Find arctan without calculator

Question

Does there exist any formula to calculate the value of $\arctan$ without a calculator? Suppose I'm writing an exam and I have to find the value of $\arctan(0.52)$, how can I go about it? This can be an approximation formula like Bhaskara I's sine approximation formula $\displaystyle \sin x^{\circ} \approx \frac{4x(180-x)}{40500-x(180-x)}$.

Any alternative methods are also appreciated.

Answer 1

As @Travis Willse answered, we just need to consider the case where $0\leq x \leq 1$.

For sure, we can use the Taylor series but we can do better using the $[2n+1,2n]$ Padé approximant $P_n$.

This write $$\tan^{-1}(x)=x\,\frac {1+\sum_{k=1}^{n}a_k\,x^{2k} } {1+\sum_{k=1}^{n}b_k\,x^{2k} }+k_n \,O(x^{2n+1})$$

I wrote in the table the first coefficients and the error $$\left( \begin{array}{ccc} n & \text{coefficient } a_k& \text{coefficient } b_k & \text{error coefficient}\\ 1 & \left\{\frac{4}{15}\right\} & \left\{\frac{3}{5}\right\} & \frac 4 {175}\\ 2 & \left\{\frac{7}{9},\frac{64}{945}\right\} & \left\{\frac{10}{9},\frac{5}{21}\right\} & \frac{64}{43659}\\ 3 & \left\{\frac{50}{39},\frac{283}{715},\frac{256}{15015}\right\} & \left\{\frac{21}{13},\frac{105}{143},\frac{35}{429}\right\}& \frac{256}{2760615} \\ \end{array} \right)$$

Using those for $x=0.52$ we should obtain $$\left\{\frac{130663}{272400},\frac{1457959438}{3040456875}, \frac{315243387069}{657415438750}\right\}$$ which are in an absolute error of $1.54\times 10^{-4}$, $5.62\times 10^{-7}$ and $2.03\times 10^{-9}$.

Another way to look at the accuracy is to consider the norm $$\Phi_n=\int_0^1 \left(\tan^{-1}(x)-P_n \right)^2\,dx$$ They are $3.46\times 10^{-6}$ , $2.06\times 10^{-9}$ and $1.36\times 10^{-12}$

For shortcut calculations in an exam, the first one is sufficient and rewriting it $$P_1=\frac{x \left(4 x^2+15\right)}{9 x^2+15}$$

Edit

In the linked paper kindly provided by @Travis Willse, equation $(11)$ $$\tan^{-1}(x) \simeq \frac x{1+\frac 9{32}x^2}$$ is particularly good and more than simple : maximum error equal to $0.005$ and infinite norm equal to $7.61\times 10^{-6}$.

Answer 2

Which approximation is best depends on the value of the argument and the precision you require. For $x \approx 0$, the first- or third-order Taylor polynomial based at $0$ is a good approximation: $$T_1(x) = x, \qquad T_3(x) = x - \frac{1}{3} x^3.$$ For example, $T_3(0.52) = 0.47951\ldots$ but $\arctan 0.52 = 0.47313\ldots$. Indeed, Taylor's Theorem implies that $T_4(0.52) = T_3(0.52)$ is within $$\frac{24}{5!}(0.52)^5 < 7.761 \cdot 10^{-3}$$ of $\arctan 0.52$.

On the other hand, a consequence of the formula for the linear approximation $T_1$ is that for large $|x|$, $\arctan \frac{1}{x} \approx x$ and so $$\arctan x = \pm \frac{\pi}{2} - \arctan \frac{1}{x} \approx \pm\frac{\pi}{2} - \frac{1}{x}.$$

Answer 3

If you are working in radians (as you really should...) there is a well known formula.

$\arctan x = \sum_\limits{n=0}^\infty \frac {x^{2n+1}}{2n+1}$

This is a formula taught in most calculus classes, but as a precalc student, you are not expected to know this yet. I should also note that it only converges if $|x|\le 1.$ However, there are tricks and workarounds such that you can work with an x in the appropriate inteval.

However, for $\arctan 0.52,$ since $\frac {\sqrt 3}{3}\approx 0.577$, $\arctan 0.52 < \frac {\pi}{3}$ (or $30^\circ$) but only slighty less.