Question

${\log }_e\left[\right(1+x{)}^{1+x}(1-x{)}^{1-x}]=a(\frac{x^2}{1.2}+\frac{x^4}{3.4}+\frac{x^6}{5.6}+...\infty )$
Find $a$

Answer 1

${\log }_e\left[\right(1+x{)}^{1+x}(1-x{)}^{1-x}]$

$={\log }_e(1+x{)}^{1+x}+{\log }_e(1-x{)}^{1-x}$

$=(1+x){\log }_e(1+x)+(1-x){\log }_e(1-x)$

$=\left[{\log }_e\right(1+x)+{\log }_e(1-x\left)\right]+x\left[\log \right(1+x)-\log (1-x\left)\right]$

$={\log }_e(1-x^2)+x{\log }_e\left(\frac{1+x}{1-x}\right)$

$=(-x^2-\frac{x^4}{2}-\frac{x^6}{3}-....)+2x(x+\frac{x^3}{3}+\frac{x^5}{5}+...)$

$=(-x^2-\frac{x^4}{2}-\frac{x^6}{3}-...)+(2x^2+\frac{2x^3}{3}+\frac{2x^5}{5}+....)$

$=(x^2+\frac{x^4}{3.2}+\frac{x^6}{5.3}+...\infty )$

$=2(\frac{x^2}{1.2}+\frac{x^4}{3.4}+\frac{x^6}{5.6}+...\infty )$

$\Rightarrow a=2$