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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3walfram alpha seems like a great guy. – Mathgrad Dec 24 '13 at 12:10
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1See this technique. – Mhenni Benghorbal Dec 24 '13 at 13:32
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@MhenniBenghorbal you helped me indirectly, to solve another doubt! thx! :) – malloc Dec 24 '13 at 13:39
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@malloc: you are welcome. – Mhenni Benghorbal Dec 24 '13 at 13:44
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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 $\arccos (-x)=\pi-\arccos(x)$),
$$\arccos \frac{n^2-1}{n^2+1}=\pi-\arccos \frac{1-n^2}{1+n^2}$$ $$=\pi-2\arctan n=\pi-2\left(\frac{\pi}{2}-\sum_{k=0}^\infty (-1)^k \frac{(\frac{1}{n})^{2k+1}}{2k+1}\right)$$
$$\arccos \frac{n^2-1}{n^2+1}=2\sum_{k=0}^\infty (-1)^k \frac{(\frac{1}{n})^{2k+1}}{2k+1}$$
Jean-Claude Arbaut
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but the expantion of $2arctg(x)$ is $\pi-\frac{2}{n}+\frac{2}{3 n^3}-\frac{2}{5 n^5}+O((\frac{1}{n})^7)$ – malloc Dec 24 '13 at 13:22
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1@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. – Jean-Claude Arbaut Dec 24 '13 at 13:24
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finally i understand.. only one thing why wolfram alpha put the $\pi$, in the expansion? – malloc Dec 24 '13 at 13:31
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@malloc You probably asked by mistake $\arccos ((1-n^2)/(1+n^2))$ to WolframAlpha, while it's really $\arccos ((n^2-1)/(n^2+1))$. And since $\arccos (-x)=\pi-\arccos (x)$... – Jean-Claude Arbaut Dec 24 '13 at 13:38
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@malloc But I get $2 x-(2 x^3)/3+(2 x^5)/5+O(x^6)$ (see here) – Jean-Claude Arbaut Dec 24 '13 at 13:48
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you must see the expansion for x to infinity http://www.wolframalpha.com/input/?i=lim+x-%3E+infinity+2arctan%28x%29 – malloc Dec 24 '13 at 13:52
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1To 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! – Jean-Claude Arbaut Dec 24 '13 at 13:54
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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 $0$.
Mathgrad
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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 :http://mathworld.wolfram.com/MaclaurinSeries.html. It is tiresome but not hard. – Mathgrad Dec 24 '13 at 12:24