Question

If $f=\frac{x}{1+x^2}+\frac{1}{3}{\left(\frac{x}{1+x^2}\right)}^3+\frac{1}{5}{\left(\frac{x}{1+x^2}\right)}^5+...$ and $g=x-\frac{2}{3}{x}^3+\frac{1}{5}{x}^5+\frac{1}{7}{x}^7+\frac{2}{9}{x}^9+...$, then $f=\frac{1}{3}cg.$ Find $c.$

Answer 1

Let $y=\frac{x}{1+x^2}$
then $f=y+\frac{y^3}{3}+\frac{y^5}{5}+...$
$=\frac{1}{2}\log \left(\frac{1+y}{1-y}\right)$
$=\frac{1}{2}\log \left(\frac{1+\frac{x}{1+x^2}}{1-\frac{x}{1+x^2}}\right)$
$=\frac{1}{2}\log \left(\frac{1+x+x^2}{1-x+x^2}\right)$ ...(1)
and $g=x-\frac{2}{3}{x}^3+\frac{1}{5}{x}^5+\frac{1}{7}{x}^7+\frac{2}{9}{x}^9+\frac{1}{11}{x}^{11}+\frac{1}{13}{x}^{13}+\frac{2}{15}{x}^{15}+...$

$=x-(1-\frac{1}{3})x^3+\frac{1}{5}{x}^5+\frac{1}{7}{x}^7-\frac{1}{3}(1-\frac{1}{3})x^9+\frac{1}{11}{x}^{11}+\frac{1}{13}{x}^{13}-\frac{1}{5}(1-\frac{1}{3})x^{15}+...$
$=(x+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+...)-(x^3+\frac{x^9}{3}+\frac{x^{15}}{5}+...\infty )$
$=\frac{1}{2}\log \left(\frac{1+x}{1-x}\right)-\frac{1}{2}\log \left(\frac{1+x^3}{1-x^3}\right)$
$=\frac{1}{2}\log [\frac{(1+x)}{(1-x)}.\frac{(1+x^3)}{(1-x^3)}]$
$=\frac{1}{2}\log \left[\frac{(1+x)(1-x)(1+x+x^2)}{(1-x)(1+x)(1-x+x^2)}\right]$
$=\frac{1}{2}\log \left(\frac{1+x+x^2}{1-x+x^2}\right)$
from (1) and (2), we get
$f\equiv g$
$\Rightarrow c=3$