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My Calculus 2 professor referenced that such a series is impossible, but why? I understand how to properly find the power series of arctan(x) centered at 0.

The Glis Jackel
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  • Cf. https://math.stackexchange.com/q/3439994/96384, https://math.stackexchange.com/q/3469364/96384, https://math.stackexchange.com/q/1069869/96384. – Torsten Schoeneberg Jun 16 '23 at 19:11
  • Your Calculus 2 professor is not very good. In fact, when I teach Calculus 2, I show them the series centred at $0$ and let them check it has radius of convergence $1$. We wonder why. Then I show them the series centered at 1 and let them check its radius of convergence is $\sqrt2$. Then we wonder what strange thing, stopping convergence, could have distance $1$ from $x=0$ and distance $\sqrt 2$ from $x=1$. The week after, we do complex numbers. – Torsten Schoeneberg Jun 17 '23 at 01:13

2 Answers2

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Impossible is nothing. By termwise integration of the Taylor series of $\frac{1}{1+x^2}$ centered at $x=1$ we easily get the Taylor series of $\arctan(x)$ centered at $x=1$. Its radius of convergence is $\sqrt{2}$ since the closest singularities to $x=1$ lie at $\pm i$.

$$\arctan(x)=\frac{\pi}{4}+\sum_{m\geq 0}\left(\frac{(-1)^m(x-1)^{4m+1}}{2^{2m+1}(4m+1)}-\frac{(-1)^m(x-1)^{4m+2}}{2^{2m+1}(4m+2)}+\frac{(-1)^m(x-1)^{4m+3}}{2^{2m+2}(4m+3)}\right)$$

Since, for instance, $\frac{\pi}{4}=\arctan\left(\frac{1}{2}\right)+\frac{1}{2}\arctan\left(\frac{3}{4}\right)$, that Taylor series can be exploited to provide fast-converging series to $\pi$ of the BBP-type, like the celebrated $$ \pi = \sum_{k\geq 0}\frac{1}{16^k}\left(\frac{4}{8k+1}-\frac{2}{8k+4}-\frac{1}{8k+5}-\frac{1}{8k+6}\right).$$

Jack D'Aurizio
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Since $$ \begin{align} \tan(\arctan(1+x)-\arctan(1)) &=\frac{(1+x)-1}{1+(1+x)\cdot1}\\ &=\frac x{2+x}\tag{1} \end{align} $$ we have $$ \begin{align} \arctan(1+x) &=\frac\pi4+\arctan\left(\frac{x}{2+x}\right)\\ &=\frac\pi4+\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\left(\frac x{2+x}\right)^{2k+1}\tag{2} \end{align} $$ Expanding $(2)$ using the binomial theorem, the coefficient of $x^n$, for $n\ge1$, is $$ \begin{align} &\frac{\left(-\frac12\right)^n}n\sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor}(-1)^{k+1}\binom{n}{2k+1}\\ &=\frac{\left(-\frac12\right)^n}n\sum_{k=0}^n\frac{(-i)^k-i^k}{2i}\binom{n}{k}\\ &=\frac{\left(-\frac12\right)^n}n\frac{(1-i)^n-(1+i)^n}{2i}\\ &=\frac{\left(-\frac1{\sqrt2}\right)^n}n\frac{\left(\frac{1-i}{\sqrt2}\right)^n-\left(\frac{1+i}{\sqrt2}\right)^n}{2i}\\ &=-\frac{\left(-\frac1{\sqrt2}\right)^n}n\sin\left(\frac{n\pi}4\right)\tag{3} \end{align} $$ Therefore, $$ \arctan(1+x)=\frac\pi4-\sum_{n=1}^\infty\frac{\sin\left(\frac{n\pi}4\right)}n\left(-\frac{x}{\sqrt2}\right)^n\tag{4} $$ Note that the series in $(2)$ converges for $-1\lt x\lt\infty$ and $(4)$ matches Jack D'Aurizio's answer.

robjohn
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