TITLE: Why is arctangent smooth?
QUESTION [5 upvotes]: I just read a proof by Spivak that $\arctan(x)$ has the $(2n+1)$-th Taylor polynomial at zero $$x - \frac{x^3}{3} + \cdots + (-1)^n\frac{x^{2n+1}}{2n+1}$$ The proof relied on the assumption that $\arctan(x)$ has $2n +1$ derivatives in order for the Taylor polynomial to exist. Spivak also made the point that the existence of a "good approximation polynomial" does not imply the existence of a Taylor polynomial if the function isn't differentiable enough.
But he never did prove that arctangent is smooth (at zero at least). Is there a simple way to see this?

REPLY [5 votes]: There's no need to assume that it has $2n+1$ derivatives. In point of fact, it is infinitely differentiable, and all of its non trivial derivatives are rational functions of the form $$\frac{P(x)}{(1+x^2)^k}$$ for some polynomial $P(x)$ and some positive integer $k$. (You can prove this by induction, using the quotient rule for the induction step, and logarithmic differentiation for the first derivative.)

REPLY [2 votes]: How does Spivak define $\arctan(x)$? My guess is you can recover the definition from $\arctan(x)=\int_0^x\frac{dt}{1+t^2}$, or rather $\frac{d}{dx}\arctan(x)=\frac{1}{1+x^2}$ and then see that the r.h.s. is nice and smooth. This is also a nice way of getting the power series coefficients of $\arctan$ as $\frac{1}{1+t^2}$ has a nice expansion and can be integrated term-by-term.

REPLY [2 votes]: The derivative of $\arctan x$ is $\frac1{1+x^2}$, which is the composition of the two smooth functions $x \mapsto 1+x^2$ and $x \mapsto 1/x$, so $\arctan x$ is smooth itself.