# How to approximate the value of $\arctan(x)$ for $x> 1$ using Maclaurin's series?

The expansion of $f(x) = \arctan(x)$ at $x=0$ seems to have interval of convergence $[-1, 1]$ $$\arctan(x) = x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}+\mathcal{O}\left(x^{13}\right)$$ Does it mean that I cannot approximate $\arctan(2)$ using this series? Also I'm getting radius of convergence $|x| < 1$ using ratio test. How do I get $|x| \leqslant 1$?

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A bit of notation: I've always thought it's $O(x^{13})$ rather than $O[x]^{13}$; with round parenthesis and the power of $x$ inside the parenthesis. –  user2468 Aug 26 '12 at 4:07
Oo... sorry. I copy pasted directly from Mathematics :D –  Monkey D. Luffy Aug 26 '12 at 4:08
About the region of convergence: (-1,1) comes from the ratio test, x=1 comes from the alternating series test and x=-1 is excluded by comparison to the Harmonic series. –  Ragib Zaman Aug 26 '12 at 4:55

Use, for $x>0$ $$\arctan(x) + \arctan\left(\frac{1}{x}\right) = \frac{\pi}{2}$$ For $x<0$ use the parity.

Added: Use $$\tan\left(\frac{\pi}{2} - \phi\right) = \frac{\sin\left(\frac{\pi}{2}-\phi\right)}{\cos\left(\frac{\pi}{2}-\phi\right) } = \frac{\cos(\phi)}{\sin(\phi)} = \frac{1}{\tan(\phi)}$$ Thus, for $x = \tan(\phi)$: $$\tan\left(\frac{\pi}{2} - \phi\right) = \frac{1}{x}, \quad \frac{\pi}{2} - \phi = \arctan\left(\frac{1}{x}\right), \quad \frac{\pi}{2} - \arctan(x) = \arctan\left(\frac{1}{x}\right)$$

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could you elaborate a bit please? –  Monkey D. Luffy Aug 26 '12 at 4:01
thanks ... I think i get that ... very clever method. could you help me with ratio test. I'm getting only $<$ using ratio. I am not getting $\leq$ –  Monkey D. Luffy Aug 26 '12 at 4:05
For the equality, you need Abel's convergence theorem. –  Potato Aug 26 '12 at 4:08

If you want to approximate the value of $\arctan(2)$, then you can expand $\arctan(x)$ at a point close to $2$. For example you can have an expansion at the point $1$, but if you approximate at a point that is closer to 2 is better. Here is the Taylor series at the point $x=1$

$$\arctan(x) = \frac{\pi}{4} +{\frac {1}{2}} \left( x-1 \right) -{\frac {1}{4}} \left( x- 1 \right) ^{2}+{\frac {1}{12}} \left( x-1 \right) ^{3}+O \left( \left( x-1 \right) ^{4} \right)\,.$$

Another expansion at the point $x=\frac{3}{2}$ is given by,

$$\arctan(x) = \arctan \left( {\frac {3}{2}} \right) + {\frac {4}{13}} \left( x-{ \frac {3}{2}} \right) -{\frac {24}{169}} \left( x-{\frac {3}{2}} \right) ^{2}+{\frac {368}{6591}} \left( x-{\frac {3}{2}} \right) ^{3} +O \left( \left( x-{\frac {3}{2}} \right) ^{4} \right)$$ It is easier, for a hand calculations, to derive the first series. Substituting $x=1$, of course you can use the Abel's theorem to assure the convergence, you get, $$\arctan(x) = \arctan(2) \approx \frac{\pi}{4}+\frac{1}{3}$$ As I said, if you use the second series with the same number of terms, you will have a smaller error.

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I meant for all cases $x>1$ ... still thanks!! –  Monkey D. Luffy Aug 26 '12 at 4:23
@MonkeyD.Luffy: No, because the problem is with radius of convergence of the series. The first series is good for $|x-1|<1$ and the second series is good for $|x-\frac{3}{2}|<1$. The boundary points can be included if they obey some conditions. See Abel's theorem. –  Mhenni Benghorbal Aug 26 '12 at 4:59