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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?

Chris
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    It's as straightforward as taking the derivatives, right? The first derivative is $1/(1+x^2)$, so the $n$-th derivative is always going to be $P(x)/(1+x^2)^{n}$, where $P(x)$ is a polynomial. So not only is it smooth, but the Taylor series expansion around $x=0$ is convergent for $|x|<1$. – mjqxxxx Jan 03 '13 at 23:42
  • @mjqxxx Can you make that precise(ish)? – Chris Jan 03 '13 at 23:43
  • The derivative of $P(x)/(1+x^2)^{n}$ is $Q(x)/(1+x^2)^{n+1}$, where $Q(x)=-2nxP(x) + (1+x^2)P'(x)$; in particular, if $P$ is a polynomial, then so is $Q$. – mjqxxxx Jan 03 '13 at 23:45
  • @mjqxxx Awesome. I hadn't noticed that. (You should write that up!) – Chris Jan 03 '13 at 23:46

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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.)

Cameron Buie
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  • This is pretty much exactly what I had imagined doing. (+1, though) – Chris Jan 03 '13 at 23:55
  • Good! That will get you where you need to go. I suspect that, rather than assuming that $\arctan(x)$ is smooth, Spivak was simply leaving proof of that fact to the reader(s). – Cameron Buie Jan 04 '13 at 00:08
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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.

Alex R.
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  • It's the inverse of the restriction of tangent to $(\frac{-\pi}{2}, \frac{\pi}{2})$. I'm thinking maybe logarithmic differentiation might help? – Chris Jan 03 '13 at 23:42
  • @user1296727: Right, so you probably derived $\frac{d}{dx}\arctan(x)=\frac{1}{1+x^2}$ which is again differentiable (and moreover smooth). – Alex R. Jan 03 '13 at 23:44
  • I know about the expansion of $\frac{1}{1+x^2}$; that's how he obtained the new expression. His point was that $P_{2n + 1,a}(x)$ equals $f$ to order $2n+1$, so that if this expansion you mention, $P(x)$, does as well, then $P(x) = P_{2n + 1,a}(x)$. – Chris Jan 03 '13 at 23:44
  • (That is, if you assume that $P_{2n + 1,a}(x)$ exists.) – Chris Jan 03 '13 at 23:45
  • @user1296727: could you clarify your notation, what does $P_{2n+1,\alpha}(x)$ mean? – Alex R. Jan 03 '13 at 23:46
  • The $(2n+1)$th Taylor polynomial of $\arctan(x)$ about $a$. – Chris Jan 03 '13 at 23:47
  • @user1296727: OK, but you just need to show that $\arctan(x)$ is smooth right? Your definition of smooth is infinitely differentiable right? – Alex R. Jan 03 '13 at 23:48
  • Yes, that's what I meant. – Chris Jan 03 '13 at 23:49
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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.

mrf
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