this goes to $arccos(1)$ which is equal to $0$.

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  • $\begingroup$ I know this, but i must expand it with maclaurin $\endgroup$
    – malloc
    Dec 24, 2013 at 12:22
  • $\begingroup$ It's just the taylor expansion of this function at $x=0$. You can use the taylor expansion formulae at $x=0$. It's here :mathworld.wolfram.com/MaclaurinSeries.html. It is tiresome but not hard. $\endgroup$
    – Mathgrad
    Dec 24, 2013 at 12:24

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