Question

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

Answer 1

let `x=1`
`pi/4 = arctan(1) = (1)-(1)^3/3+(1)^5/5-(1)^7/7...`
`pi/4 = 1-1/3+1/5-1/7...`
`pi = 4(1-1/3+1/5-1/7...)`

Answer 2

Using the specified TS approximation we get `~~= 2.90 \ \ \` (3sf)

Explanation:

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

`arctanx ~~ x - x^3/3 + x^5/5 - x^7/7`

Using the well known result:

`tan (pi/4) =1 => arctan1=pi/4`

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

`pi/4 ~~ 1 - 1/3 + 1/5 - 1/7`

`\ \ \ \ = (3*5*7 - 5*7 + 3*7 - 3.5)/(3*5*7)`

`\ \ \ \ = (105 - 35 + 21-15)/(105)`

`\ \ \ \ = (76)/(105)`

Thus:

`pi ~~ 4 * (76)/(105)`

`\ \ \ \ = 304/105`

`\ \ \ \ ~~= 2.90 \ \ \` (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`.