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.
Best Answer
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 $$\left\{-\frac{3}{5},\frac{3}{25},\frac{6}{125},-\frac{24}{625},\frac{12}{15625}, \frac{156}{15625},-\frac{1992}{546875},-\frac{672}{390625},\frac{3408}{1953125} \right\}$$