I'm trying to solve the following problem from Apostol, Calculus, Volume I (p. 284) and could use some help:
Prove:
$$\arctan x = \sum_{k=0}^{n-1} \dfrac{(-1)^k x^{2k+1}}{2k+1} + E_{2n} (x), \qquad |E_{2n} (x)| \leq \dfrac{x^{2n+1}}{2n+1} \quad \text{if } 0 \leq x \leq 1.$$
Getting the Taylor expansion of $\arctan x$ is straightforward (and is worked out as an example in the text). My question: how do I prove the requested bound on the error term?
This exercise is in the section immediately following the proof and examples of this Theorem (and some slight variations) for estimating the error term which states that:
If the $(n+1)$st derivative of $f$ satisfies
$$ m \leq f^{(n+1)}(t) \leq M$$
then
$$ m \dfrac{(x-a)^{n+1}}{(n+1)!} \leq E_n (x) \leq M \dfrac{(x-a)^{n+1}}{(n+1)!} \quad \text{if } x > a,$$
and
$$ m \dfrac{(a-x)^{n+1}}{(n+1)!} \leq (-1)^{n+1} E_n (x) \leq M \dfrac{(a-x)^{n+1}}{(n+1)!} \quad \text{if } x < a.$$
In order to apply this I need to get bounds $m$ and $M$ on the $(n+1)st$ derivative of $\arctan x$, which I cannot seem to arrive at.
Updated :Taking successive derivatives, I find:
$$\begin{align*}
f'(x) &= \dfrac{1}{1+x^2}\\
f''(x) &= \dfrac{-2x}{(1+x^2)^2}\\
f^{(3)} (x) &= \dfrac{6 (x^2 – 1/3)}{(1+x^2)^3}\\
f^{(4)} (x) &= \dfrac{-24 x (x^2 – 1)}{(1+x^2)^4}\\
f^{(5)} (x) &= \dfrac{120 (x^4 – 2x^2 + 1/5)}{(1+x^2)^5}
\end{align*}$$
I should add, from the formula for $|E_{2n} (x)|$ that we are trying to prove it seems the bound $M$ must be $(2n)!$.
So, it seems clear the $n!$ term I want is there, but I can't seem to figure out how to explicitly bound this. It seems that it should be rather obvious from here, but I'm having trouble.
Any help is appreciated (hints or full solutions are equally welcome).
Best Answer
We know
$$\begin{align*} \arctan (x) & = \int_0^x \dfrac{1}{1+t^2} dt \\ & = \int_0^x \left[1 - t^2 + t^4 - \cdots + (-1)^{n-1} t^{2(n-1)} + \dfrac{(-1)^{n} t^{2n}}{1+t^2} \right] dt\\ & = \sum_{k = 0}^{n-1} \dfrac{(-1)^k x^{2k+1}}{2k+1} + (-1)^{n}\int_0^x \dfrac{t^{2n}}{1+t^2} dt\\ & = \sum_{k = 0}^{n-1} \dfrac{(-1)^k x^{2k+1}}{2k+1} + E_{2n}(x) \end{align*}$$
Hence, we have $E_{2n} (x) = \displaystyle{(-1)^{n}\int_0^x \dfrac{t^{2n}}{1+t^2} dt}$ so we are just trying to bound the integral. So, we have
$$\begin{align*} |E_{2n} (x)| & = \left|(-1)^{n}\int_0^x \dfrac{t^{2n}}{1+t^2} dt\right|\\ & = \int_0^x \dfrac{t^{2n}}{1+t^2} dt \qquad (\text{the integrand } \geq 0 \text{ for } t \in [0,1])\\ & \leq \int_0^x t^{2n} dt \qquad \qquad \left(t \in [0,1] \implies t^{2n} \geq \dfrac{t^{2n}}{1+t^2}\right)\\ & = \dfrac{x^{2n+1}}{2n+1} \end{align*}$$
Which is the bound we were looking for.