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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).

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Why don't you try a proof of the bound $[0,n!]$ with induction on $n$ for the degree of the polynomials in $$\frac{d^n}{dx^n}\arctan x$$? – Pedro Tamaroff Apr 4 '12 at 13:04 Can you elaborate a bit? I'm not sure how I could set up such a proof. To make the inductive step I would need to be able to say something about the $n$th derivative, but I can't seem to see how to do that. – user23784 Apr 4 '12 at 20:39
You have shown all polynomials have a leading term $(-1)^n n!$ for the first few cases. Since derivation by the quotient rules affects the degree of both numerator and denominator in a linear fashion, you can address this change and show that the denominator will always be of greater degree than the numerator, thus you can bound the derivatives by $n!$. Understood? – Pedro Tamaroff Apr 4 '12 at 21:45
Right, but I still need to show that the absolute value of that ratio of polynomials is less than 1. I don't see that that immediately follows from the degree of the polynomials in the numerator and denominator for $0 < x < 1$. Quite likely I'm missing something obvious. – user23784 Apr 4 '12 at 21:54
You're missing the fact the polynomials are monic, which means the leading power has $1$ for a coefficent. Thus, the fact that $\operatorname{deg}(P)\leq\operatorname{deg}(Q)$ suffices. – Pedro Tamaroff Apr 4 '12 at 22:12
up vote 4 down vote accepted

We know

$$\begin{align*} \arctan (x) & = \int_0^x \dfrac{1}{1+t^2} dt \\ & = \int_0^x \left[1 - t^2 + t^4 - \cdots + (-1)^n t^{2n-2} + \dfrac{(-1)^{n} t^{2n}}{1+t^2} \right] dt\\ & = \sum_{k = 0}^{n-1} \dfrac{(-1)^k x^{2k+1}}{2k+1} + \int_0^x \dfrac{t^{2n}}{1+t^2} dt\\ & = \sum_{k = 0}^{n-1} \dfrac{(-1)^k x^{2k+1}}{2k+1} + E_{2n}(x) \end{align*}$$

Hence, we have $E_{2n} (x) = \displaystyle{\int_0^x \dfrac{t^{2n}}{1+t^2} dt}$ so we are just trying to bound the integral. So, we have

$$\begin{align*} |E_{2n} (x)| & = \left|\int_0^x \dfrac{t^{2n}}{1+t^2} dt\right|\\ & = \int_0^x \dfrac{t^{2n}}{1+t^2} dt \qquad (\text{the integrand } \geq 0 \text{ for } t \in [0,1])\\ & \leq \int_0^x t^{2n} dt \qquad \qquad \left(t \in [0,1] \implies t^{2n} \geq \dfrac{t^{2n}}{1+t^2}\right)\\ & = \dfrac{x^{2n+1}}{2n+1} \end{align*}$$

Which is the bound we were looking for.

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This will clearly work. I thought you specifically needed to use the bounds on $f^{(n)}$, but this is enough as it stands. – Pedro Tamaroff Apr 5 '12 at 1:23 Sorry if it seemed like I needed something else. This should have been apparent to me, but wasn't... trying to get a handle on these remainder term problems has been tricky, but I think I'm getting there. Thanks for helping out though! I appreciate it. – user23784 Apr 5 '12 at 1:27
I've only just looked at this: @user23784, a most impressive answer!!! – emjay Aug 22 '14 at 5:42

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.

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Ok, thanks. I was hoping to go the Lagrange formula route since this problem is in the section of exercises immediately after the Lagrange formula section, so I'm assuming the book intends that solution. Also, the book has not defined an alternating series, as yet. I could look up that proof and use that, but it seems to not be in the spirit of the Exercise. – user23784 Apr 4 '12 at 2:20
Almost certainly, the bound involved here is from the alternating series. The hint is that the form of the error bound is exactly the same as the term after the truncated series. – copper.hat Apr 4 '12 at 6:54

Although non-rigorous, the following trick is useful.



We have that $0<f'\leq1$



and successively




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

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 $n!$

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Cool trick, thanks. The book has not yet defined complex numbers, so I feel this would be a bit of a cheat to use, but still very slick. – user23784 Apr 4 '12 at 2:22
@rar I'll continue here. Focus on the independent constant. See how it is at most $1$ in absolute value. – Pedro Tamaroff Apr 5 '12 at 0:06
I see that it is at most 1 in absolute value in this case, but I don't see how that can follow from the degree of the polynomials. – user23784 Apr 5 '12 at 0:35
Ok, I think I have a much simpler way to do this that will avoid the induction all together. I'll post it as an answer once I write it up and make sure it holds together. Before you put any more effort into trying to explain the induction to me... – user23784 Apr 5 '12 at 0:51
T: I've made a note of this little trick for future reference!!! – emjay Aug 22 '14 at 5:43

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