Bound for error term in Taylor expansion of $\arctan x$.

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).

Solution 1:

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.

Solution 2:

The Lagrange formula for the remainder tends to be not useful for getting good estimates of the error.

The Taylor series for $\arctan x$ is most easily obtained by term by term integration of the series for $\frac{1}{1+x^2}$.

Note that if $-1<x<0$ or $0<x<1$, we get an alternating series. The error when you truncate at a certain point is less (in absolute value) than the (absolute value of) the first "neglected" term. That seems to be exactly what you are being asked to show.

The result about alternating series is standard, and found in most calculus books. If you need a proof, one can easily be supplied.

The same is true at $x=\pm 1$, but because of the missing power of $x$, these values should get special treatment.

Solution 3:

Although non-rigorous, the following trick is useful.

Write

$$f'(x)=\dfrac{1}{1+x^2}=\frac{1}{2i}\left(\frac{1}{x-i}-\frac{1}{x+i}\right)$$

We have that $0<f'\leq1$

Then

$$f'(x)=-\frac{1}{2i}\left(\frac{1}{(x-i)^2}-\frac{1}{(x+i)^2}\right)$$

and successively

$$f''(x)=2!\frac{1}{2i}\left(\frac{1}{(x-i)^3}-\frac{1}{(x+i)^3}\right)$$

$$f'''(x)=-3!\frac{1}{2i}\left(\frac{1}{(x-i)^4}-\frac{1}{(x+i)^4}\right)$$

$$\cdots=\cdots$$

$$f^{(n)}(x)=(-1)^n n!\frac{1}{2i}\left(\frac{1}{(x-i)^{n+1}}-\frac{1}{(x+i)^{n+1}}\right)$$

Moreover $$\frac{1}{2i}\left(\frac{1}{(x-i)^{n+1}}-\frac{1}{(x+i)^{n+1}}\right)$$ will give a rational function in $x$ with monic polynomials where the degree in the denominator is $2n$ and in the denominator $n-1$, thus you can bound your derivatives by $0$ and $n!$