How to use maclaruin series of arctanx to calculate $\pi/4$ with 0,01 error?

Question

question is how many should we expand the series to calculate $pi/4$?

maclaurin series of arctanx is $\sum _{n=0}^{\infty }\dfrac {\left( -1\right) ^{n}x^{2n+1}} {\left( 2n+1\right) } $ and

$arctan(1)=\sum _{n=0}^{-\infty }\dfrac {\left( -1\right) ^{n}} {2n+1}=\dfrac {\pi } {4} $

since the reminder $R_{n}\left( x\right) =\dfrac {f^{n+1}\left( a\right) \left( x-a\right) ^{n+1}} {\left( n+1\right) !} $

so $R_{n}\left( x\right)=\dfrac {1} {2\left( n+1\right) +1}=\dfrac {1} {2n+3} $

error=$\left| \dfrac {1} {2n+3}\right| \leq 0,01 $

$ 2n+3\geq 100\\ n\geq \dfrac {97} {2}\Rightarrow n\geq 48$

is this correct?

Answer 1

Because of $|f^{(n+1)}(a)|\le 1$: Basically you are right, but it's $n\ge 49$.

Answer 2

here https://en.wikipedia.org/wiki/Approximations_of_%CF%80

pi / 4 = 4 arctan(1/5) - arctan(1/239)