# How to, using Taylor series approximation , estimate the value of π, when arctan(x) ≈ x-x3/3+x5/5-x7/7 ?

Mar 28, 2018

let $x = 1$
$\frac{\pi}{4} = \arctan \left(1\right) = \left(1\right) - {\left(1\right)}^{3} / 3 + {\left(1\right)}^{5} / 5 - {\left(1\right)}^{7} / 7. . .$
$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7.} . .$
$\pi = 4 \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7.} . .\right)$

Apr 7, 2018

Using the specified TS approximation we get $\approx = 2.90 \setminus \setminus \setminus$ (3sf)

#### Explanation:

Using the given Taylor Series approximation (truncation) we have:

$\arctan x \approx x - {x}^{3} / 3 + {x}^{5} / 5 - {x}^{7} / 7$

Using the well known result:

$\tan \left(\frac{\pi}{4}\right) = 1 \implies \arctan 1 = \frac{\pi}{4}$

So, substituting $x = 1$ into the given TS we have:

$\frac{\pi}{4} \approx 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7}$

$\setminus \setminus \setminus \setminus = \frac{3 \cdot 5 \cdot 7 - 5 \cdot 7 + 3 \cdot 7 - 3.5}{3 \cdot 5 \cdot 7}$

$\setminus \setminus \setminus \setminus = \frac{105 - 35 + 21 - 15}{105}$

$\setminus \setminus \setminus \setminus = \frac{76}{105}$

Thus:

$\pi \approx 4 \cdot \frac{76}{105}$

$\setminus \setminus \setminus \setminus = \frac{304}{105}$

$\setminus \setminus \setminus \setminus \approx = 2.90 \setminus \setminus \setminus$ (3sf)

A (not so) interesting fact is that using this particular Taylor Series and method to approximate $\pi$ tp 3 significant figures (or 2 decimal places) requires $147$ terms of the sequence, as is therefore a particularly inefficient method to estimate $\pi$.