Power series expansion of $\arctan{\left(\frac{1-x^2}{2+x^2}\right)}$

Question

I'm looking for help with power series expansion (around $x_0=0$, in real domain) of the following function:

$f(x) = x\arctan{\left(\frac{1-x^2}{2+x^2}\right)}$.

Obviously the actual problem is $g(x) = \arctan{\left(\frac{1-x^2}{2+x^2}\right)}$, then $f(x)=xg(x)$ so what we really need is an expansion for $g$. I've already tried the standard approach of differentiating $g$, then expanding the derivative into power series and applying the term-by-term integration theorem, but what I get is $g'(x) = \frac{-6x}{2x^4 +2x^2 +5}$ and I'm having a hard time here trying to expand it into a series. Of course there's $\frac{1}{2x^4 +2x^2 +5} = \frac{1}{1 + \left(\frac{2}{5}x^2(\frac{2x+1}{5})\right)}$, but this leaves me with $\sum_{n=1}^{\infty}\frac{2^n}{5^{n+1}}\left(x^2(2x^2+1)\right)^n$ and I'm stuck. Any hints would be appreciated.

Answer 1

The first thing I should do is to make $x^2=t$ and to consider $$f(t)=\tan ^{-1}\left(\frac{1-t}{2+t}\right)$$ So, as you thought about it, differentiate $$f'(t)=-\frac{3}{2 t^2+2t+5}$$ Computing the complex roots of the denominator and using partial fraction decomposition $$f'(t)=\frac{i}{2 \left(t+\left(\frac{1}{2}-\frac{3 i}{2}\right)\right)}-\frac{i}{2 \left(t+\left(\frac{1}{2}+\frac{3 i}{2}\right)\right)}$$ $$f'(t)=\sum_{n=0}^\infty -\frac{i}{2} \left(-\frac{1}{5}-\frac{3 i}{5}\right)^{n+1} t^n-\sum_{n=0}^\infty -\frac{i}{2} \left(-\frac{1}{5}+\frac{3 i}{5}\right)^{n+1} t^n$$ $$f'(t)=\sum_{n=0}^\infty \frac{i}{2}\left(\left(-\frac{1}{5}+\frac{3 i}{5}\right)^{n+1} -\left(-\frac{1}{5}-\frac{3 i}{5}\right)^{n+1}\right)t^n$$ Using de Moivre, this write $$f'(t)=\sum_{n=0}^\infty -\left(\frac{5}{2}\right)^{-\frac{n+1}{2}} \sin \left((n+1) \left(\pi -\tan ^{-1}(3)\right)\right) t^n$$ Expanding the sine $$f'(t)=\sum_{n=0}^\infty \left(-\sqrt{\frac{2}{5}}\right)^{n+1} \sin \left((n+1) \tan ^{-1}(3)\right)\, t^n=\sum_{n=0}^\infty a_n t^n$$ Do not worry about the $a_n$'s; they are rational numbers generating the sequence $$\left\{-\frac{3}{5},\frac{6}{25},\frac{18}{125},-\frac{96}{625},\frac{12}{3125}, \frac{936}{15625},-\frac{1992}{78125},-\frac{5376}{390625},\frac{30672}{1953125},- \frac{7584}{9765625},\cdots\right\}$$ So, integrating termwise $$f(t)=\tan ^{-1}\left(\frac{1}{2}\right)+\sum_{n=0}^\infty \frac {a_n}{n+1} t^{n+1}$$ Back to $x$ $$x \tan ^{-1}\left(\frac{1-x^2}{x^2+2}\right)=\tan ^{-1}\left(\frac{1}{2}\right)x+\sum_{n=0}^\infty \frac {a_n}{n+1} x^{2n+3}=\tan ^{-1}\left(\frac{1}{2}\right)x+\sum_{n=0}^\infty b_n x^{2n+3}$$ The $b_n$'s generate the sequence $$\left\{-\frac{3}{5},\frac{3}{25},\frac{6}{125},-\frac{24}{625},\frac{12}{15625}, \frac{156}{15625},-\frac{1992}{546875},-\frac{672}{390625},\frac{3408}{1953125} \right\}$$

Answer 2

For $|x|<\sqrt{2}\,$, we have \begin{align*} \arctan\frac{1-x^2}{2+x^2}&=\arctan\frac{1}{2}+\sum_{n=1}^{\infty}\frac{(-1)^n}{2^n} \Biggl[\sum_{k=1}^{n} \binom{n-1}{k-1} \frac{1}{k}\biggl(\frac{3}{2}\biggr)^k \sum_{\ell=0}^{k-1}(-1)^\ell\binom{\ell}{k-\ell-1} \biggl(\frac{4}{5}\biggr)^{\ell+1}\Biggr]x^{2n}. \end{align*}

References

1. Feng Qi and Mark Daniel Ward, Closed-form formulas and properties of coefficients in Maclaurin's series expansion of Wilf's function, arXiv (2021), available online at https://arxiv.org/abs/2110.08576v1.
2. https://math.stackexchange.com/a/4332741/945479.
3. https://math.stackexchange.com/a/4331451/945479.

Answer 3

For $n\ge1$, by the Faa di Bruno formula and some properties of the partial Bell polynomials $B_{n,k}$, we obtain \begin{align*} \biggl(\arctan\frac{1-t}{2+t}\biggr)^{(n)} &=-3\biggl(\frac{1}{2t^2+2t+5}\biggr)^{(n-1)}\\ &=-3\sum_{k=0}^{n-1} \frac{(-1)^kk!}{(2t^2+2t+5)^{k+1}} B_{n-1,k}\bigl(4t+2,4,0,\dotsc,0\bigr)\\ &=-3\sum_{k=0}^{n-1} \frac{(-1)^kk!}{(2t^2+2t+5)^{k+1}} 4^k B_{n-1,k}\biggl(t+\frac12,1,0,\dotsc,0\biggr)\\ &\to-3\sum_{k=0}^{n-1} \frac{(-1)^kk!}{5^{k+1}} 4^k B_{n-1,k}\biggl(\frac12,1,0,\dotsc,0\biggr), \quad t\to0\\ &=-3\sum_{k=0}^{n-1} \frac{(-1)^kk!}{5^{k+1}} 4^k \frac{1}{2^{n-k-1}}\frac{(n-1)!}{k!}\binom{k}{n-k-1}\frac1{2^{2k-n+1}}\\ &=-3(n-1)!\sum_{k=0}^{n-1} (-1)^k\frac{2^k}{5^{k+1}} \binom{k}{n-k-1}, \end{align*} where we used \begin{equation}\label{Bell-x-1-0-eq} B_{n,k}(x,1,0,\dotsc,0) =\frac{1}{2^{n-k}}\frac{n!}{k!}\binom{k}{n-k}x^{2k-n}. \end{equation} Consequently, we arrive at \begin{equation*} \arctan\frac{1-t}{2+t} =\arctan\frac12+\frac35\sum_{n=0}^\infty\Biggl[\sum_{k=0}^{n-1} (-1)^{k+1} \biggl(\frac{2}{5}\biggr)^k \binom{k}{n-k-1}\Biggr]\frac{t^n}{n} \end{equation*} and \begin{equation*} \arctan\frac{1-x^2}{2+x^2} =\arctan\frac12+\frac35\sum_{n=0}^\infty\Biggl[\sum_{k=0}^{n-1} (-1)^{k+1} \biggl(\frac{2}{5}\biggr)^k \binom{k}{n-k-1}\Biggr]\frac{x^{2n}}{n}. \end{equation*}

References

1. F. Qi and B.-N. Guo, Explicit formulas for special values of the Bell polynomials of the second kind and for the Euler numbers and polynomials, Mediterr. J. Math. 14 (2017), no. 3, Article 140, 14 pages; available online at https://doi.org/10.1007/s00009-017-0939-1.
2. F. Qi, D.-W. Niu, D. Lim, and Y.-H. Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, J. Math. Anal. Appl. 491 (2020), no. 2, Article 124382, 31 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124382.