Approximate arctan(1/2) to within 1/33

Question

So I found the derivative of arctan(x) to be 1 / (1+x^2)

I can't use a calculator, but I want to find arctan(1/2) to within 1/33.

I was thinking about the alternating Taylor Series for arctan (which I forgot how to do).

Also how would I go about approximating after the series?

Answer 1

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Note that $\ds{\arctan\pars{\root{3} \over 3} = {\pi \over 6} \approx 0.5236}$ and $\ds{\verts{{1/2 \over \pi/6} - 1} \sim 10^{-2}}$. Then, it will be useful to expand $\ds{\arctan\pars{x}}$ 'around' $\ds{x = {\root{3} \over 3}}$.

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For example, up to the second order of $\ds{\pars{\half - {\root{3} \over 3}}}$: \begin{align} \color{#f00}{\arctan\pars{\half}} & = \arctan\pars{{\root{3} \over 3} + \bracks{\half - {\root{3} \over 3}}} \\[1cm] & \approx \arctan\pars{\root{3} \over 3} + \arctan'\pars{\root{3} \over 3}\pars{\half - {\root{3} \over 3}} \\[5mm] & + \half\,\arctan''\pars{\root{3} \over 3}\pars{\half - {\root{3} \over 3}}^{2} \\[1cm] & = {\pi \over 6} + {3 \over 4}\pars{\half - {\root{3} \over 3}} - {3\root{3} \over 16}\pars{\half - {\root{3} \over 3}}^{2} = \color{#f00}{{\pi \over 6} - \pars{{23 \over 64}\,\root{3}- {9 \over 16}}} \equiv \xi \\[5mm] & \approx \color{#f00}{0.4636} \end{align} with corrections of order $\ds{10^{-4}}$.

Note that $\ds{\tan\pars{\xi} \approx 0.49999\color{#f00}{4259547465}}$.

Answer 2

Here is a discussion about the Taylor series of $arctan(x)$. Plug in $\frac1 2$ for $x$.

This answer is also already discussed here.

Answer 3

If you have a look here, the Taylor series you are looking for is $$\tan^{-1}( x) = \sum^{\infty}_{n=0} \frac{(-1)^n}{2n+1} x^{2n+1}\quad\text{ for }|x| \le 1\!$$ and so you want to know how many terms $m$ are required such that $$\Big|\tan^{-1}(\frac 12) -\sum^{m}_{n=0} \frac{(-1)^n}{(2n+1) 2^{2n+1}}\Big| \leq \epsilon$$

The series being alternating and and decreasing, so the $m^{\text{th}}$ remainder is such that $$\frac{1}{(2m+3) 2^{2m+3}}\leq \epsilon\implies {(2m+3) 2^{2m+3}}\geq \frac 1\epsilon$$ Since the lhs varies quite fast with $m$, just using inspection, you should be able to quickly find the smallest value of $m$ which satisfies the last inequality.