I'm asked to approximate $\pi$ with an error of $10^{-3}$ using that $\tan(\pi /4) = 1$ and the Maclaurin series of $\arctan$ that converges for $-1 < x \leq 1$.
$$ \arctan(x) = \sum_{i=1}^\infty (-1)^{i+1} \frac{x^{2i-1}}{2i-1} $$
For the Lagrange form of the remainder, we have that if $\vert f^{(n+1)}(x) \vert \leq M$ in the interval $[0,1]$ then
$$R_n(1) \leq \left\vert \frac{M \; x^{n+1}}{(n+1)!} \right\vert =\left\vert \frac{M }{(n+1)!} \right\vert $$
Where $R_n(x)$ is the remainder or the error. But I have problems findings this $M$, every time I derivate $f(x)=\arctan(x)$ the values of $f^{(n)}(x)$ at the interval $[0,1]$ get bigger and I can't find a general expression for $f^{(n)}(x)$ . Is there any way to find this $M_n$ such that $ \vert f^{(n)}(x)\vert \leq M_n $ ?
I find out using the computer that for an error of $10^{-m}$ I need $10^m$ terms of the sum, there must be a relation in some part.
$\endgroup$ 91 Answer
$\begingroup$You can use the alternating series theorem to bound the error. The error is of the sign of the first neglected term and smaller than it. Since the $501^{st}$ term is $+\frac 1{1001}$ you are there.
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