approximation of $\pi$ by $\arctan$

Question

Determinate the order n of the Maclaurin polynomial for $f(x)=4tan^{-1}x$ so that the remainader term $|R_{n}(1)|<0.000005$.
Here $R_{n}(1)=\frac{f^{(n+1)}(c)}{(n+1)!}$ for some c between 0 and 1 How can I estimate $f^{(n+1)}(c)$, it seem that the derivatives of f have no pattern

Answer 1

Since you have the alternating series $$4\sum_{n=0}^\infty\frac{(-1)^n}{2n+1},$$ you can use the Alternating Series Remainder Theorem. The remainder of your series is bounded by $|a_{n+1}|$ where $a_n=\dfrac{(-1)^n}{2n+1}.$

So all you need to do is solve the inequality $$\frac{4}{2n+3}\le\frac{1}{200000},$$ giving $n\ge 200000.$ (Yes it is a VERY slowly converging series!)

Of course, this is the series centered around $x=0.$ It will change (drastically) if you are wanting to center the polynomial around $x=1.$