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How to find the value of $\tan^{-1}x$ where $x$ from $0$ to $2\pi$ in form of fraction instead of decimal value?

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Are you sure you're reading the question correctly? The inverse tangent function gives angles as output, not input. – Austin Mohr Oct 19 '11 at 4:08
Also, how would you compute $\tan^{-1}x$ if $x$ is given in decimal? Is there any reason you can't convert the fraction to decimal? – marty cohen Oct 19 '11 at 4:41
@AustinMohr, the question is actually $\int_0^{2\Pi} \! \frac{1}{1+x^{2}}\, \mathrm{d} x$, it meant, I have to find that value right? – DRN Oct 19 '11 at 5:40
@martycohen, i didnt get what u meant,sorry.. – DRN Oct 19 '11 at 5:44
up vote 1 down vote accepted

if you are denoting by $\tan^{-1} x$ the usual $\arctan$ function. You can use for $x\in (-1,1)$ its taylor series representation, which is

$$ \arctan(x)= \sum_{k=0}^{\infty}(-1)^k \frac{x^{1+2k}}{1+2 k} = x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots $$ Note that, strictly speaking in general you will not get a "fraction" for $\arctan(x)$. But for the other hand,
given an $\varepsilon$ error tolerance you can always build using this formula a rational number $\varepsilon$ close to $\arctan(x)$ for any fixed $x\in (-1,1)$.

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The OP might consider the continued fraction expansion for $\arctan z$ as well. As $\arctan z = z F(\frac{1}{2},1; \frac{3}{2};z)$, one may apply the Gauss Hypergeometric continued fraction. Namely, we have $\arctan z = z/(1+z/(3+4z^2/(5+9z^2/(7+16z^2/\ldots$. – A Walker Oct 19 '11 at 5:13
@AWalker: Right; the beauty of this is that the CF has a wider domain of applicability than the series... – J. M. Oct 19 '11 at 10:40
i still did not get the idea.. how to use that series in form of 0 to $2\Pi$ – DRN Oct 19 '11 at 14:04
@Norlyda Just like the Taylor series, this continued fraction expansion has partial estimations that converge to $\arctan z$. For example, $z=1$ gives the estimates $1, \frac{3}{4}, \frac{19}{24}, \frac{40}{51}, \frac{436}{555}, \frac{161}{205}\ldots$. The latter approximates $\pi/4$ with an error of roughly $10^{-4}$. – A Walker Oct 19 '11 at 16:37

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