-1
$\begingroup$

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

  • 3
    Are you sure you're reading the question correctly? The inverse tangent function gives angles as *output*, not input.2011-10-19
  • 0
    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?2011-10-19
  • 0
    @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?2011-10-19
  • 0
    @martycohen, i didnt get what u meant,sorry..2011-10-19

1 Answers 1

1

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 $x\in (-1,1)$.

  • 1
    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 $\arctan z = z/(1+z/(3+4z^2/(5+9z^2/(7+16z^2/\ldots$.2011-10-19
  • 0
    @AWalker: Right; the beauty of this is that the CF has a wider domain of applicability than the series...2011-10-19
  • 0
    i still did not get the idea.. how to use that series in form of 0 to $2\Pi$2011-10-19
  • 0
    @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