Question

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

Answer 1

Consider, ${\log }_e\left[\right(1+x{)}^{1+x}.(1-x{)}^{1-x}]$
$=(1+x){\log }_e(1+x)+(1-x){\log }_e(1-x)$
$=(1+x)(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-...\infty )+(1-x)(-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}-...\infty )$
$=2(-\frac{x^2}{2}-\frac{x^4}{4}-\frac{x^6}{6}-.....)+2(x^2+\frac{x^4}{3}+.....)$
$=2(\frac{x^2}{1.2}+\frac{x^4}{3.4}+\frac{x^6}{5.6}+...)$
Comparing with given expression,
$a=2$