Question

If $f=\frac{x}{1+x^2}+\frac{1}{3}{\left(\frac{x}{1+x^2}\right)}^3+\frac{1}{5}{\left(\frac{x}{1+x^2}\right)}^5+...$ to $\infty$ and $g=x-\frac{2}{3}{x}^3+\frac{1}{5}{x}^5+\frac{1}{7}{x}^7-\frac{2}{9}{x}^9+...$, then $f=d\times g$. Find 4d.

Answer 1

$f=\frac{x}{1+x^2}+\frac{1}{3}{\left(\frac{x}{1+x^2}\right)}^3+\frac{1}{5}{\left(\frac{x}{1+x^2}\right)}^5+...$

We know, $Log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}...$

$\therefore f=\frac{1}{2}\left[\log (1+\frac{x}{1+x^2})-\log (1-\frac{x}{1+x^2})\right]$
$=\frac{1}{2}\left[\log \left(\frac{1+x+x^2}{1+x^2}\right)-\log \left(\frac{1+x^2-x}{1+x^2}\right)\right]$
$\Rightarrow f=\frac{1}{2}\log \left(\frac{x^2+x+1}{x^2-x+1}\right)$
Now, $g=x-\frac{2}{3}{x}^3+\frac{1}{5}{x}^5+\frac{1}{7}{x}^7-\frac{2}{9}{x}^9+...$
$g=x-(1-\frac{1}{3})x^3+\frac{1}{5}{x}^5+\frac{1}{7}{x}^7-\frac{1}{3}(1-\frac{1}{3})x^9+...$
$g=(x+\frac{1}{3}{x}^3+\frac{1}{5}{x}^5+\frac{1}{7}{x}^7+......)-(x^3+\frac{{\left(x^3\right)}^3}{3}+\frac{{\left(x^3\right)}^5}{3}+.....)$
We know, $\frac{1}{2}\left(Log\right(1+x)-Log(1-x\left)\right)=x+\frac{x^3}{3}+\frac{x^5}{5}....$
$\therefore g=\frac{1}{2}[\log (1+x)-\log (1-x)]-\frac{1}{2}[\log (1+x^3)-\log (1-x^3)]$
$=\frac{1}{2}\log \left(\frac{1+x}{1-x}\right)-\frac{1}{2}\log \left(\frac{1+x^3}{1-x^3}\right)$
$\Rightarrow g=\frac{1}{2}\log \left(\frac{(1+x)(1-x^3)}{(1-x)(1+x^3)}\right)$
$\Rightarrow g=\frac{1}{2}\log \left(\frac{x^2+x+1}{x^2-x+1}\right)$
$\Rightarrow g=f$

Comparing with given value ,
$d=1$
$\Rightarrow 4d=4$