Maclaurin expansion of arctan: convergence?

Question

In my textbook, the Maclaurin series expansion of $\arctan{x}$ is found by integrating a geometric series, that is, by noting that

$\frac{d}{dx}(\arctan(x)) = \frac{1}{x^2+1}$

then rewriting the latter as a geometric series over which one can then integrate. What bothers me is that the geometric series is only convergent when $|x| < 1$, but $\arctan(x)$ is defined for all $x$. This question of convergence is dismissed by the author, but I'm curious as to what's really going on here. Is the series expansion still valid outside the radius of convergence, and if so, why?

The series for $\arctan x$ also converges to $\arctan x$ at $\pm 1$. But it diverges for $|x|\gt 1$. — André Nicolas, Dec 22 '13 at 03:41
Related : http://math.stackexchange.com/questions/29649/why-is-arctanx-x-x3-3x5-5-x7-7-dots — lab bhattacharjee, Dec 22 '13 at 03:42

score 3 · Answer 1 · edited Dec 22 '13 at 04:38

3

If you do the integral using partial fractions you get $\frac{i}{2}\ln \frac{x+i} {x-i}$ which suggests something dodgy at $x=\pm i$ restricting the possible radius of convergence for the Maclaurin expansion to the unit circle.

If you try to solve $$ \tan \; \theta = i,$$ then writing $z=\sin \theta$ you get $$\frac{z}{\sqrt{1-z^2}} = i.$$ Squaring, this simplifies to $$ 0 = -1.$$

edited Dec 22 '13 at 04:38

Avi Steiner

4,209

answered Dec 22 '13 at 03:58

David Holden

18,040

score 2 · Accepted Answer · answered Dec 22 '13 at 04:33

This is actually an important property of MacLaurin series expansions and highlights the difference between the MacLaurin series expansion of a function and the function itself. When you calculate the Taylor series expansion $a_0+a_1x+ a_2x^2+\cdots$ of a function $f$, what you're actually doing is finding a sequence of polynomials $$a_0,\; a_0+a_1x, \;a_0+a_1x + a_2x^2,\ldots$$ approximating $f$. But how well do they approximate $f$? In the case of $f(x)=\arctan x$, they approximate $f$ really well when $|x|\leq 1$, but not very well at all otherwise.

Maclaurin expansion of arctan: convergence?

2 Answers2