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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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    @martycohen, i didnt get what u meant,sorry..2011-10-19

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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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    @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 $10^{-4}$.2011-10-19